Haloalkanes and Haloarenes MCQs for NEET

NEET Chemistry For Haloalkanes And Haloarenes Multiple Choice Questions

Question 1. The given compound

Haloalkanes And Haloarenes Allylic Halide

is an example of ______

  1. Allylic halide
  2. Vinylic halide
  3. Benzylic halide
  4. Aryl halide

Answer: 1. Allylic halide

The compound where the halogen group is attached to a sp³ hybridised carbon atom next to the carbon which is adjacent to the carbon-carbon double bond is known as allylic halide.

Hence, the given molecule is an alkyl halide.

Question 2. The correct sequence of bond enthalpy of the ‘C — X’ bond is

  1. CH3 – Cl > CH3 – F > CH3 – Br > CH3 -I
  2. CH3 – F < CH3– Cl < CH3– Br < CH3 – I
  3. CH3 – F > CH3— Cl > CH3— Br > CH – I
  4. CH3 – F < CH3– Cl > CH3 – Br > CH3 – I

Answer: 3. CH3 – F > CH3 – Cl > CH3 – Br > CH – I

The correct order of bond enthalpy is CH3– F > CH3 – Cl > CH3 – Br > CH3 – I

Question 3. The reaction of C6H5CH=CHCH3 with HBr produces

Haloalkanes And Haloarenes Benzyl Carbocation

Answer: 3

Haloalkanes And Haloarenes More Stable Benzyl Carbocation

Question 4. In the replacement reaction Haloalkanes And Haloarenes Tertiary Halide . The reaction will be most favourable if M happens to be

  1. Na
  2. K
  3. Rb
  4. Li

Answer: 3. Rb

Tertiary halide shows SN1 mechanism i.e., ionic mechanism. In the given reaction negative ions will attack on carbocation. Thus greater the tendency of ionisation (greater ionic character in the M – F bond) more favourable the reaction. The most ionic bond is Rb – F in the given examples thus most favourable reaction will be with Rb-F.

Read and Learn More NEET MCQs with Answers

Question 5. When chlorine is passed through propene at 400°C, which of the following is formed?

  1. PVC
  2. Allyl chloride
  3. Propyl chloride
  4. 1, 2-Dichloroethane

Answer: 2. Allyl chloride

Haloalkanes And Haloarenes Allyl Chloride

At 400°C temperature, substitution occurs instead of addition.

Question 6. Which of the following is suitable to synthesize chlorobenzene?

Haloalkanes And Haloarenes Synthesize Chlorobenzene

Answer: 1

Haloalkanes And Haloarenes

Arenes react with halogens in the presence of a Lewis acid like anhydrous FeCl3, FeBr3 or AICI3 to yield haloarenes, for example,

Question 7. The incorrect statement regarding chirality is

  1. SN¹ reaction yields a 1:1 mixture of both enantiomers
  2. The product obtained by the SN² reaction of haloalkane having chirality at the reactive site shows an inversion of configuration
  3. Enantiomers are superimposable mirror images of each other
  4. A racemic mixture shows zero optical rotation.

Answer: 3. Enantiomers are superimposable mirror images of each other

Enantiomers are non-superimposable mirror images of each other. Enantiomers possess identical physical properties namely, melting point, boiling point, refractive it dex, etc. Then, only differ with respect to the rotation of plane polarised 1ight. If one of the enantiomers is dextrorotatory, then the other will be laevorotatory.

Question 8. The major product formed in the dehydrohalogenation reaction of 2-bromopentane is pent-2-ene. This product formation is based on the?

  1. Huckel’s Rule
  2. Saytzeff s Rule
  3. Hund’s Rule
  4. Hofmann Rule

Answer: 2. Saytzeff s Rule

It is an example of B-elimination, as the major product is 2-pentene (more substituted) not 1-pentene, hence it follows the Saytzeffb rule

Haloalkanes And Haloarenes 2 Pentene not 1 pentene

Question 9. The elimination reaction of 2-bromopentane to form pent-2-ene is

  1. β-Elimination reaction
  2. Follows Zaitsev rule
  3. Dehydrohalogenation reaction
  4. Dehydration reaction.
  1. (1), (2), (3)
  2. (1), (3), (4)
  3. (2), (3), (4)
  4. (1), (2), (4)

Answer: 1. (1), (2), (3)

Haloalkanes And Haloarenes Beta Elimination Reaction

Question 10. The hydrolysis reaction that takes place at the slowest rate, among the following is

Haloalkanes And Haloarenes Aryl Halides

Answer: 1

Aryl halides are less reactive as compared to alkyl halides as the halogen atom in these compounds is firmly attached and cannot be replaced by nucleophiles such as OH, NH2 etc. In chlorobenzene, the electron pair of the chlorine atom is in conjugation with n-electrons of the benzene ring. Thus C-Cl bond acquires a double bond character and is difficult to break

Question 11. The compound A on treatment with Na gives B, and with PCl5 gives C. B and C react together to give diethyl ether. A, B and C are in the order

  1. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, \mathrm{C}_2 \mathrm{H}_6, \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}\)
  2. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}\)
  3. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}, \mathrm{C}_2 \mathrm{H}_6, \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
  4. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}\)

Answer: 4. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}\)

The compound A on treatment with Na gives B, and with PCl5 gives C. B and C react together to give diethyl ether.

Haloalkanes And Haloarenes Dieethyl Ether

Question 12. The compound C7H8 undergoes the following reactions:

⇒ \(\mathrm{C}_7 \mathrm{H}_8 \underrightarrow{3 \mathrm{Cl}_2 / \Delta}\) A \(\underrightarrow{\mathrm{Br}_2 / \mathrm{Fe}}\) B \(\underrightarrow{\mathrm{Zn} / \mathrm{HCl}}\) C

The product C is

  1. m-bromotoluene
  2. ο-bromotoluene
  3. 3-bromo-2,4,6-trichlorotoluene
  4. p-bromotoluene.

Answer: 1. The product C is

Haloalkanes And Haloarenes m bromotoluence

Question 13. Identify A and predict the type of reaction.

Haloalkanes And Haloarenes Bromoanisole

Answer: 4

m-Bromoanisole gives only the respective meta-substituted aniline. This is a substitution reaction which goes by an elimination-addition pathway.

Haloalkanes And Haloarenes m Bromoanisole

Question 14. Consider the reaction, \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{NaCN} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}+\mathrm{NaBr}\)

This reaction will be the fastest in

  1. Ethanol
  2. Methanol
  3. N, N’ -dimethylformamide (DMF)
  4. Water.

Answer: 3. N, N’ -dimethylformamide (DMF)

The reaction \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{NaCN} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}+\mathrm{NaBr}\) follows S mechanism which is favoured by polar aprotic solvent i.e., N, N’-dimethylformamide (DMF),

Haloalkanes And Haloarenes Dimethylformamide

Question 15. Which of the following biphenyls is optically active?

Haloalkanes And Haloarenes Biphenyl Is Optically Active

Answer: 4

o-Substituted biphenyls are optically active as both the rings are not in one plane and their mirror images are non-superimposable.

Question 16. For the following reactions:

Haloalkanes And Haloarenes Saturated Compound Is Converted To Unsaturated Compound

Which of the following statements is correct?

  1. (1) is elimination, (2) and (3) are substitution reactions.
  2. (1) is a substitution, (2) and (3) are addition reactions.
  3. (1) and (2) are elimination reactions and (3) is an addition reaction.
  4. (1) is elimination, (2) is substitution and (3) is addition reaction.

Answer: 4. (1) is the elimination, (2) is substitution and (3) is addition reaction.

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{KOH} \longrightarrow \mathrm{CH}_3 \mathrm{CH} =\mathrm{CH}_2\) + \(\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}\)

The saturated compound is converted into the unsaturated compound by the removal of a group of atoms hence, it is an elimination reaction.

Haloalkanes And Haloarenes Saturated Compound Is Converted To Unsaturated Compound Then It Is Elimination Reaction

-Br group is replaced by – OH group hence, it is a substitution reaction.

Haloalkanes And Haloarenes Br Is Unsaturated Compound Into Saturated Compound

The addition of Br2 converts an unsaturated compound into a saturated compound hence, it is an addition reaction.

Question 17. Two possible stereo-structures of CH3CHOHCOOH, which are optically active, are called

  1. Atropisomers
  2. Enantiomers
  3. Mesomers
  4. Diastereomers.

Answer: 2. Enantiomers

Haloalkanes And Haloarenes Enantiomers

Question 18. In an SN1 reaction on chiral centres, there is

  1. Inversion more than retention leads to partial racemisation
  2. 100% retention
  3. 100% inversion
  4. 100% racemisation.

Answer: 1. Inversion more than retention leading to partial racemisation

In the case of optically active alkyl halides, SN1 reaction is accompanied by racemisation. The carbocation formed in the slow step being sp² hybridised is planar and an attack of nucleophiles may take place from either side resulting in a mature of products, one having the same configuration and the other having an inverted configuration.

The isomer corresponding to inversion is present in slight excess because SN1 also depends upon the degree of shielding of the front side of the reacting carbon.

Question 19. Which of the following compounds will undergo racemisation when solution of KOH hydrolysis?

Haloalkanes And Haloarenes Racemisation Solutiin Of KOH Hydrolysis

  1. (1) and (2)
  2. (2) and (4)
  3. (3) and (4)
  4. (1) and (4)

Answer: None

Due to chirality Haloalkanes And Haloarenes Chiraliry Racemisationonly compound (4) will undergo racemisation.

Hence, all the given options are incorrect.

Question 20. Given:

Haloalkanes And Haloarenes Pair Of Conformers

1 and 2 are

  1. Identical
  2. A pair of conformers
  3. A pair of geometrical isomers
  4. A pair of optical isomers.

Answer: 2. A pair of conformers

1 and 2 are staggered and eclipsed conformers.

Question 21. Which of the following acids does not exhibit optical isomerism?

  1. Maleic acid
  2. α-Amino acids
  3. Lactic acid
  4. Tartaric acid

Answer: 1. Maleic acid

Maleic acid shows geometrical isomerism and not optical isomerism.

Haloalkanes And Haloarenes Maleic Acid

Question 22. Consider the reactions :

  1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br}\) \(\underrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}\) \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{HBr}\)
  2. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br}\) \(\underrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}}\) \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{Br}^{-}\)

The mechanisms of reactions (1) and (2) are respectively

  1. SN1 and SN2
  2. SN1 and SN1
  3. SN2 and SN2
  4. SN2 and SN1

Answer: 3. SN2 and SN2

If the reaction is SN1, there will be the formation of carbocation and the rearrangement takes place. In these reactions there is no rearrangement hence both are SN2 mechanisms.

Question 23. Which of the following compounds undergoes nucleophilic substitution reaction most easily?

Haloalkanes And Haloarenes Nucleophilic Substitution Reaction

Answer: 1

Electron withdrawing groups like – NO2 facilitate nucleophilic substitution reactions in chlorobenzene.

Question 24. Which one is most reactive towards SN1 reaction?

  1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{Br}\)
  2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{Br}\)
  3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{C}\left(\mathrm{CH}_3\right)\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{Br}\)
  4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}\)

Answer: 3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{C}\left(\mathrm{CH}_3\right)\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{Br}\)

SN1 reactions proceed via the formation of a carbocation intermediate.

The more stable is the carbocation more reactive the alkyl aryl halide towards SN1.

In C5H5C+(CH3)(C6H6) carbocation, the two phenyl rings by their +R effect and -CH, by its +I effect diminish the positive charge and make it stable.

Haloalkanes And Haloarenes Alhyl Halide Towards

Question 25. The correct order of increasing reactivity of

Haloalkanes And Haloarenes Nucleophilic Compound

C—X bond towards nucleophiles in the following compounds is

  1. 1 < 2 < 4 < 3
  2. 2 < 3 < 1 < 4
  3. 4 < 3 < 1 < 2
  4. 3 < 2 < 1 < 4

Answer: 1. 1 < 2 < 4 < 3

The order of reactivity is dependent on the stability of the intermediate carbocation formed by the cleavage of the C-X bond. The 3° carbocation (formed from 3) will be more stable than its 2° counterpart (formed from 4) which in turn will be more stable than the arenium ion (formed from 1).

Also, the aryl halide has a double bond character in the C-X bond which makes the cleavage more difficult. However, in spite of all the stated factors, 2 will be more reactive than I due to the presence of the electron-withdrawing -NO2 group. C-Xbond becomes weak and undergoes a nucleophilic substitution reaction.

Question 26. In the following reaction, Haloalkanes And Haloarenes Ether Product the product ‘X’ is

  1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OCH}_2 \mathrm{C}_6 \mathrm{H}_5\)
  2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}\)
  3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3\)
  4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C}_6 \mathrm{H}_5\)

Answer: 3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3\)

Haloalkanes And Haloarenes Mg And Ether

Question 27. Which of the following reactions is an example of a nucleophilic substitution reaction?

  1. 2RX + 2Na → R – R + 2NaX
  2. RX + H2 → KH + HX
  3. RX + Mg → RMgX
  4. RX + KOH→ ROH + KX

Answer: 4. RX + KOH → ROH + KX

Question 28. How many stereoisomers does this molecule have? CH3CH = CHCH2CHBrCH3

  1. 8
  2. 2
  3. 4
  4. 6

Answer: 3. 4

The given compound may be written as

Haloalkanes And Haloarenes Geometrical Isomerism

Both geometrical isomerism (cis-trans form) and optical isomerism is possible in the given compound.

No. of optical isomer = 2n = 21 = 2

(where n = no. of asymmetric carbon)

Hence total no. of stereoisomers = 2 + 2 = 4

Question 29. In an SN2 substitution reaction of the type \(R-\mathrm{Br}+\mathrm{Cl}^{-}\) \(\underrightarrow{\mathrm{DMF}}\) \(\mathrm{R}-\mathrm{Cl}+\mathrm{Br}^{-}\) which one of the following has the highest relative rate?

Haloalkanes And Haloarenes Primary And Secondary Alkyl Halides

Answer: 2

In an SN2 substitution reaction of the type \(R-\mathrm{Br}+\mathrm{Cl}^{-}\) \(\underrightarrow{\mathrm{DMF}}\) \(\mathrm{R}-\mathrm{Cl}+\mathrm{Br}^{-}\)

SN2 mechanism is followed in the case of primary and secondary alkyl halides i.e. SN2 reaction is favoured by small groups on the carbon atoms attached to halogen so, CH3 – X > R – CH2 – X > R2CH – X > R3C – X. Primary is more reactive than secondary and tertiary alkyl halides

Question 30. If there is no rotation of plane polarised light by a compound in a specific solvent, though to be chiral, it may mean that

  1. The compound is certainly meso
  2. There is no compound in the solvent
  3. The compound may be a racemic mixture
  4. The compound is certainly a chiral.

Answer: 1. The compound is certainly meso

Meso compound does not rotate plane polarised light. The compound which contains tetrahedral atoms with four different groups but the whole molecule is achiral, is known as meso compound. It possesses a plane of symmetry and is optically inactive.

One of the asymmetric carbon atoms turns the plane of polarised light to the right and the other to the left and to the same extent so that the rotation due to the upper half is compensated by the lower half, i.e., internally compensated, and finally, there is no rotation of plane polarised light

Question 31. CH3 – CHCl – CH2 – CH3 has a chiral centre. Which one of the following represents its R-configuration?

Haloalkanes And Haloarenes R Configuration

Answer: 2

Haloalkanes And Haloarenes R-Configuration

Question 32. Which of the following is not chiral?

  1. 2-Hydroxypropanoic acid
  2. 2-Butanol
  3. 2,3-Dibromopentane
  4. 3-Bromopentane

Answer: 4. 3-Bromopentane

Due to the absence of an asymmetric carbon atom.

Question 33. Which of the following undergoes nucleophilic substitution exclusively by SN1 mechanism?

  1. Ethyl chloride
  2. Isopropyl chloride
  3. Chlorobenzene
  4. Benzyl chloride

Answer: 4. Benzyl chloride

SN1 reaction is favoured by heavy (bulky) grouPs on the carbon atom attached to halogens and the nature of carbonium ion in the substrate is

Benzyl > Aliyl > Tertiarv > Secondary > Primary > Methyl halides.

Question 34. The chirality of the compound

Haloalkanes And Haloarenes Chirality Compound

  1. R
  2. S
  3. E
  4. Z

Answer: 1. R

Haloalkanes And Haloarenes 1 bromo 1 chloroethane

The lowest priority atom is always away from the viewer. Priority is seen on the basis of atomic no. and if atomic no are the same then on the basis of atomic mass.

If clockwise then it is R, if anticlockwise then it is S.

Name of the moiecule is, (R) 1-bromo-1-chioroethane.

Question 35. Which of the following is least reactive in a nucleophilic substitution reaction?

  1. (CH3)3C – Cl
  2. CH2 = CHCl
  3. CH3CH2Cl
  4. CH2 = CHCH2Cl

Answer: 2. CH2 = CHCl

The non-reactivity of the chlorine atom in vinyl chloride can be explained from the molecular orbital point of view as follows. If the chlorine atom has sp² hybridisation, the C – Cl bond will be a σ-bond and the two lone pairs of electrons will occupy the other two sp² orbitals. This would leave a p orbital containing a lone pair, and this orbital could now conjugate with the π-bond of the ethylenic link.

Thus two M’O’s will be required to accommodate these four π-electrons ‘Furthermore’ since chlorine is more electronegative than carbon, the electrons will tend to be found in the vicinity of the chlorine atom. Nevertheless, the chlorine atom has now lost full control of the lone pair and so, is less negative than it would have been had there been no conjugation.

Since two carbon atoms have acquired a share in their pair each carbon atom acquires a small negative charge. Hence, owing to the delocalisation of bonds (through conjugation) the vinyl chloride molecule has an increased stability. Before the chlorine atom can be placed by some other group the lone pair must be localised again on the chlorine atom. This requires energy, and so the chlorine is more firmly bound.

Haloalkanes And Haloarenes Chlorine Is More Firmly Bound

Question 36. Which of the following pairs of compounds are enantiomers?

Haloalkanes And Haloarenes Enantiomers Compounds

Answer: 1

These two are non-superimposable mirror images of each other, so they are enantiomers.

Question 37. The reactivity order of halides for dehydrohalogenation is

  1. R-F>R-Cl>R-Br>R-I
  2. R-I>R-Br>R-Cl>R-F
  3. R-I>R-Cl>R-Br>R-F
  4. R-F>R-I>R-Br>R-Cl

Answer: 2. R-I>R-Br>R-Cl>R-F

I > Br > Cl > F → atomic radii

F, Ci, Br, I belong to the same group orderly. Atomic radii go on increasing as the nuclear charge increases in preceding downwards in a group. The decreasing order of bond length is C – I > C – Br > C – Cl > C – F.

The order of bond dissociation energy R- F >R- Cl>R- Br> R – I. During dehydrohalogenation C – I bond breaks more easily than the C – F bond. So reactivity order of halides is, R – I > R – Br > R – Cl > R – F.

Question 38. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}\) \(\underrightarrow{\mathrm{NaCN}}\) x \(\underrightarrow{{\mathrm{Ni}}/{\mathrm{H}_2}}\) Y \(\underrightarrow{\text { acetic anhydride }}\) Z

Z in the above reaction sequence is

  1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NHCOCH}_3\)
  2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2\)
  3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CONHCH}_3\)
  4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CONHCOCH}_3\)

Answer: 1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NHCOCH}_3\)

Question 39. CH3-CH2-CH-CH3 obtained by chlorination of n-butane will be

  1. Meso form
  2. Racemic mixture
  3. d-form
  4. l-form

Answer: 2. Racemic mixture

Chlorination of n-butane takes place via free radical formation. i.e \(\mathrm{Cl}_2 \longrightarrow \dot{\mathrm{Cl}}+\dot{\mathrm{Cl}}\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\) \(\underrightarrow{\dot{\mathrm{C}}}\) \(\mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{HCH}_2 \mathrm{CH}_3+\mathrm{HCl}\)

sp² hybrid planar shape intermolecular and Cl may attack from either side to give

Haloalkanes And Haloarenes Racemic Mixture

Question 40. An organic compound A(C4H9Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative then, A is

  1. t-butyl chloride
  2. s-butyl chloride
  3. iso-butyl chloride
  4. n-butyl chloride.

Answer: 1. t-butyl chloride

An organic compound A(C4H9Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative

Wurtz reaction: It involves the reaction of alkyl halides with Na in ether to form higher alkanes.

2R – X + 2Na → R – R + 2NaX

In the given problem, \(2 \mathrm{C}_4 \mathrm{H}_9 \mathrm{Cl}+2 \mathrm{Na}\) \(\underrightarrow{\text { Ether }}\) \(\mathrm{C}_4 \mathrm{H}_9-\mathrm{C}_4 \mathrm{H}_9+2 \mathrm{NaCl}\)

Compound A is t-butyl chloride, in this compound all – CH, groups have primary hydrogen only and are able to give only, one chloro derivative.

⇒ \(\left(\mathrm{CH}_3\right)_3 \mathrm{CC}\left(\mathrm{CH}_3\right)_3\) \(\underrightarrow{\mathrm{Cl}_2}\) \( \mathrm{CH}_2 \mathrm{Cl}\left(\mathrm{CH}_3\right)_2 \mathrm{C}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)

Question 41. A compound of molecular formula C7H16 shows optical isomerism, compound will b

  1. 2,3-dimethyl pentane
  2. 2,2-dimethylbutane
  3. 2-methyl hexane
  4. None of these.

Answer: 1. 2,3-dimethyl pentane

Organic compounds exhibit the property of enantiomerism (optical isomerism) only when their molecules are chiral. Most chiral compounds have a chiral centre which is an atom bonded to four different atoms or groups.

Haloalkanes And Haloarenes Enantiomerism

2,3-Dimethylpentane has one chiral C-atom and does not have any symmetric element.

Question 42. Which of the following compounds is not chiral?

  1. \(\mathrm{CH}_3 \mathrm{CHDCH}_2 \mathrm{Cl}\)
  2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHDCl}\)
  3. \(\mathrm{DCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}\)
  4. \(\mathrm{CH}_3 \mathrm{CHClCH}_2 \mathrm{D}\)

Answer: 3. \(\mathrm{DCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}\)

Haloalkanes And Haloarenes Two Identiacal H Atoms

The above compound has no chiral ‘C’-atom’ All the ‘C’ atoms are attached to two identical ‘H’ atoms, so they are not symmetrical.

Question 43. Replacement of Cl of chlorobenzene to give phenol requires drastic conditions. But chlorine of 2,4-dinitrochlorobenzene is readily replaced because

  1. NO2 donates e at meta position
  2. NO2 withdraws e from ortho/para positions
  3. NO2 makes ring electrons rich at ortho and para
  4. NO2 withdraws e from meta position.

Answer: 2. NO2 withdraws e from ortho/para positions

Haloalkanes And Haloarenes Ortholpara Positions

Withdrawal of electrons by -NO2 groups from ortho para positions causes easier removal of -Cl atom due to the development of positive charge on o- and p- positions.

Question 44. The alkyl halide is converted into an alcohol by

  1. Elimination
  2. Dehydrohalogenation
  3. Addition
  4. Substitution.

Answer: 4. Substitution.

⇒ \(\underset{\mathrm{Ethyl bromide (aqueous)}}{{\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}}}+\mathrm{KOH} \longrightarrow \underset{\mathrm{Ethyl alochol}}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}+\mathrm{KBr}\)

Question 45. The following reaction is described as

Haloalkanes And Haloarenes Biomolecular Reaction

Answer: 1

SN2 reactions are bimolecular reactions where the rate of reaction depends on the concentration of both substrate and nucleophile. When OH attacks the substrate from the opposite side of the leaving group i.e., Br a transition state results, in which both OH and Br are partially bonded to a carbon atom.

Question 46. The reaction of t-butyl bromide with sodium methoxide produces

  1. Sodium t-butoxide
  2. t-butyl methyl ether
  3. Isobutane
  4. Isobutylene

Answer: 4. Isobutylene

Isobuythylene is obtained.

Haloalkanes And Haloarenes Isobuthylene

Thus, the reaction produces isobutylene.

Question 47. Grignard reagent is prepared by the reaction between

  1. Magnesium and alkane
  2. Magnesium and aromatic hydrocarbon
  3. Zinc and alkyl halide
  4. Magnesium and alkyl halide

Answer: 4. Magnesium and alkyl halide

Grignard reagent is prepared by heating an alkyl halide with dry magnesium powder in dry ether.

⇒ \(R-X+\mathrm{Mg}\) \(\underrightarrow{\text { Dry ether }}\) \(\underset{\mathrm{(Grignard Reagent )}}{{R-\mathrm{Mg}-X}}\)

Question 48. Chlorobenzene reacts with Mg in dry ether to give a compound (A) which further reacts with ethanol to yield

  1. Phenol
  2. Benzene
  3. Ethylbenzene
  4. Phenyl ether.

Answer: 2. Benzene

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}\) \(\underrightarrow{\mathrm{Mg}}\) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{MgCl}\) \(\underrightarrow{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}}\) \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OMgCl}\)

Question 49. Benzene reacts with n-propyl chloride in the presence of anhydrous AlCl3 to give

  1. 3-propyl-1-chlorobenzene
  2. n-propylbenzene
  3. No reaction
  4. Isopropylbenzene.

Answer: 4. Isopropylbenzene.

Haloalkanes And Haloarenes Isopropybenzene

Question 50. Which chloro derivative of benzene among the following would undergo hydrolysis most readily with aqueous sodium hydroxide to furnish the corresponding hydroxy derivative?

Haloalkanes And Haloarenes Chloro Derivate Of Benzene

Answer: 1

Cl in 2, 4,6- trinitrochloro benzene is activated by three -NO2 groups at o-and p-positions and hence undergoes hydrolysis most readily.

Question 51. Which of the following is an optically active compound?

  1. 1-Butanol
  2. 1-Propanol
  3. 2-Chlorobutane
  4. 4-Hydroxyheptane

Answer: 3. 2-Chlorobutane

Haloalkanes And Haloarenes Chiral carbon

2-Chlorolrutane contains a chiral carbon atom and hence it is an optically active compound.

Question 52. Trichloroacetaldehyde, CCl3CHO reacts with chlorobenzene in the presence of sulphuric acid and produces

Haloalkanes And Haloarenes Trichloroacetaldehyde

Answer: 2

It gives D.D.T

(p,p di chlorodiphenyltrichloroethane).

Haloalkanes And Haloarenes p p Dichlorodiphenyltrichloroethane

Question 53. Industrial preparation of chloroform employs acetone and

  1. Phosgene
  2. Calcium hypochlorite
  3. Chlorine gas
  4. Sodium chloride.

Answer: 3. Chlorine gas

Haloalkanes And Haloarenes Phosgene

Question 54. Phosgene is a common name for

  1. Phosphoryl chloride
  2. Thionyl chloride
  3. Carbon dioxide and phosphine
  4. Carbonyl chloride.

Answer: 4. Carbonyl chloride.

MCQ on Coordination Compounds for NEET

NEET Chemistry For Coordination Compounds Multiple Choice Questions

Question 1. The correct order of the stoichiometries of AgCl formed when AgNO3 in excess is treated with the complexes: \(\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3, \mathrm{CoCl}_3 .5 \mathrm{NH}_3, \mathrm{CoCl}_3 \cdot 4 \mathrm{NH}_3\) respectively is

  1. 3AgCl, 1AgCl, 2AgCl
  2. 3AgCl, 2AgCl, 1AgCl
  3. 2AgCl, 3AgCl, 2AgCl
  4. 1AgCl, 3AgCl, 2AgCl

Answer: 2. 3AgCl, 2AgCl, 1AgCl

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3+3 \mathrm{AgNO}_3 \rightarrow 3 \mathrm{AgCl} \downarrow\) + \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left(\mathrm{NO}_3\right)_3\)

⇒ \({\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2+2 \mathrm{AgNO}_3 \rightarrow 2 \mathrm{AgCl} \downarrow+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]\left(\mathrm{NO}_3\right)_2}\)

⇒ \({\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgCl} \downarrow+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_3}\)

Question 2. Cobalt(3) chloride forms several octahedral complexes with ammonia. Which of the following will not give a test for chloride ions with silver nitrate at 25 °C?

  1. \(\mathrm{CoCl}_3 \cdot 5 \mathrm{NH}_3\)
  2. \(\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3\)
  3. \(\mathrm{CoCl}_3 \cdot 3 \mathrm{NH}_3\)
  4. \(\mathrm{CoCl}_3 \cdot 4 \mathrm{NH}_3\)

Answer: 3. \(\mathrm{CoCl}_3 \cdot 3 \mathrm{NH}_3\)

For octahedral complexes, the coordination number is 6.

Hence, \(\mathrm{CoCl}_3 \cdot 3 \mathrm{NH}_3 \text { i.e., }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\) will not ionise and will not give test for Cl ion with silver nitrate.

Question 3. An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetra aquachromium(3) chloride. The number of moles of AgCl precipitated would be

  1. 0.003
  2. 0.01
  3. 0.001
  4. 0.002

Answer: 3. 0.001

An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetra aquachromium(3) chloride.

⇒ \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}+\mathrm{AgNO}_3 \rightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_3+\mathrm{AgCl}\)

No. of millimoles of solution = 100 mL x 0.01 M

=1 millimole

= 10-3 mole

So, a mole of AgCl = 0.001

Question 4. Which of the following will exhibit maximum ionic conductivity?

  1. \(\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)
  2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\)
  3. \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}_2\)
  4. \(\left[\mathrm{Ni}(\mathrm{CO})_4\right]\)

Answer: 1. \(\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)

Read and Learn More NEET MCQs with Answers

Ionic conductance increases with increasing the number of ions, produced after decomposition.

Coordination Compounds Ions Conductance Increases

Question 5. A coordination complex compound of cobalt has the molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three-mole ions in an aqueous solution. On reaching this solution with an excess of AgNO3 solution, we get two moles of AgCl precipitate. The ionic formula for this complex would be

  1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}_2\)
  2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]\left[\mathrm{Cl}\left(\mathrm{NO}_2\right)\right]\)
  3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right) \mathrm{Cl}\right]\left[\left(\mathrm{NH}_3\right) \mathrm{Cl}\right]\)
  4. \(\left(\mathrm{Co}\left(\mathrm{NH}_3\right)_5\right]\left[\left(\mathrm{NO}_2\right)_2 \mathrm{Cl}_2\right]\)

Answer: 1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}_2\)

A coordination complex compound of cobalt has the molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three-mole ions in an aqueous solution. On reaching this solution with an excess of AgNO3 solution, we get two moles of AgCl precipitate.

As the complex gives two moles of AgCl ppt. with AgNO3 solution, the complex must have two ionisable Cl atoms. Hence, the probable complex, which gives three mole ions may be \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right] \mathrm{Cl}_2\)

⇒ \({\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right] \mathrm{Cl}_2 \rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right]^{2+}+2 \mathrm{Cl}^{-}}\)

Question 6. Which complex compound is most stable?

  1. \(\left[\mathrm{CoCl}_2(e n)_2\right] \mathrm{NO}_3\)
  2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]_2\left(\mathrm{SO}_4\right)_3\)
  3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{Br}\right]\left(\mathrm{NO}_3\right)_2\)
  4. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_3\right)_3\right]\)

Answer: 1. \(\left[\mathrm{CoCl}_2(e n)_2\right] \mathrm{NO}_3\)

Chelating ligands (for example, en) form more stable complexes than unidentate ligands.

Question 7. Ethylene diamine tetraacetate (EDTA) ion is

  1. Tridentate ligand with three “N” donor atoms
  2. Hexadentate ligand with four “O” and two “N” donor atoms
  3. Unidentate ligand
  4. Bidentate ligand with two “N” donor atoms.

Answer: 2. Hexadentate ligand with four “O” and two “N” donor atoms

Ethylenediaminetetraacetate ion (EDTA4-) is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion.

Question 8. The correct increasing order of trans-effect of the following species is

  1. \(\mathrm{NH}_3>\mathrm{CN}^{-}>\mathrm{Br}^{-}>\mathrm{C}_6 \mathrm{H}_5^{-}\)
  2. \(\mathrm{CN}^{-}>\mathrm{C}_6 \mathrm{H}_5^{-}>\mathrm{Br}^{-}>\mathrm{NH}_3\)
  3. \(\mathrm{Br}^{-}>\mathrm{CN}^{-}>\mathrm{NH}_3>\mathrm{C}_6 \mathrm{H}_5^{-}\)
  4. \(\mathrm{CN}^{-}>\mathrm{Br}^{-}>\mathrm{C}_6 \mathrm{H}_5^{-}>\mathrm{NH}_3\)

Answer: 2. \(\mathrm{CN}^{-}>\mathrm{C}_6 \mathrm{H}_5^{-}>\mathrm{Br}^{-}>\mathrm{NH}_3\)

The intensity of the trans-effect (as measured by the increase in the rate of substitution of the trans ligand) follows the sequence: \(\mathrm{CN}^{-}>\mathrm{C}_6 \mathrm{H}_5^{-}>\mathrm{Br}^{-}>\mathrm{NH}_3\)

Question 9. The sum of the coordination number and oxidation number of the metal M in the complex [M(en)2(C2O4)lCl (where en is ethylenediamine) is

  1. 6
  2. 7
  3. 8
  4. 9

Answer: 4. 9

[M(en)2(C2O4)]Cl:

Oxidation number of metal = +3

Coordination number of metal = 6

∴ Sum of oxidation number and coordination number = 3 + 6 = 9

Question 10. The anion of acetylacetone (acac) forms Co(acac)3 chelate with Co3+. The rings of the chelate are

  1. Five membered
  2. Four membered
  3. Six membered
  4. Three membered.

Answer: 3. Six membered

The ligand acetylacetone forms a six-membered chelate ring in the complex [Co(acac)3].

Question 11. Which of the following statements is true?

  1. Silicon exhibits 4 coordination numbers in its compound.
  2. The bond energy of F2 is less than Cl2
  3. Mn(3) oxidation state is more stable than Mn(2) in aqueous state.
  4. Elements of 15th gp show only +3 and +5 oxidation states.

Answer: 2. Bond energy of F2 is less than Cl2.

The bond energy of F2 is less than Cl2 due to interelectronic repulsions in small-sized F-atoms. Silicon exhibits coordination number 6. In an aqueous state, Mn(2) is more stable.

⇒ \(\mathrm{Mn} \rightleftharpoons \mathrm{Mn}^{2+}+2 e^{-}\)

The common oxidation states of t5th group elements are -3, +3 and +5

Question 12. Coordination number of Ni in [Ni(C2O4)3]4is

  1. 3
  2. 6
  3. 4
  4. 2

Answer: 2. 6

(C2O42-) → bidentate ligand.

3 molecules attached from two sides with Ni make coordination number 6.

Question 13. The coordination number and oxidation state of Cr in K3[Cr(C2O4)3] are respectively

  1. 3 and + 3
  2. 3 and 0
  3. 6 and + 3
  4. 4 and + 2

Answer: 3. 6 and + 3

Coordination Compounds Coodination Number And Oxidation State

The number of atoms of the ligands that are directly bound to the central metal is known as the coordination number. It is six here.

Oxidation state : Let oxidation state of Cr be x + 3 (+1) + x+3(-2) =0+ 3 +x- 6= 0 = x=+3

Question 14. Which of the following ligands is expected to be bidentate?

  1. \(\mathrm{CH}_3 \mathrm{NH}_2\)
  2. \(\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{N}\)
  3. \(\mathrm{Br}\)
  4. \(\mathrm{C}_2 \mathrm{O}_4{ }^{2-}\)

Asnwer: 4. \(\mathrm{C}_2 \mathrm{O}_4{ }^{2-}\)

When a ligand has two groups that are capable of bonding to the central atom, it is said to be bidentate. Thus, the only ligand, which is expected to be bidentate is (C2O42-) as

Question 15. Homoleptic complex from the following complexes is

  1. Pentaamminecarbonatocobalt(3) chloride
  2. Triamminetriaquachromium(3) chloride
  3. Potassium trioxalatoaluminate(3)
  4. Diamminechloridonitrito-N-platinum(2)

Answer: 3. Potassium trioxalatoaluminate(3)

Complexes in which a metal is bound to only one kind of ligand are known as homoleptic complexes. Potassium trioxalatoaluminate(3), K3[AlC2O4)3] is an example of a homoleptic complex.

Question 16. The IUPAC name of the complex [Ag(H2O)2][Ag(CN)2] is

  1. Dicyanidosilver(2) diaquaargentate(2)
  2. Diaquasilver (2) dicyanidoargentate(2)
  3. Dicyanidosilver(1) diaquaargentate(1)
  4. Diaquasilver (1) dicyanidoargentate(1).

Answer: 4. Diaquasilver (1) dicyanidoargentate(1).

Question 17. The name of complex ion, [Fe(CN)6]3- is

  1. Hexacyanitoferrate(3) ion
  2. Tricyanoferrate(3) ion
  3. Hexacyanidoferrate(3) ion
  4. Hexacyanoiron(3) ion.

Answer: 3. Hexacyanidoferrate(3) ion

Question 18. The correct IUPAC name for [CrF2(en)2]Cl is

  1. Chlorodifluoridoethylenediaminechromium (3) chloride
  2. Difluoridobis(ethylenediamine)chromium (3) chloride
  3. Difluorobis-(ethylenediamine)chromium (3) chloride
  4. Chlorodifluoridobis(ethylenediamine) chromium(3).

Answer: 2. Difluoridobis(ethylenediamine)chromium (3) chloride

Question 19. The hypothetical complex chlorodiaquatriammine cobalt(3) chloride can be represented as

  1. \(\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_3\left(\mathrm{H}_2 \mathrm{O}\right)_2\right] \mathrm{Cl}_2\)
  2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{Cl}_3\right]\)
  3. \(\left[\mathrm{Co}\left(\mathrm{NH}_2\right)_3\left(\mathrm{H}_2 \mathrm{O}\right)_2 \mathrm{Cl}\right]\)
  4. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{H}_2 \mathrm{O}\right)_3\right] \mathrm{Cl}_3\)

Answer: 1. \(\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_3\left(\mathrm{H}_2 \mathrm{O}\right)_2\right] \mathrm{Cl}_2\)

Chlorodiaquatriamminecobalt(3) chloride can be represented as \(\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_3\left(\mathrm{H}_2 \mathrm{O}\right)_2\right] \mathrm{Cl}_2\)

Question 20. IUPAC name of [Pt(NH3)3(Br)(NO2)Cl]Cl is

  1. Triamminebromochloronitroplatinum(4) chloride
  2. Triamminebromonitrochloroplatinum(4) chloride
  3. Triamminechlorobromonitroplatinum(4) chloride
  4. Triamminenitrochlorobromoplatinum(4) chloride.

Answer: 1. Triamminebromochloronitroplatinum(4) chloride

The ligands are named in the alphabetic order according to the latest IUPAC system. So, the name of \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3 \mathrm{Br}\left(\mathrm{NO}_2\right) \mathrm{Cl}\right] \mathrm{Cl}\) is triamminebromochloronitroplatinum(4) chioride. (The oxidation no. of ‘Pt’ is +4)

Question 21. The formula of dichloro bis(urea)copper(2) is

  1. \(\left[\mathrm{Cu}\left\{\mathrm{O}=\mathrm{C}\left(\mathrm{NH}_2\right)_2\right\} \mathrm{Cl}\right] \mathrm{Cl}\)
  2. \(\left[\mathrm{CuCl}_2\right]\left\{\mathrm{O}=\mathrm{C}\left(\mathrm{NH}_2\right)_2\right\}\)
  3. \(\left[\mathrm{Cu}\left\{\mathrm{O}=\mathrm{C}\left(\mathrm{NH}_2\right)_2\right\} \mathrm{Cl}_2\right.\)
  4. \(\left[\mathrm{CuCl}_2\left\{\mathrm{O}=\mathrm{C}\left(\mathrm{NH}_2\right)_2\right\}_2\right]\)

Answer: 4. \(\left[\mathrm{CuCl}_2\left\{\mathrm{O}=\mathrm{C}\left(\mathrm{NH}_2\right)_2\right\}_2\right]\)

The formula of dichloro bis(urea)copper(2) is [CuCl2{(NH2)2Co}2]

Question 22. The type of isomerism shown by the complex [CoCl2(en)2] is

  1. Geometrical isomerism
  2. Coordination isomerism
  3. Ionization isomerism
  4. Linkage isomerism.

Answer: 1. Geometrical isomerism

[CoCl2(en)2], exhibits geometrical isomerism, as the coordination number of Co is 6 and this compound has octahedral geometry.

Coordination Compounds Octahedral geometry

Question 23. Number of possible isomers for the complex [Co(en)2Cl2]Cl will be (en = ethylenediamine)

  1. 1
  2. 3
  3. 4
  4. 2

Answer: 2. 3

Possible isomers of [Co(en)2Cl2]Cl

Coordination Compounds Possible Iosmers

Question 24. The complexes [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6] are the examples of which type of isomerism?

  1. Linkage isomerism
  2. Ionization isomerism
  3. Coordination isomerism
  4. Geometrical isomerism

Answer: 3. Coordination isomerism

Coordination isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in the complex example, \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]\) and \(\left.\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Co}(\mathrm{CN})_6\right]\)

Question 25. The complex, [Pt(py)(NH2)BrCl] will have how many geometrical isomers?

  1. 3
  2. 4
  3. 0
  4. 2

Answer: 1. 3

⇒ \(\left[\mathrm{Pt}(p y)\left(\mathrm{NH}_3\right) \mathrm{BrCl}\right]\) can have three isomers.

Coordination Compounds Three Geometrical Isomersims

Question 26. The existence of two different coloured complexes with the composition of [Co(NH3)4Cl2]+ is due to

  1. Linkage isomerism
  2. Geometrical isomerism
  3. Coordination isomerism
  4. Ionization isomerism.

Answer: 2. Geometrical isomerism

Coordination Compounds Existance Of Two Different Coloured Complexes Of Two Geometrical Isomerisms

Question 27. Which one of the following complexes is not expected to exhibit isomerism?

  1. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2\right]^{2+}\)
  2. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\)
  3. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\)
  4. \(\left[\mathrm{Ni}(e n)_3\right]^{2+}\)

Answer: 3. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\)

Compounds having tetrahedral geometry do not exhibit isomerism due to the presence of symmetry elements. Here, [Ni(NH3)2Cl2] has a tetrahedral geometry.

Question 28. Which of the following does not show optical isomerism?

  1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]^0\)
  2. \(\left[\mathrm{Co}(\text { en }) \mathrm{Cl}_2\left(\mathrm{NH}_3\right)_2\right]^{+}\)
  3. \(\left[\mathrm{Co}(e n)_3\right]^{3+}\)
  4. \(\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right]^{+}\)(en = ethylenediamine)

Answer: 1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]^0\)

Optical isomerism is shown by

  1. Complexes of the type \({\left[M(A A)_2 Y_2\right]}\), containing one symmetrical bidentate ligand i.e., \(\left[\mathrm{Co}(e n) \mathrm{Cl}_2\left(\mathrm{NH}_3\right)_2\right]^{+} \text {. }\)
  2. Complexes of the type \(\left[M(A A)_3\right]\), containing a symmetrical bidentate ligand i.e., \(\left[\mathrm{Co}(e n)_3\right]^{3+}\).
  3. Complexes of the type \(\left[M(A A)_2 X_2\right]\), i.e. \(\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right]^{+}\).

However, complexes of the type \(\left[M A_3 B_3\right]\) show geometrical isomerism, known as fac-mer isomerism.

∴ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\) exhibits fac-mer isomerism.

Coordination Compounds fac mer isomerism

Question 29. Which of the following will give a pair of enantiomorphs?

  1. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Co}(\mathrm{CN})_6\right]\)
  2. \(\left[\mathrm{Co}(\text { en })_2 \mathrm{Cl}_2\right] \mathrm{Cl}\)
  3. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{PtCl}_6\right]\)
  4. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_2\) (en \(=\mathrm{NH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2\))

Answer: 2. \(\left[\mathrm{Co}(\text { en })_2 \mathrm{Cl}_2\right] \mathrm{Cl}\)

Either a pair of crystals, molecules or compounds that are mirror images of each other but are not identical, and that rotate the plane of polarised light equally, but in opposite directions are called enantiomorphs.

Coordination Compounds Opposite Directions Of Enantiopmorphs

Question 30. [CO(NH3)4(NO2)2]Cl exhibits

  1. Linkage isomerism, geometrical isomerism and optical isomerism
  2. Linkage isomerism, ionization isomerism and optical isomerism
  3. Linkage isomerism, ionization isomerism and geometrical isomerism
  4. Ionization isomerism, geometrical isomerism and optical isomerism.

Answer: 3. Linkage isomerism, ionization isomerism and geometrical isomerism

Ionization isomerism arises when the coordination compounds give different ions in solution.

⇒ \({\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right)_2\right] \mathrm{Cl} \rightleftharpoons\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right)_2\right]^{+}+\mathrm{Cl}^{-}}\)

⇒ \({\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right) \mathrm{Cl}\right] \mathrm{NO}_2 \rightleftharpoons\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right) \mathrm{Cl}\right]^{+}+\mathrm{NO}_2^{-}}\)

Linkage isomerism occurs in complex compounds which contain ambidentate ligands like \(\mathrm{NO}_2^{-}, \mathrm{SCN}^{-}, \mathrm{CN}^{-}, \mathrm{S}_2 \mathrm{O}_3^{2-}\) and \(\mathrm{CO}\).

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right)_2\right] \mathrm{Cl} \text { and }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4(\mathrm{ONO})_2\right] \mathrm{Cl}\) are linkage isomers NO2 is linked through N or through O. Octahedral complexes of the type Ma4b2 exhibit geometrical isomerism

Coordination Compounds Octahedral Complexes Of Ehibit Geometrical Isomerism

Question 31. Which one of the following is expected to exhibit optical isomerism? (en = ethylenediamine)

  1. cis- \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\)
  2. trans- \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\)
  3. cis- \(\left[\mathrm{Co}(\text { en })_2 \mathrm{Cl}_2\right]^{+}\)
  4. trans- \(\left[\mathrm{Co}(\text { en })_2 \mathrm{Cl}_2\right]^{+}\)

Answer: 3. cis- \(\left[\mathrm{Co}(\text { en })_2 \mathrm{Cl}_2\right]^{+}\)

Optical isomerism is not shown by square planar complexes.

Octahedral complexes of general formulae, \(\left[M a_2 b_2 c_2\right]^{n \pm},[M a b c d e f],\left[M(A A)_3\right]^{n \pm},\left[M(A A)_2 a_2\right]^{n \pm}\) (where AA = symmetrical bidentate ligand), \(\left[M(A A)_2 a b\right]^{n \pm} \text { and }\left[M(A B)_3\right]^{n \pm}\)

(where AB = unsymmetrical ligands) show optical isomerism does not show optical isomerism (superimposable mirror image). But cis-form shows optical isomerism.

Coordination Compounds Optical Isomerism

Coordination Compounds cis form shows Optical Isomerism

Question 32. Which of the following coordination compounds would exhibit optical isomerism?

  1. Pentaamminenitrocobalt(3) iodide
  2. Diamminedichloroplatinum(2)
  3. trans-Dicyanobis(ethylenediamine) chromium(3) chloride
  4. tris-(Ethylenediamine)cobalt(3) bromide

Answer: 4. tris-(Ethylenediamine)cobalt(3) bromide

⇒ \(\left[\mathrm{Co}(e n)_3\right]^{3+}\)

Coordination Compounds tris Dichyanobis Chromium Chloride

Question 33. Which of the following will give a maximum number of isomers?

  1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]\)
  2. \(\left[\mathrm{Ni}(\text { en })\left(\mathrm{NH}_3\right)_4\right]^{2+}\)
  3. \(\left[\mathrm{Ni}\left(\mathrm{C}_2 \mathrm{O}_4\right)(e n)_2\right]^{2-}\)
  4. \(\left[\mathrm{Cr}(\mathrm{SCN})_2\left(\mathrm{NH}_3\right)_4\right]^{+}\)

Answer: 4. \(\left[\mathrm{Cr}(\mathrm{SCN})_2\left(\mathrm{NH}_3\right)_4\right]^{+}\)

[Cr(SCN)2(NH3)4]+ shows linkage, geometrical and optical isomerism.

Question 34. Which complex compound will give four isomers?

  1. \(\left[\mathrm{Fe}(e n)_3\right] \mathrm{Cl}_3\)
  2. \(\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right] \mathrm{Cl}\)
  3. \(\left[\mathrm{Fe}\left(\mathrm{PPh}_3\right)_3 \mathrm{NH}_3 \mathrm{ClBr}\right] \mathrm{Cl}\)
  4. \(\left[\mathrm{Co}\left(\mathrm{PPh}_3\right)_3 \mathrm{Cl}\right] \mathrm{Cl}_3\)

Answer: 3. \(\left[\mathrm{Fe}\left(\mathrm{PPh}_3\right)_3 \mathrm{NH}_3 \mathrm{ClBr}\right] \mathrm{Cl}\)

[Fe(PPh3)3NH3ClBr]Cl can give two optical and two geometrical isomers. While other complexes do not form geometrical isomers.

Question 35. The total number of possible isomers for the complex compound [Cu2(NH3)4] [Pt2Cl4] are

  1. 5
  2. 6
  3. 3
  4. 4

Answer: 4. 4

The isomers of the complex compound \(\left[\mathrm{Cu}^{\mathrm{II}}\left(\mathrm{NH}_3\right)_4\right] \left[\mathrm{Pt}^{11} \mathrm{Cl}_4\right]\) are:

  1. \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}\right]\left[\mathrm{Pt}\left(\mathrm{NH}_3\right) \mathrm{Cl}_3\right]\)
  2. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}\right]\left[\mathrm{Cu}\left(\mathrm{NH}_3\right) \mathrm{Cl}_3\right]\)
  3. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{CuCl}_4\right]\)

So, the total no. of isomers is = 4

Question 36. The number of geometrical isomers of the complex [Co(NO2)3(NH3)3] is

  1. 4
  2. 0
  3. 2
  4. 3

Answer: 3. 2

Possible geometrical isomers are :

Coordination Compounds Two Geometrical Iosmers Of the Complex

Question 37. The number of geometrical isomers for [Pt(NH3)2Cl2] is

  1. 3
  2. 4
  3. 1
  4. 2

Answer: 4. 2

Coordination Compounds Two cis form and trans form

Question 38. The order of energy absorbed which is responsible for the colour of complexes

  1. \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2(e n)_2\right]^{2+}\)
  2. \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4(\text { en })\right]^{2+}\) and
  3. \(\left[\mathrm{Ni}(e n)_3\right]^{2+}\) is
  1. (1) > (2) > (3)
  2. (C) > (2) > (1)
  3. (3) > (1) > (2)
  4. (2)> (1) >  (3)

Answer: 3. (3) > (1) > (2)

Chelating ligand increases the stability of the complex compounds and the higher the number of chelating ligands, the higher the stability. The stronger is the strength of the ligand, the greater the energy absorbed by the complex.

Hence, the order is : \(\left[\mathrm{Ni}(e n)_3\right]^{2+}>\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2(e n)_2\right]^{2+}>\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4(e n)\right]^{2+}\)

Question 39. Match List-1 with List-2

Coordination Compounds Match The Columns

Choose the correct answer from the options given below.

  1. (1) – (D), (2) – (A), (3) – (B), (4) – (C)
  2. (1) – (D), (2) – (B), (3) – (A), (4) – (C)
  3. (1) – (B), (2) – (D), (3) – (C), (4) – (A)
  4. (1) – (A), (2) – (C), (3) – (D), (4) – (B)

Answer: 1. (1) – (D), (2) – (A), (3) – (B), (4) – (C)

⇒ \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\):

Coordination Compounds Number Of Unpaired Electrons

Spin only magnetic moment = \(\sqrt{n(n+2)}\)

where n = number of unpaired electrons.

Hence, n = 1.

μ = \(\sqrt{1(1+2)}=\sqrt{3}=1.73\) B.M

Coordination Compounds Spin Magnetic Moment

n = 5, \(\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92\) B.M.

Coordination Compounds Spin Magnetic Moment

n = 4, \(\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\) B.M

Question 40. Which of the following is the correct order of increasing the field strength of ligands to form coordination compounds?

  1. \(\mathrm{SCN}^{-}<\mathrm{F}^{-}<\mathrm{C}_2 \mathrm{O}_4^{2-}<\mathrm{CN}^{-}\)
  2. \(\mathrm{SCN}^{-}<\mathrm{F}^{-}<\mathrm{CN}^{-}<\mathrm{C}_2 \mathrm{O}_4^{2-}\)
  3. \(\mathrm{F}^{-}<\mathrm{SCN}^{-}<\mathrm{C}_2 \mathrm{O}_4^{2-}<\mathrm{CN}^{-}\)
  4. \(\mathrm{CN}^{-}<\mathrm{C}_2 \mathrm{O}_4^{2-}<\mathrm{SCN}^{-}<\mathrm{F}^{-}\)

Answer: 1. \(\mathrm{SCN}^{-}<\mathrm{F}^{-}<\mathrm{C}_2 \mathrm{O}_4^{2-}<\mathrm{CN}^{-}\)

According to the spectrochemical series, the order of increasing field strength is : \(\mathrm{SCN}^{-}<\mathrm{F}^{-}<\mathrm{C}_2 \mathrm{O}_4^{2-}<\mathrm{CN}^{-}\)

Question 41. What is the correct electronic configuration of the central atom in K4[Fe(CN)6] based on the crystal field

Answer: 3

In K4[Fe(CN)6] complex, Fe is in +2 oxidation state.

Coordination Compounds Strong Field Ligand

As CN is a strong field ligand, it causes pairing of electrons therefore, electronic configuration of \(\mathrm{Fe}^{2+}\) in \(\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]\) is \(t_{2 \mathrm{~g}}^6 e_{g^6}^0\)

Question 42. Aluminium chloride in acidified aqueous solution forms a complex ‘A’, in which the hybridisation state of ‘A’ is ‘B’. What are ‘A’ and ‘B’ respectively?

  1. \(\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}, s p^3 d^2\)
  2. \(\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}, s p^3\)
  3. \(\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}, d s p^2\)
  4. \(\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}, d^2 s p^3\)

Answer: 1. \(\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}, s p^3 d^2\)

Question 43. The crystal field stabilisation energy (CFSE) for will be CoCl6]4- is 18000 cm-1 The CFSE for [CoCl4]2-

  1. 6000 cm-1
  2. 16000 cm-1
  3. 18000 cm-1
  4. 8000 cm-1

Answer: 4. 8000 cm-1

⇒ \(\Delta_t=\frac{4}{9} \Delta_o=\frac{4}{9} \times 18000=8000 \mathrm{~cm}^{-1}\)

Question 44. The geometry and magnetic behaviour of the complex [Ni(CO)4] are

  1. Square planar geometry and diamagnetic
  2. Tetrahedral geometry and diamagnetic
  3. Square planar geometry and paramagnetic
  4. Tetrahedral geometry and paramagnetic.

Answer: 2. Tetrahedral geometry and diamagnetic

Ni(28) : [Ar]3d84s2

CO is a strong field ligand, so, unpairedelectrons get paired.

Coordination Compounds Geometry And Mganetic Electrons

Thus, the complex is sp3 hybridised with tetrahedral geometry and is diamagnetic in nature.

Question 45. Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co3+ is

  1. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Co}(e n)_3\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  2. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{Co}(e n)_3\right]^{3+}\)
  3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{Co}(e n)_3\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)
  4. \(\left[\mathrm{Co}(en)_3\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(2017)\)

Answer: 4. \(\left[\mathrm{Co}(en)_3\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(2017)\)

Increasing order of crystal field splitting energy is \(\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\text { en }\)

Thus, increasing order of crystal field splitting energy for the given complexes is : \({\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{Co}(e n)_3\right]^{3+}}\)

As, E = \(\frac{h c}{\lambda}\)

Thus, increasing order of wavelength of absorption is : \(\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)

Question 46. Pick out the correct statement with respect to [Mn(CN)6]3-.

  1. It is sp³d2 hybridised and tetrahedral.
  2. It is d2sp³ hybridised and octahedral.
  3. It is dsp2 hybridised and square planar.
  4. It is sp³d2 hybridised and octahedral.

Answer: 2. It is d2sp³ hybridised and octahedral.

[Mn(CN)6]3-: Letoxidation state of Mn be x.

x+6x(-1)=-3 + x=+3

Electronic configuration of Mn : [Ar]4s2 3d5

Electronic configuration of Mn3+ : [Ar]3d4

CN is a strong field ligand thus, it causes the pairing of electrons in the 3d-orbital.

Coordination Compounds Octahedral

Thus, [Mn(CN)6]3- has d²sp³ hybridization and has octahedral geometry.

Question 47. Jahn-Teller effect is not observed in high spin complexes of

  1. d7
  2. d8
  3. d4
  4. d9

Answer: 2. d8

Jahn-Teller distortion is usually significant for asymmetrically occupied eg orbitals since they are directed towards the ligands and the energy gain is considerably greater. In the case of unevenly occupied t2g, orbitals, the fan-teller distortion is very weak since the t2g, set does not point directly at the ligands and therefore, the energy gain is much less.

High spin complexes:

Coordination Compounds Jahn Teller Distortion

Question 48. The hybridization involved in complex [Ni(CN)4]2- is (At. No. Ni = 28)

  1. sp³
  2. d²sp²
  3. d²sp³
  4. dsp²

Answer: 4. dsp²

[Ni(CN)4]2-: Oxidation number of Ni = +2

Electronic configuration of Ni2+: [Ar]3d84s0

Coordination Compounds Hybridization Is Involved

The pairing of electrons in the d-orbital takes place due to the presence of a strong field ligand (CN).

Question 49. Among the following complexes the one which shows zero crystal field stabilization energy (CFSE) is

  1. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)
  2. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)
  3. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  4. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)

Answer: 2. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)

H2O is a weak field ligand, hence Δ0 < pairing energy.

CFSE=(-0.4x+0.6y)Δ0

where r and y are no. of electrons occupying t2g and eg orbitals respectively.

For [Fe(H2O)6]3+ complex ion,

Fe3+ (3d5) = t³2g e²g=-0.4×3+0.6 x2= 0.0or 0 Δ0

Question 50. A magnetic moment at 1.73 BM will be shown by one of the following

  1. \(\mathrm{TiCl}_4\)
  2. \(\left[\mathrm{CoCl}_6\right]^{4-}\)
  3. \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}\)
  4. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\)

Answer: 3. \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}\)

Oxidation state of Cu in I Cu(NH3)4)2+ is + 2

Coordination Compounds Oxidation Stage Of Cu

It has one unpaired electron (n = 1)

μ = \(\sqrt{n(n+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.73 \mathrm{BM}\)

Question 51. Crystal field splitting energy for high spin d4 octahedral complex is

  1. – 1.2 Δ0
  2. – 0.6 Δ0
  3. -0.8 Δ0
  4. – 1.6 Δ0

Answer: 2. – 0.6 Δ0

CFSE = (- 0.4 x + 0.6 y) Δ0

where, x = No. of electrons occupying t2g orbitals

y = no. of electrons occupying eg orbitals

= (- 0.4 x 3 + 0.6 x 1)Δ0 [High spin d4 =t2g3eg1)

= (- 1.2 + 0.6)Δ0 = -0.6 Δ0

Question 52. Which among the following is a paramagnetic complex?

  1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  2. \(\left[\mathrm{Pt}(e n) \mathrm{Cl}_2\right]\)
  3. \(\left[\mathrm{CoBr}_4\right]^{2-}\)
  4. \(\mathrm{Mo}(\mathrm{CO})_6\)

(Atomic number Mo = 42, Pt = 78)

Answer: 3. \(\left[\mathrm{CoBr}_4\right]^{2-}\)

Co2+ in [CoBr4]2- has 3d74s0 configuration and Br is a weak field ligand. Thus, it has 3 unpaired electrons and hence, paramagnetic.

Question 53. Which is diamagnetic?

  1. \(\left[\mathrm{CoF}_6\right]^{3-}\)
  2. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\)
  3. \(\left[\mathrm{NiCl}_4\right]^{2-}\)
  4. \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\)

Answer: 2. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\)

In [Ni(CN)4]2- all orbitals are doubly occupied, hence, it is diamagnetic

Coordination Compounds Diamagnetic

CN is a strong field ligand and causes the pairing of 3d-electrons of Ni2+

Question 54. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?

  1. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}\)
  2. \(\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}\)
  3. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  4. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)

Answer: 1. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}\)

⇒ \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}: s p^3 d^2\) (outer), octahedral, paramagnetic

⇒ \(\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}: s p^3 d^2\) (outer), octahedral, diamagnetic

⇒ \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}: d^2 s p^3\) (inner), octahedral, paramagnetic

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}: d^2 s p^3\) (inner), octahedral, diamagnetic

Question 55. The low spin complex of d6-cation in an octahedral field will have the following energy

  1. \(\frac{-12}{5} \Delta_o+P\)
  2. \(\frac{-12}{5} \Delta_o+3 P\)
  3. \(\frac{-2}{5} \Delta_o+2 P\)
  4. \(\frac{-2}{5} \Delta_o+P\)

0 = crystal field splitting energy in an octahedral field, P = Electron pairing energy)

Answer: 2. \(\frac{-12}{5} \Delta_o+3 P\)

CFSE = (-0.4x + 0.6y)Δ0 + zP

where x = number of electrons occupying t2g orbital

y = number of electrons occupying eg orbital

z = number of pairs of electrons

For low spin d6 complex electronic configuration

= \(t_{2 g}{ }^6 e_g^0\) or \(t_{2 g}{ }^{2,2,2} e_g^0\)

∴ x=6, y=0, z=3

CFSE = \((-0.4 \times 6+0 \times 0.6) \Delta_0+3 P\)

= \(\frac{-12}{5} \Delta_o+3 P\)

Question 56. A red precipitate is obtained when an ethanol solution of dimethylglyoxime is added to ammoniacal Ni(2). Which of the following statements is not true?

  1. The red complex has a square planar geometry.
  2. The complex has symmetrical H-bonding.
  3. The red complex has a tetrahedral geometry.
  4. Dimethylglyoxime functions as a bidentate ligand.

Coordination Compounds Dimethylglyoxime

Answer: 3. Red complex has a tetrahedral geometry.

[Ni(dmg)2] is square planar in structure not tetrahedral.

Question 57. Of the following complex ions, which is diamagnetic in nature?

  1. \(\left[\mathrm{NiCl}_4\right]^{2-}\)
  2. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\)
  3. \(\left[\mathrm{CuCl}_4\right]^{2-}\)
  4. \(\left[\mathrm{CoF}_6\right]^{3-}\)

Answer: 2. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\)

Number of unpaired electrons = 2

Hence, [NiCl4]2- is paramagnetic

Coordination Compounds Its A Paramagnetic

Number of unpaired electrons = 0, so it is diamagnetic in nature.

Coordination Compounds Diamaganetic nature

No. of unpaired electron = 1, so it is paramagnetic.

Coordination Compounds Unpaired Electrons So Its A Symmetrical

No. of unpaired electrons = 4, so it is paramagnetic.

Question 58. The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2+ are d4, d5, d6 and d7 respectively. Which one of the following will exhibit minimum paramagnetic behaviour?

  1. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  2. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  3. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  4. \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

(Atomic number Cr = 24, Mn = 25, Fe = 26, Co = 27)

Answer: 3. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2+ are d4, d5, d6 and d7 respectively.

∴ \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}: \mathrm{Mn}^{2+}=3 d^6\)

Number of unpaired electrons =5

∴ \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} ; \mathrm{Fe}^{2+}=3 d^6\)

Number of unpaired electrons =4

∴  \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}: \mathrm{Co}^{2+}=3 d^7\)

Number of unpaired electrons =3

∴ \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}: \mathrm{Cr}^{2+}=3 d^4\)

Number of unpaired electrons =4

Minimum paramagnetic behaviour is shown by \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

Question 59. Which of the following complex compounds will exhibit the highest paramagnetic behaviour?

  1. \(\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  2. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  4. \(\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}\)

(Atomic number Ti = 22, Cr = 24, Co = 27, Zn = 30)

Answer: 2. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)

Ti\(:[\mathrm{Ar}] 3 d^2 4 s^2, \mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 d^1 4 s^0\) (1 unpaired electron)

Cr:\([\mathrm{Ar}] 3 d^4 4 s^2, \mathrm{Cr}^{34}:[\mathrm{Ar}] 3 d^3 4 s^0\) (3 unpaired electrons)

Co: \([\mathrm{Ar}] 3 d^7 4 s^2, \mathrm{Co}^{3+} ;[\mathrm{Ar}] 3 d^6 4 s^0\)

(No unpaired electrons because of pairing)

Zn:\([\mathrm{Ar}] 3 d^{10} 4 s^2, \mathrm{Zn}^{2+}:[\mathrm{Ar}] 3 d^{10}\) (No unpaired electrons)

∴ \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\) exhibits the highest paramagnetic behaviour as it contains 3 unpaired electrons.

Question 60. Which of the following complex ions is not expected to absorb visible light?

  1. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\)
  2. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  3. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  4. \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

Answer: 1. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\)

A transition metal complex absorbs visible light only if it has unpaired electrons.

Coordination Compounds Transition Metal Complex

No unpaired electron so does not absorb visible light.

Question 61. Crystal field stabilization energy for high spin d4 octahedral complex is

  1. – 1.8 Δ0
  2. – 1.6 Δ0 + P
  3. – 1.2 Δ0
  4. – 0.6 Δ0

Answer: 4. – 0.6 Δ0

Coordination Compounds d orbitals In Symmetrical Filed Of Ligands

Question 62. Out of \(\mathrm{TiF}_6^{2-}, \mathrm{CoF}_6^{3-}, \mathrm{Cu}_2 \mathrm{Cl}_2\) and \(\mathrm{NiCl}_4^{2-}\) (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28) the colourless species are

  1. \(\mathrm{Cu}_2 \mathrm{Cl}_2\) and \(\mathrm{NiCl}_4{ }^{2-}\)
  2. \(\mathrm{TiF}_6{ }^{2-}\) and \(\mathrm{Cu}_2 \mathrm{Cl}_2\)
  3. \(\mathrm{CoF}_6{ }^{3-}\) and \(\mathrm{NiCl}_4{ }^{2-}\)
  4. \(\mathrm{TiF}_6{ }^{2-}\) and \(\mathrm{CoF}_6{ }^{3-}\)

Answer: 2. \(\mathrm{TiF}_6{ }^{2-}\) and \(\mathrm{Cu}_2 \mathrm{Cl}_2\)

A species is coloured when it contains unpaired d-electrons which are capable of understanding d-d transition on adsorption of light of a particular wavelength.

In \(\mathrm{TiF}_6{ }^{2-}, \mathrm{Ti}^{4+}: 3 d^0,\) colourless

In \(\mathrm{CoF}_6{ }^{3-}, \mathrm{Co}^{3+}: 3 d^6\), coloured

In \(\mathrm{Cu}_2 \mathrm{Cl}_2, \mathrm{Cu}^{+}: 3 d^{10},\) colourless

In \(\mathrm{NiCl}_4^{2-}, \mathrm{Ni}^{2+}: 3 d^8,\) coloured

Thus, \(\mathrm{TiF}_6^{2-}\left(3 d^0\right)\) and \(\mathrm{Cu}_2 \mathrm{Cl}_2\left(3 d^{10}\right)\) with empty and fully filled d-orbitals appear colourless as they are not capable of undergoing d-d transition

Question 63. Which of the following complex ions is expected to absorb visible light?

  1. \(\left[\mathrm{Ti}(e n)_2\left(\mathrm{NH}_3\right)_2\right]^{4+}\)
  2. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  3. \(\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}\)
  4. \(\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_3\left(\mathrm{NH}_3\right)_3\right]^{3+}\)

[Atomic number Zn = 30, Sc = 21, Ti = 22, Cr = 24]

Answer: 2. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)

⇒ \(\mathrm{Ti}^{4+} \rightarrow 3 d^0, \mathrm{Cr}^{3+} \rightarrow 3 d^3\)

⇒ \(\mathrm{Zn}^{2+} \rightarrow 3 d^{10}, \mathrm{Sc}^{3+} \rightarrow 3 d^0\)

Question 64. Which of the following complexes exhibits the highest paramagnetic behaviour?

  1. \(\left[\mathrm{Co}(o x)_2(\mathrm{OH})_2\right]^{-}\)
  2. \(\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  3. \(\left[\mathrm{V}(\mathrm{gl} \text { ) })_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}\)
  4. \(\left[\mathrm{Fe}(e n)(b p y)\left(\mathrm{NH}_3\right)_2\right]^{2+}\)

where gly = glycine, en = ethylenediamine and bpv = bipyridyl moities. (At. nos. Ti = 22, V = 23, USB Fe = 26, Co = 27)

Answer: 1. \(\left[\mathrm{Co}(o x)_2(\mathrm{OH})_2\right]^{-}\)

Question 65. In which of the following coordination entities the magnitude of Δ0 (CFSE in the octahedral field) will be maximum?

  1. \(\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}\)
  2. \(\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}\)
  3. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)
  4. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)

(Atomic number Co = 27)

Answer: 1. \(\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}\)

When the ligands are arranged ln order of the magnitude of crystal field splitting, the arrangement, thus, obtained is called a spectrochemical series.

Arranged in increasing field strength as \(\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{Cl}^{-}<\mathrm{NO}_3^{-}<\mathrm{F}^{-}<\mathrm{OH}^{-}<\mathrm{C}_2 \mathrm{O}_4{ }^{2-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\text { en }\) \(<\mathrm{NO}_2^{-}<\mathrm{CN}^{-}<\mathrm{CO}\)

It has been observed that ligands before H2O are weak field ligands while ligands after H2O are strong field ligands.

Coordination Compounds Strong And Weak Field Ligands

CFSE in an octahedral field depends upon the nature of ligands.

The stronger the ligands larger the value of Δoct.

Question 66. The d electron configurations of \(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}\) and \(\mathrm{Ni}^{2+}\) are \(3 d^4, 3 d^5, 3 d^6\) and \(3 d^8\) respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour?

  1. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  2. \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  3. \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  4. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

(Atomic number Cr = 24, Mn = 25, Fe = 26, Ni = 28)

Answer: 2. \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

Coordination Compounds Minimum Paramagnetic Behaviour

The greater the number of unpaired electrons, the higher is the paramagnetism. Hence, [Ni(H2O)6]2+ will exhibit the minimum paramagnetic behaviour.

Question 67. [Cr(H2O)6]Cl3 (Atomic number of Cr = 24) has a magnetic moment of 3.83 B.M. The correct distribution of 3d electrons in the chromium of the complex is

  1. \(3 d_{x y}^1, 3 d_{y z}^1, 3 d_{z^2}^1\)
  2. \(3 d_{\left(x^2-y^2\right)}^1, 3 d_{z^2}^1, 3 d_{x z}^1\)
  3. \(3 d_{x y}^1, 3 d_{\left(x^2-y^2\right)}^1, 3 d_{y z}^1\)
  4. \(3 d_{x y}^1, 3 d_{y z}^1, 3 d_{x z}^1\)

Answer: 4. \(3 d_{x y}^1, 3 d_{y z}^1, 3 d_{x z}^1\)

Magneti moment = \(\sqrt{n(n+2)}\)

3.83 = \(\sqrt{n(n+2)} \Rightarrow(3.83)^2=n(n+2)\)

14.6689 = \(n^2+2 n\)

On solving the equation, n=3

Coordination Compounds magnetic Moment

Question 68. Which one of the following is an inner orbital complex as well as diamagnetic in behaviour?

  1. \(\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}\)
  2. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)
  3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{34}\)
  4. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}\)

(Atomic number : Zn = 30, Cr = 24, Co = 27, Ni = 28)

Answer: 3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{34}\)

Coordination Compounds Electron Pair Fom Six Ligands

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} ; \mathrm{Co}(27):[\mathrm{Ar}]^{18} 3 d^7 4 s^2\)

electron pair from six ligands \(\left(\mathrm{NH}_3\right)\)

⇒ \(d^2 s p^3 \rightarrow\) inner octahedral complex and diamagnetic.

⇒ \(\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+} \rightarrow s p^3 d^2\) (outer)and diamagnetic.

⇒ \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+} \rightarrow d^2 s p^3\) (inner) and paramagnetic.

⇒ \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+} \rightarrow s p^3 d^2\) (outer)and paramagnetic.

Question 69. Among \(\left[\mathrm{Ni}(\mathrm{CO})_4\right],\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-},\left[\mathrm{NiCl}_4\right]^{2-}\) species, the hybridisation states at the Ni atom are, respectively

  1. \(s p^3, d s p^2, d s p^2\)
  2. \(s p^3, d s p^2, s p^3\)
  3. \(s p^3, s p^3, d s p^2\)
  4. \(d s p^2, s p^3, s p^3\)

[Atomic number of Ni = 28]

Answer: 2. \(s p^3, d s p^2, s p^3\)

Coordination Compounds Weak Field Ligand

Question 70. CN is a strong field ligand. This is due to the fact that

  1. It carries a negative charge
  2. It is a pseudohalide
  3. It can accept electrons from metal species
  4. It forms high-spin complexes with metal species.

Answer: 2. It is a pseudohalide

Cyanide ion is a strong field ligand because it is a pseudohalide ion. Pseudohalide ions are stronger coordinating ligands and they have the ability to form σ-bond (from the pseudohalide to the metal) and π bond (from the metal to pseudohalide).

Question 71. Considering H2O as a weak field ligand, the number of unpaired electrons in [Mn(H2O)6]2+ will be (atomic number of Mn = 25)

  1. Three
  2. Five
  3. Two
  4. Four.

Answer: 2. Five

Mn (25):3d54s2

Coordination Compounds Spin Complex

In the presence of a weak field ligand, there will be no pairing of electrons. So, it will form a high spin complex, i.e., the number of unpaired electrons = 5.

Question 72. In an octahedral structure, the pair of d orbitals involved in d²sp³ hybridisation is

  1. \(d_{x^2-y^2}, d_{z^2}\)
  2. \(d_{x z}, d_{x^2-y^2}\)
  3. \(d_z, d_{x z}\)
  4. \(d_{x y}, d_{y z}\)

Answer: 1. \(d_{x^2-y^2}, d_{z^2}\)

In the formation of d²sp³ hybrid orbitals, two (n – 1) d orbitals of eg set i.e. (n -1)d²z and (n – 1)dx²-y²  orbitals), one ns and three np(npx npy and npz) orbitals combine together and form six d²sp³ hybrid orbitals.

Question 73. The number of unpaired electrons in the complex ion [CoF6]3- is

  1. 2
  2. 3
  3. 4
  4. Zero.

(Atomic number Co = 27)

Answer: 3. 4

⇒ \(\left[\mathrm{CoF}_6\right]^{3-}\)

Coordination Compounds Four Unpaired Electrons

Thus, the number of unpaired electrons = 4

Question 74. The atomic numbers of Cr and Fe are respectively 24 and 26, which of the following is paramagnetic with the spin of an electron?

  1. \(\left[\mathrm{Cr}(\mathrm{CO})_6\right]\)
  2. \(\left[\mathrm{Fe}(\mathrm{CO})_5\right]\)
  3. \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\)
  4. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)

Answer: 4. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\)

Odd electrons, ions and molecules are paramagnetic.

In Cr(CO)6 molecule 12 electrons are contributed by the CO group and it contains no odd electron.

Cr: \(3 d^6 4 s^1\)

∴ \(\mathrm{Fe}(\mathrm{CO})_5\) molecule also does not contain odd electron.

Fe: \(3 d^6 4 s^2\)

In \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\) ion, \(\mathrm{Fe}(+2): 3 d^6 4 s^0\)

∴ No odd electrons,

In \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}\) ion, \(\mathrm{Cr}(+3): 3 d^3 4 s^0\)

This ion contains odd electrons so it is paramagnetic.

Question 75. Which statement is incorrect?

  1. \(\mathrm{Ni}(\mathrm{CO})_4\) – tetrahedral, paramagnetic
  2. \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\) – square planar, diamagnetic
  3. \(\mathrm{Ni}(\mathrm{CO})_4\) – tetrahedral, diamagnetic
  4. \(\left[\mathrm{NiCl}_4\right]^{2-}\) – tetrahedral, paramagnetic

Answer: 1. \(\mathrm{Ni}(\mathrm{CO})_4\) – tetrahedral, paramagnetic

In Ni(CO)4 complex, Ni(0) will have 3d10 configuration

Coordination Compounds Tetrahedral And Diagmagnetic

Hence, [Ni(CO)4] will have tetrahedral geometry and diamagnetic as there are no unpaired electrons.

Question 76. Iron carbonyl, Fe(CO)5 is

  1. Tetranuclear
  2. Mononuclear
  3. Trinuclear
  4. Dinuclear.

Answer: 2. Mononuclear

Based on the number of metal atoms present in a complex, they are classified as :

Example, \(\mathrm{Fe}(\mathrm{CO})_5:\) mononuclear; \(\mathrm{Co}_2(\mathrm{CO})_8\): dinuclear ; \(\mathrm{Fe}_3(\mathrm{CO})_{12}\) : trinuclear

Question 77. An example of a sigma-bonded organometallic compound is

  1. Grognards reagent
  2. Ferrocene
  3. Cobaltocene
  4. Ruthenocene.

Answer: 1. Grognard reagent

In sigma-bonded organometallic complexes, the metal atom and carbon atom of the ligand are joined together with a sigma bond, i.e., the ligand contributes one electron and is therefore, called one electron donor, for example, Grignard’s reagent R-Mg-X

Question 78. Which of the following has the longest C—O bond length? (Free C—O bond length in CO is 1.128 Å)

  1. \(\left[\mathrm{Fe}(\mathrm{CO})_4\right]^{2-}\)
  2. \(\left[\mathrm{Mn}(\mathrm{CO})_6\right]^{+}\)
  3. \(\mathrm{Ni}(\mathrm{CO})_4\)
  4. \(\left[\mathrm{Co}(\mathrm{CO})_4\right]^{-}\)

Answer: 1. \(\left[\mathrm{Fe}(\mathrm{CO})_4\right]^{2-}\)

The greater the negative charge on the carbonyl complex, the more easy it would be for the metal to permit its electrons to participate in the back bonding, the higher the M-C bond order and simultaneously there would be a larger reduction in the C-O bond order. Thus, [Fe(CO)4]2- has the lowest C-O bond order means the longest bond length.

Question 79. Which of the following carbonyls will have the strongest C – O bond?

  1. \(\mathrm{Mn}(\mathrm{CO})_6^{+}\)
  2. \(\mathrm{Cr}(\mathrm{CO})_6\)
  3. \(\mathrm{V}(\mathrm{CO})_6^{-}\)
  4. \(\mathrm{Fe}(\mathrm{CO})_5\)

Answer: 1. \(\mathrm{Mn}(\mathrm{CO})_6^{+}\)

The presence of a positive charge on the metal carbonyl would resist the flow of the metal electron charge to π orbitals of CO. This would increase the CO bond order and hence, CO in a metal carbonyl cation would absorb at a higher frequency compared to its absorption in a neutral metal carbonyl.

Question 80. Which of the following does not have a metal-carbon bond?

  1. \(\mathrm{Al}\left(\mathrm{OC}_2 \mathrm{H}_5\right)_3\)
  2. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{MgBr}\)
  3. \(\mathrm{K}\left[\mathrm{Pt}\left(\mathrm{C}_2 \mathrm{H}_4\right) \mathrm{Cl}_3\right]\)
  4. \(\mathrm{Ni}(\mathrm{CO})_4\)

Answer: 1. \(\mathrm{Al}\left(\mathrm{OC}_2 \mathrm{H}_5\right)_3\)

AI(OC2H5)3 contains bonding through O and thus it does not have a retail-carbon bond.

Question 81. Among the following which is not the π-bonded organometallic compound?

  1. \(\mathrm{K}\left[\mathrm{PtCl}_3\left(\eta^2-\mathrm{C}_2 \mathrm{H}_4\right)\right]\)
  2. \(\mathrm{Fe}\left(\eta^5-\mathrm{C}_5 \mathrm{H}_5\right)_2\)
  3. \(\mathrm{Cr}\left(\eta^6-\mathrm{C}_6 \mathrm{H}_6\right)_2\)
  4. \(\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}\)

Answer: 4. \(\left(\mathrm{CH}_3\right)_4 \mathrm{Sn}\)

π-bonded organometallic compounds include organometallic compounds of alkenes, alkynes and some other carbon-containing compounds having electrons in their p-orbitals.

Question 82. Which of the following organometallic compounds is σ and π-bonded?

  1. \(\left[\mathrm{Fe}\left(\eta^5-\mathrm{C}_5 \mathrm{H}_5\right)_2\right]\)
  2. \(\mathrm{K}\left[\mathrm{PtCl}_3\left(\eta^2-\mathrm{C}_2 \mathrm{H}_4\right)\right]\)
  3. \(\left[\mathrm{Co}(\mathrm{CO})_5 \mathrm{NH}_3\right]^{2+}\)
  4. \(\mathrm{Fe}\left(\mathrm{CH}_3\right)_3\)

Answer: 3. \(\left[\mathrm{Co}(\mathrm{CO})_5 \mathrm{NH}_3\right]^{2+}\)

[Co(CO)5NH3]2+: In this complex, Co-atom is attached with NH3 through o bonding and with CO through dative π-bond.

Question 83. Shape of Fe(CO)5 is

  1. Octahedral
  2. Square planar
  3. Trigonal bipyramidal
  4. Square pyramidal.

Answer: 3. Trigonal bipyramidal

Question 84. In metal carbonyl having general formula M(CO)x where M = metal, x = 4 and the metal is bonded to

  1. Carbon and oxygen
  2. C ≡O
  3. Oxygen
  4. Carbon.

Answer: 4. Carbon.

In M(CO)4, metal is bonded to the ligands via carbon atoms with both σ and πbond character. Both metal-to-ligand and ligand-to-metal bonding are possible.

Question 85. Which of the following complexes is used to be as an anticancer agent?

  1. mer- \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\)
  2. cis- \(\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2\right]\)
  3. cis- \(\mathrm{K}_2\left[\mathrm{PtCl}_2 \mathrm{Br}_2\right]\)
  4. \(\mathrm{Na}_2 \mathrm{CoCl}_4\)

Answer: 2. cis- \(\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2\right]\)

Question 86. Copper sulphate dissolves in excess of KCN to give

  1. \(\mathrm{Cu}(\mathrm{CN})_2\)
  2. \(\mathrm{CuCN}\)
  3. \(\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}\)
  4. \(\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}\)

Answer: 3. \(\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}\)

First cupric cyanide is formed which decomposes to give cuprous cyanide and cyanogen gas. Cuprous cyanide dissolves in excess of potassium cyanide to form a complex, potassium cyanide \(\left[\mathrm{K}_3 \mathrm{Cu}(\mathrm{CN})_4\right]\)

Coordination Compounds Potassium Cyanide

Question 87. Which of the following is considered to be an anticancer species?

Coordination Compounds Anticancer Species

Answer: 3

cis-platin is cls-[PtCl2NH3)2] is used as an anticancer agent.

Question 88. In the silver plating of copper, K[Ag(CN)2] is used instead of AgNO3. The reason is

  1. A thin layer of Ag is formed on Cu
  2. More voltage is required
  3. Ag+ ions are completely removed from the solution
  4. Less availability of Ag+ ions, as Cu cannot displace Ag from [Ag(CN)2] ion.

Answer: 4. Less availability of Ag+ ions, as Cu cannot displace Ag from [Ag(CN)2]

Copper being more electropositive readily precipitates silver from their salt (Ag+) solution

⇑ \(\mathrm{Cu}+2 \mathrm{AgNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{Ag}\)

In \(\mathrm{K}\left[\mathrm{Ag}(\mathrm{CN})_2\right]\) solution, a complex anion \(\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}\) is

Question 89. CuSO4 when reacts with KCN forms CuCN, which is insoluble in water. It is soluble in excess of KCN, due to the formation of the following complex

  1. \(\mathrm{K}_2\left[\mathrm{Cu}(\mathrm{CN})_4\right]\)
  2. \(\mathrm{K}_3\left[\mathrm{Cu}(\mathrm{CN})_4\right]\)
  3. \(\mathrm{CuCN}_2\)
  4. \(\mathrm{Cu}\left[\mathrm{KCu}(\mathrm{CN})_4\right]\)

Answer: 2. \(\mathrm{K}_3\left[\mathrm{Cu}(\mathrm{CN})_4\right]\)

Copper sulphate reacts with potassium cyanide giving a white precipitate of cuprous cyanide and cyanogen gas. The cuprolls cyanide dissolves in excess of KCN forming potassium cuprocyanide K3[Cu(CN)2].

⇒ \(2 \mathrm{CuSO}_4+4 \mathrm{KCN} \rightarrow 2 \mathrm{CuCN}+(\mathrm{CN})_2+2 \mathrm{~K}_2 \mathrm{SO}_4\)

⇒ \(\mathrm{CuCN}+3 \mathrm{KCN} \rightarrow \mathrm{K}_3\left[\mathrm{Cu}(\mathrm{CN})_4\right]\)

Question 90. Hypo is used in photography to

  1. Reduce AgBr grains to metallic silver
  2. Convert metallic silver to silver salt
  3. Remove undecomposed silver bromide as a soluble complex
  4. Remove reduced silver

Answer: 3. Remove undecomposed silver bromide as a soluble complex

Undecomposed AgBr forms a soluble complex with hypo and the reaction is given as:

⇒ \(\mathrm{AgBr}+2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \rightarrow \underset{\mathrm{ Soluble complex}}{\mathrm{Na}_3\left[\mathrm{Ag}\left(\mathrm{S}_2 \mathrm{O}_3\right)_2\right]}+\mathrm{NaBr}\)

 

NEET MCQ’s on d & f Block Elements

NEET Chemistry For d And f Block Elements Multiple Choice Questions

Question 1. Sc(Z=21) is a transition element but Zn (Z = 30) is not because

  1. Both Sc3+ and Zn2+ ions are colourless and form white compounds
  2. In the case of Sc, 3d orbitals are partially filled but in Zn, these are filled
  3. The last electron is assumed to be added to the 4s level in the case of Zn
  4. Both Sc and Zn do not exhibit variable gn oxidation states.

Answer: 2. In the case of Sc, 3d orbitals are partially filled but in Zn, these are filled

Sc (Z = 21) has incompletely filled 3d-orbitals in its ground state (3d1), it is considered as a transition element but Zn (Z = 30) has completely filled d-orbitals (3d10) in its ground state and its common oxidation state (+2), thus, it is not considered as a transition element.

Question 2. Which of the following ions has electronic configuration [Ar] 3d6?

  1. \(\mathrm{Ni}^{3+}\)
  2. \(\mathrm{Mn}^{3+}\)
  3. \(\mathrm{Fe}^{3+}\)
  4. \(\mathrm{Co}^{3+}\)

(Atomic nunbers Mn = 25, Fe = 26, Co = 27, Ni = 28)

Answer: 4. \(\mathrm{Co}^{3+}\)

The electronic configuration ofthe given ions is \(\mathrm{Ni}^{3+}:[\mathrm{Ar}] 3 d^0 4 s^0, \mathrm{Mn}^{3+}:[\mathrm{Ar}] 3 d^4 4 s^0\); \(\mathrm{Fe}^{3+}:[\mathrm{Ar}] 3 d^5 4 s^0, \mathrm{Co}^{3+}:[\mathrm{Ar}] 3 d^6 4 s^0\)

Question 3. Among the following series of transition metal ions, the one where all metal ions have 3d² electronic configuration is

[Atomic number Ti = 22, V = 23, Cr = 24, Mn = 25]

  1. \(\mathrm{Ti}^{3+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{4+}\)
  2. (\(\mathrm{Ti}^{+}, \mathrm{V}^{4+}, \mathrm{Cr}^{6+}, \mathrm{Mn}^{7+}\)
  3. \(\mathrm{Ti}^{4+}, \mathrm{V}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Mn}^{3+}\)
  4. \(\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}\)

Answer: 4. \(\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}\)

⇒ \({ }_{22} \mathrm{Ti}: 3 d^2 4 s^2 ; \mathrm{Ti}^{2+}: 3 d^2\)

⇒ \({ }_{23} \mathrm{~V}: 3 d^3 4 s^2 ; \mathrm{V}^{3+}: 3 d^2\)

⇒ \({ }_{24} \mathrm{Cr}: 3 d^4 4 s^2 ; \mathrm{Cr}^{4+}: 3 d^2\)

⇒ \({ }_{25} \mathrm{Mn}: 3 d^5 4 s^2 ; \mathrm{Mn}^{5+}: 3 d^2\)

Read and Learn More NEET MCQs with Answers

Question 4. Which of the following configurations is correct for iron?

  1. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^7\)
  2. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^5\)
  3. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5\)
  4. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^6\)

Answer: 4. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^6\)

Question 5. Which of the following has more unpaired d-electrons?

  1. \(\mathrm{N}^{3+}\)
  2. \(\mathrm{Fe}^{2+}\)
  3. \(\mathrm{Zn}^{+}\)
  4. \(\mathrm{Cu}^{+}\)

Answer: 2. \(\mathrm{Fe}^{2+}\)

Question 6. The electronic configuration of transition elements is exhibited by

  1. \(n s^1\)
  2. \(n s^2 n p^5\)
  3. \(n s^2(n-1) d^{1-10}\)
  4. \(n s^2(n-1) d^{10}\)

Answer: 3. \(n s^2(n-1) d^{1-10}\)

The general electronic configuration of transition elements is ns2 (n – 1)d1-10.

Question 7. The electronic configurations of four elements are given below. Which element does not belong to the same family as others?

  1. \([\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^2\)
  2. \([\mathrm{Kr}] 4 d^{10} 5 s^2\)
  3. \([\mathrm{Ne}] 3 s^2 3 p^5\)
  4. \([\mathrm{Ar}] 3 d^{10} 4 s^2\)

Anwer: 3. \([\mathrm{Ne}] 3 s^2 3 p^5\)

It is the electronic configuration of a p-block element whereas other configurations are those of d-block elements

Question 8. The stability of Cu2+ is more than Cu+ salts in an aqueous solution due to

  1. Hydration energy
  2. Second ionisation enthalpy
  3. First ionisation enthalpy
  4. Enthalpy of atomization.

Answer: 1. Hydration energy

The stability of \(\mathrm{Cu}_{(a q)}^{2+}\) rather than \(\mathrm{Cu}_{(\mathrm{aq})}^{+}\) is due to the much more negative \(\Delta_{\text {ind }} H^{\circ}\) of \(\mathrm{Cu}_{(\text {app }}^{2+}\) than \(\mathrm{Cu}_{(\text {aqp })}^{+}\), which more than compensates for the second ionisation enthalpy of \(\mathrm{Cu}\). \(\mathrm{V}_2 \mathrm{O}_4\) dissolves in acids to give \(\mathrm{VO}^{2+}\) salts.

Question 9. Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of

  1. Having similar chemical properties
  2. Belonging to the same group
  3. Diagonal relationship
  4. Lanthanoid contraction.

Answer: 4. Lanthanoid contraction.

The atomic and ionic radii of Zt and Hf are almost identical due to the poor shielding effect of 4f-electrons, which leads to lanthanoid contraction.

Question 10. Identify the incorrect statement.

  1. \(\mathrm{Cr}^{2+}\left(d^4\right)\) is a stronger reducing agent than \(\mathrm{Fe}^{2+}\left(d^6\right)\) in water.
  2. Transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.
  3. Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
  4. The oxidation states of chromium in \(\mathrm{CrO}_4^{2-}\) and \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) are not the same.

Answer: 4. The oxidation states of chromium in \(\mathrm{CrO}_4^{2-}\) and \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) are not the same.

The oxidation states of Cr in \(\mathrm{CrO}_4^{2-}\) and \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) is same i.e., +6.

Question 11. The calculated spin-only magnetic moment of Cr2+ ion is

  1. 3.87 BM
  2. 4.90 BM
  3. 5.92 BM
  4. 2.84 BM

Answer: 2. 4.90 BM

Cr: \(3 d^5 4 s^1, \mathrm{Cr}^{2+}: 3 d^4\) has four unpaired electrons.

μ \(=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24} \approx 4.90\) B.M.

Question 12. Match the metal ions given in Column A with the spin magnetic moments of the ions given in Column B and assign the correct code:

d And f Block Elements Metal Ions Abd Spin Magnetic Moments

  1. 1-D, 2-E, 3-B, 4-A
  2. 1-A, 2-B, 3-C, 4-D
  3. 1-D, 2-A, 3-B, 4-C
  4. 1-C, 2-E, 3-A, 4-B

Answer: 1. 1-D, 2-E, 3-B, 4-A

⇒ \(\mathrm{Co}^{3+}:[\mathrm{Ar}] 3 d^6\), unpaired \(e^{-}(n)=4\)

Spin magnetic moment \((\mu)=\sqrt{4(4+2)}=\sqrt{24}\) B.M.

⇒ \(\mathrm{Cr}^{3+}:[\mathrm{Ar}] 3 d^3\), unpaired \(e^{-}(n)=3\)

Spin magnetic moment \((\mu)=\sqrt{3(3+2)}=\sqrt{15}\) B.M.

⇒ \(\mathrm{Fe}^{3+}:[\mathrm{Ar}] 3 d^5\), unpaired \(e^{-}(n)=5\)

Spin magnetic moment \((\mu)=\sqrt{5(5+2)}=\sqrt{35}\) B.M.

⇒ \(\mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 d^s\), unpaired \(e^{-}(n)=2\)

Spin magnetic moment \((\mu)=\sqrt{2(2+2)}=\sqrt{8}\) B.M.

Question 13. Magnetic moment 2.84 B.M. is given by (Atomic number Ni = 28, Ti = 22, Cr = 24, Co = 27)

  1. \(\mathrm{Cr}^{2+}\)
  2. \(\mathrm{Co}^{2+}\)
  3. \(\mathrm{Ni}^{2+}\)
  4. \(\mathrm{Ti}^{3+}\)

Answer: 3. \(\mathrm{Ni}^{2+}\)

Magnetic moment \((\mu)=\sqrt{n(n+2)}\)

2.84 B.M. corresponds to 2 unpaired electrons.

Cr2+; 3d4, 4 unpaired electrons

⇒ \(\mathrm{Co}^{2+}: 3 d^p, 3\) unpaired electrons

⇒ \(\mathrm{Ni}^{2+}: 3 d^8, 2\) unpaired electrons

⇒ \(\mathrm{Ti}^{3+}: 3 d^1, 1\) unpaired electron

Question 14. Which of the following processes does not involve the oxidation of iron?

  1. Formation of Fe(CO)5 from Fe.
  2. Liberation of H2 from steam by iron at high temperature.
  3. Rusting of iron sheets.
  4. Decolourisation of blue CuSO4 solution by iron.

Answer: 1. Formation of Fe(CO)5 from Fe.

The oxidation number of Fe in Fe(CO)5 is zero.

Question 15. Which of the following statements about the interstitial compounds is incorrect?

  1. They are much harder than the pure metal.
  2. They have higher melting points than the pure metal.
  3. They retain metallic conductivity.
  4. They are chemically reactive.

Answer: 4. They are chemically reactive.

Interstitial compounds are generally chemically inert.

Question 16. Identify the alloy containing a non-metal as a constituent in it.

  1. Invar
  2. Steel
  3. Bell metal
  4. Bronze

Answer: 2. Steel

Invar ⇒ Ni(metal) + Fe(metal)

Steel ⇒ C(non-metal) + Fe(metal)

Betrl ⇒ Cu(metal) + Sn(metal) + F

Bronze ⇒ Cu(rnetal) + Sn(metal)

Question 17. The catalytic activity of transition metals and their compounds is ascribed mainly to

  1. Their magnetic behaviour
  2. Their unfilled d-orbitals
  3. Their ability to adopt variable oxidation states
  4. Their chemical reactivity.

Answer: 3. Their ability to adopt variable oxidation states

Question 18. Which one of the following does not correctly represent the correct order of the property indicated against it?

  1. Ti < V < Cr < Mn; increasing number of oxidation states
  2. Ti3+ < V3+ < Cr3+ < Mn3+; increasing magnetic moment
  3. Ti < V < Cr < Mn; increasing melting points
  4. Ti < V < Mn < Cr; increasing 2nd ionization enthalpy

Answer: 3. Ti < V < Cr < Mn; increasing melting points

d And f Block Elements Melting Point Order And Enthalpy Order

Hence, the given order is correct.

Magnetic moment \((\mu)=\sqrt{n(n+2)}\) B.M.

For \(\mathrm{Ti}^{3+} n=1, \mu=\sqrt{1(1+2)}=\sqrt{3}\) B.M.

For \(\mathrm{V}^{3+} n=2, \mu=\sqrt{2(2+2)}=\sqrt{8}\) B.M.

For \(\mathrm{Cr}^{3+} n=3, \mu=\sqrt{3(3+2)}=\sqrt{15}\) B.M.

For \(\mathrm{Mn}^{3+} n=4, \mu=\sqrt{4(4+2)}=\sqrt{24}\) B.M.

Thus, magnetic moment order : \(\mathrm{Ti}^{3+}<\mathrm{V}^{3+}<\mathrm{Cr}^{3+}<\mathrm{Mn}^{3+}\)

Question 19. Four successive members of the first series of transition metals are listed below. For which one of them does the standard potential (E°M²/M) value have a positive sign?

  1. Co (Z = 27)
  2. Ni (Z = 28)
  3. Cu(Z=29)
  4. Fe (Z = 26)

Asnwer: 3. Cu(Z=29)

d And f Block Elements Standard Potential Values

Question 20. For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order?

  1. Mn > Fe > Cr > Co
  2. Fe > Mn > Co > Cr
  3. Co > Mn > Fe > Cr
  4. Cr > Mn > Co > Fe

(Atomic numbers Cr = 24, Mn = 25, Fe = 26, Co = 27)

Answer: 1. Mn > Fe > Cr > Co

Spin correlation and exchange energy give an electronic configuration special stability which is greatest for half-filled electronic configurations. Mn2+ (d5) gets stabilisation due to half-filled configuration’ In Fe2+ (d6) the placing of one extra electron in a subshell destabilises. Placing 2 electrons in Co2+ (d7) destabilises it more. Cr2+ (d6) has one vacant subshell. Fe2+ gets more -stabilisation compared to Cr2+ through exchange energy So, the order is as follows: Mn > Fe > Cr > Co.

Question 21. Which of the following ions will exhibit colour in aqueous solutions?

  1. \(\mathrm{La}^{3+}(Z=57)\)
  2. \(\mathrm{Ti}^{3+}(Z=22)\)
  3. \(\mathrm{Lu}^{3+}(Z=71)\)
  4. \(\mathrm{Sc}^{3+}(Z=21)\)

Answer: 2. \(\mathrm{Ti}^{3+}(Z=22)\)

Ions which have unpaired electrons exhibit colour in aqueous solution. Ti3+ has an outer electronic configuration of 4s03d1, i.e., 1 unpaired electron. Thus, its solution will be coloured. Others are colourless due to empty or completely filled outermost orbitals.

Question 22. Which of the following pairs has the same size?

  1. \(\mathrm{Fe}^{2+}, \mathrm{Ni}^{2+}\)
  2. \(\mathrm{Zr}^{4+}, \mathrm{Ti}^{4+}\)
  3. \(\mathrm{Zr}^{4+}, \mathrm{Hf}^{4+}\)
  4. \(\mathrm{Zn}^{2+}, \mathrm{Hf}^{4+}\)

Answer: 3. \(\mathrm{Zr}^{4+}, \mathrm{Hf}^{4+}\)

Hf+ and Zr++ belong to group 4B. However, Hf4+ has the same size as Zr4+ due to the addition of 14 lanthanide elements before it in which electrons are added into the f subshell which poorly shields the outer electrons and contraction in size occurs.

Question 23. Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?

  1. \(3 d^5 4 s^1\)
  2. \(3 d^5 4 s^2\)
  3. \(3 d^2 4 s^2\)
  4. \(3 d^3 4 s^2\)

Answer: 2. \(3 d^5 4 s^2\)

The greater the number of valence electrons, the more will be the number of oxidation states exhibited by the element.

3d54s1 can show a maximum of 6 oxidation states.

3d54s2, carr shows a maximum of 7 oxidation states.

3d54s2 can show a maximum of 4 oxidation states.

3d34s2 can show a maximum of 5 oxidation states

Question 24. The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is

  1. Mn > Cr > Ti > V
  2. Ti > V > Cr > Mn
  3. Cr > Mn > V > Ti
  4. V > Mn > Cr > Ti

Answer: 3. Cr > Mn > V > Ti

The electronic configuration of the given elements are

Mn: \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^2\)

Cr: \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1\)

Ti: \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^2 4 s^2\)

V: \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3 4 s^2\)

In general, ionization potential (both lst and 2nd) increases from left to right across the period due to an increase in effective nuclear charge. On this basis, the second IP values should exhibit the trend: Mn>Cr>V>Ti

But the actual observed order is: Cr > Mn > V > Ti Practically, only chromium is exceptional and the rest others show the normal trend. This exceptional behaviour of chromium is due to the stable configuration (3d5) that it achieves after the loss of the first electron.

Question 25. In which of the following pairs are both the ions coloured in aqueous solution?

(Atomic number Sc = 21, Ti = 22, Ni = 28, Cu = 29,Co = 27)

  1. \(\mathrm{Ni}^{2+}, \mathrm{Cu}^{+}\)
  2. \(\mathrm{Ni}^{2+}, \mathrm{Ti}^{3+}\)
  3. \(\mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}\)
  4. \(\mathrm{Sc}^{3+}, \mathrm{Co}^{2+}\)

Answer: 2. \(\mathrm{Ni}^{2+}, \mathrm{Ti}^{3+}\)

Sc: \([\mathrm{Ar}] 3 d^1 4 s^2, \mathrm{Sc}^{3+}:[\mathrm{Ar}]\) Colourless

Ti: \([\mathrm{Ar}] 3 d^2 4 s^2, \mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 d^1\) Coloured

Ni: \([\mathrm{Ar}] 3 d^3 4 s^2, \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 d^8\) Coloured

Cu: \([\mathrm{Ar}] 3 d^{10} 4 s^1, \mathrm{Cu}^{+}:[\mathrm{Ar}] 3 d^{10}\) Colourless

Co: \([\mathrm{Ar}] 3 d^7 4 s^2, \mathrm{Co}^{2+}:[\mathrm{Ar}] 3 d^7\) Coloured

⇒ \(\mathrm{Ti}^{3+}, \mathrm{Ni}^{2+}\) and \(\mathrm{Co}^{2+}\) are coloured due to presence of unpaired electrons.

Question 26. Four successive members of the first-row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionisation enthalpy?

  1. Vanadium (Z = 23)
  2. Chromium (Z = 24)
  3. Manganese (Z = 25)
  4. Iron (Z = 26)

Answer: 3. Manganese (Z = 25)

  • \(\mathrm{V}^{2+}(23):[\mathrm{Ar}] 3 d^3 4 s^0\)
  • \(\mathrm{Cr}^{2+}(24):[\mathrm{Ar}] 3 d^4 4 s^0\)
  • \(\mathrm{Mn}^{2+}(25):[\mathrm{Ar}] 3 d^5 4 s^0\)
  • \(\mathrm{Fe}^{2+}(26):[\mathrm{Ar}] 3 d^5 4 s^1\)

Question 27. The aqueous solution containing which one of the following ions will be colourless?

(Atomic number : Sc = 21, Fe = 26,Ti = 22, Mn = 25)

  1. \(\mathrm{Sc}^{3+}\)
  2. \(\mathrm{Fe}^{2+}\)
  3. \(\mathrm{Ti}^{3+}\)
  4. \(\mathrm{Mn}^{2+}\)

Answer: 1. \(\mathrm{Sc}^{3+}\)

  • If the transition metal ion has an unpaired electron then it shows colour.\(\mathrm{Sc}^{3+}:[\mathrm{Ar}] 3 d^0 4 s^0\)
  • \(\mathrm{Fe}^{2+}:[\mathrm{Ar}] 3 d^5 4 s^1\)
  • \(\mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 d^1 4 s^0\)
  • \(\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 d^5 4 s^0\)

Sc3+ does not contain unpaired electrons, hence it will not undergo d – d transition and will not show colour.

Question 28. Which one of the following characteristics of the transition metals is associated with their catalytic activity?

  1. High enthalpy of atomization
  2. Paramagnetic behaviour
  3. Colour of hydrated ions
  4. Variable oxidation states

Answer: 4. Variable oxidation states

The transition elements, on account of their variable valency, are able to form unstable intermediate compounds very readily.

Question 29. The basic character of the transition metal monoxides follows the order

(Atomic numbers Ti = 22, V = 23, Cr = 24, Fe = 26)

  1. VO > CrO > TiO > FeO
  2. CrO > VO > FeO > TiO
  3. TiO > FeO > VO > CrO
  4. TiO > VO > CrO > FeO

Answer: 4. TiO > VO > CrO > FeO

The order of basicity of transition metal monoxides is, TiO > VO > CrO > FeO.

Question 30. Which of the following shows a maximum number of oxidation states?

  1. Cr
  2. Fe
  3. Mn
  4. V

Answer: 3. Mn

Each of the elements in groups 3 B to 7 B can show the maximum oxidation state equal to its group number. Mn in group seven and shows a maximum oxidation state of +7 in KMnO4.

Question 31. Which ion is colourless?

  1. \(\mathrm{Cr}^{4+}\)
  2. \(\mathrm{Sc}^{3+}\)
  3. \(\mathrm{Ti}^{3+}\)
  4. \(\mathrm{V}^{3+}\)

Answer: 2. \(\mathrm{Sc}^{3+}\)

⇒ \({ }_{21} \mathrm{Sc}:[\mathrm{Ar}] 3 d^1 4 s^2\)

In Sc3+ there are no unpaired ‘d’ electrons, therefore it is colourless in its solution.

Question 32. Bell metal is an alloy of

  1. Cu + Zn
  2. Cu + Sn
  3. Cu + Pb
  4. Cu + Ni

Answer: 2. Cu + Sn

Bell metal ⇒ Cu = 80%, Sn = 20%

It is used for making belts, utensils, etc.

Question 33. In which of the following compounds does transition metal have a zero oxidation state?

  1. \(\mathrm{NOClO}_4\)
  2. \(\mathrm{NH}_2 \mathrm{NH}_2\)
  3. \(\mathrm{CrO}_5\)
  4. \(\left[\mathrm{Fe}(\mathrm{CO})_5\right]\)

Answer: 4. \(\left[\mathrm{Fe}(\mathrm{CO})_5\right]\)

In iron carbonyl, the oxidation number of ‘Fe’ is zero.

⇒ \(\left[\mathrm{Fe}(\mathrm{CO})_5\right]: x+5 \times 0=0 \Rightarrow x=0\)

Question 34. Which one of the following ionic species will impart colour to an aqueous solution?

  1. \(\mathrm{Zn}^{2+}\)
  2. \(\mathrm{Cu}^{+}\)
  3. \(\mathrm{Ti}^{4+}\)
  4. \(\mathrm{Cr}^{3+}\)

Answer: 4. \(\mathrm{Cr}^{3+}\)

⇒ \(\mathrm{Cr}^{3+}(24): 1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^3\)

As Cr3+ ion has three unpaired electrons in its valence shell, so it imparts colour to an aqueous solution.

Question 35. A transition element 10 has a configuration [Ar]3d4 in its +3 oxidation state. Its atomic number is

  1. 22
  2. 19
  3. 25
  4. 26

Answer: 3. 25

The metal atom will have three more electrons.

Therefore, the atomic number of the metal = 18 + 4 + 3 =25

Question 36. Amongst \(\mathrm{TiF}_6^{2-}, \mathrm{CoF}_6^{3-}, \mathrm{Cu}_2 \mathrm{Cl}_2 \text { and } \mathrm{NiCl}_4^{2-} \text {, }\) which are the colourless species? (Atomic number of Ti = 22, Co = 27, Cu = 29, Ni = 28)

  1. \(\mathrm{CoF}_6^{3-}\) and \(\mathrm{NiCl}_4^{2-}\)
  2. \(\mathrm{TiF}_6^{2-}\) and \(\mathrm{Cu}_2 \mathrm{Cl}_2\)
  3. \(\mathrm{Cu}_2 \mathrm{Cl}_2\) and \(\mathrm{NiCl}_4^{2-}\)
  4. \(\mathrm{TiF}_6^{2-}\) and \(\mathrm{CoF}_6^{3-}\)

Answer: 2. \(\mathrm{TiF}_6^{2-}\) and \(\mathrm{Cu}_2 \mathrm{Cl}_2\)

In TiF62- titanium is in +4 oxidation state. In Cu2Cl2, the copper is in +1 state. Thus, in both cases, the transition from one d-orbital to another is not possible.

Ti: \([\mathrm{Ar}] 3 d^2 4 s^2 \rightarrow \mathrm{Ti}^{4+}:[\mathrm{Ar}] 3 d^0 4 s^0\)

Cu:\([\mathrm{Ar}] 3 d^{10} 4 s^1 \rightarrow \mathrm{Cu}^{+}:[\mathrm{Ar}] 3 d^{10} 4 s^0\)

Question 37. The mercury is the only metal which is liquid at 0°C. This is due to its

  1. High Vapour Pressure
  2. Weak Metallic Bond
  3. High Ionization Energy
  4. Both (2) And (3).

Answer: 4. Both (2) And (3).

The very high ionisation energy of Hg makes it difficult for electrons to participate in metallic bonding.

Question 38. Which of the following statements is incorrect?

  1. All the transition metals except scandium form MO oxides which are ionic.
  2. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc2O3 to Mn2O7
  3. Basic character increases from V2O3 to V2O4 to V2O5
  4. V2O4 dissolves in acids to give VO43- salts.
  5. CrO is basic but Cr2O3 is amphoteric.

Choose the correct answer from the options given below:

  1. 3 and 4 only
  2. 2 and 3 only
  3. 1 and 5 only
  4. 2 and 4 only

Answer: 1. 3 and 4 only

In vanadium, there is a gradual change of basic character from the basic V2O3 to less basic V2O4 and to amphoteric V2O5

Question 39. In the neutral or faintly alkaline medium, KMnO4 oxidises iodide into iodate. The change in the oxidation state of manganese in this reaction is from

  1. +7 to +4
  2. +6 to +4
  3. +7 to +3
  4. +6 to +5

Answer: 1. +7 to +4

In the neutral or faintly alkaline medium, KMnO4 oxidises iodide into iodate.

Reaction of MnO4 with I in neutral or faintly alkaline solution: \(\stackrel{+7}{2 \mathrm{MnO}_4^{-}}+\mathrm{H}_2 \mathrm{O}+\stackrel{-1}{\mathrm{I}^{-}} \longrightarrow \stackrel{+4}{\mathrm{MnO}_2}+2 \mathrm{OH}^{-}+\stackrel{+5}{\mathrm{IO}_3^{-}}\)

Question 40. The manganate and permanganate ions are tetrahedral, due to

  1. The π-bonding involves the overlap of d-orbitals of oxygen with d- d-orbitals of manganese
  2. The π-bonding involves the overlap of p-orbitals of oxygen with d-orbitals of manganese
  3. There is no π-bonding
  4. The π-bonding involves the overlap of the p-orbitals of oxygen with the p-orbitals of manganese.

Answer: 2. The π-bonding involves the overlap of p-orbitals of oxygen with d-orbitals of manganese

d And f Block Elements Manganate And Permanganate Ions

In manganate and permanganate ions, π-bonding takes place by the overlap of p -p-orbitals of oxygen with the d-orbitals of manganese.

Question 41. When neutral or faintly alkaline KMnO4 is treated with potassium iodide, the iodide ion is converted into ‘X’. ‘X’ is

  1. \(\mathrm{I}_2\)
  2. \(\mathrm{IO}_4^{-}\)
  3. \(\mathrm{IO}_3^{-}\)
  4. \(\mathrm{IO}^{-}\)

Answer: 3. \(\mathrm{IO}_3^{-}\)

In neutral or faintly alkaline solutions: \(2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O}+\mathrm{I}^{-} \longrightarrow 2 \mathrm{MnO}_2+2 \mathrm{OH}^{-}+\mathrm{IO}_3^{-}\)

Question 42. Which one of the following ions exhibits d-d transition and paramagnetism as well?

  1. \(\mathrm{CrO}_4^{2-}\)
  2. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\)
  3. \(\mathrm{MnO}_4^{-}\)
  4. \(\mathrm{MnO}_4^{2-}\)

Answer: 4. \(\mathrm{MnO}_4^{2-}\)

In \(\mathrm{CrO}_4^{2-}, \mathrm{Cr}^{6+}(n=0)\) diamagnetic

In \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}, \mathrm{Cr}^{6+}(n=0)\) diamagnetic

In \(\mathrm{MnO}_4^{-}, \mathrm{Mn}^{7+}(n=0)\) diamagnetic

In \(\mathrm{MnO}_4^{2-}, \mathrm{Mn}^{6+}(n=1)\) paramagnetic

In \(\mathrm{MnO}_4^{2-}\), one unpaired electron (n) is present in the d-orbital so, the d-d transition is possible.

Question 43. Name the gas that can readily decolourise acidified KMnO4 solution.

  1. \(\mathrm{SO}_2\)
  2. \(\mathrm{NO}_2\)
  3. \(\mathrm{P}_2 \mathrm{O}_5\)
  4. \(\mathrm{CO}_2\)

Answer: 1. \(\mathrm{SO}_2\)

SO, readily decolourises the pink-violet colour of acidified KMnO4 solution.

⇒ \({2 \mathrm{KMnO}_4}+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+\underset{\mathrm{(Colourless)}}{2 \mathrm{MnSO}_4}+2 \mathrm{H}_2 \mathrm{SO}_4\)

Question 44. Which one of the following statements is correct when SO2 is passed through acidified K2Cr3O73 solution?

  1. SO2 is reduced.
  2. Green Cr2(SO4)3 is formed.
  3. The solution turns blue.
  4. The solution is decolourised.

Answer: 2. Green Cr2(SO4)3 is formed.

⇒ \(\begin{aligned}
\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4+3 \mathrm{SO}_2 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4 \\
+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O} \\
(\text { Green) }
\end{aligned}\)

Question 45. Assuming complete ionisation, same moles of which of the following compounds will require the least amount of acidified KMnO4 for complete oxidation?

  1. \(\mathrm{FeSO}_3\)
  2. \(\mathrm{FeC}_2 \mathrm{O}_4\)
  3. \(\mathrm{Fe}\left(\mathrm{NO}_2\right)_2\)
  4. \(\mathrm{FeSO}_4\)

Answer: 4. \(\mathrm{FeSO}_4\)

⇒ \(\mathrm{KMnO}_4\left(\mathrm{Mn}^{7+}\right)\) changes to \(\mathrm{Mn}^{2+}\) i.e., number of electrons involved per mole of \(\mathrm{KMnO}_4\) is 5

1. For \(\mathrm{FeSO}_3\), \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \quad(\mathrm{No}\), of \(e^{-}\)s involved =1)

⇒ \(\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}\) (No. of electrons involved=2)

Total number of \(e^{-} s\) involved =1+2=3

2. For \(\mathrm{FeC}_2 \mathrm{O}_4\), \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}\)(No. of es involved =1)

⇒ \(\mathrm{C}_2 \mathrm{O}_4^{2-} \longrightarrow 2 \mathrm{CO}_2(\mathrm{No}\), of \(e^{-}s\) involved =2)

Total number of \(e^{-} \mathrm{s}\) involved =1+2=3

3. For \(\mathrm{Fe}\left(\mathrm{NO}_2\right)_2\), \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}(\) No. of \(e^{-}\)s involved=1)

⇒ \(2 \mathrm{NO}_2^{-} \longrightarrow 2 \mathrm{NO}_3^{-}\)(No. of \(e^{-}\)s involved =4)

Total number of \(e^{-}\)s involved =1+4=5

For \(\mathrm{FeSO}_4\), \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}\) (Number of \(e^{-}\)s involved =1)

Total number of \(e^{-}\)s involved =1

⇒ As \(\mathrm{FeSO}_4\) requires the least number of electrons thus, it will require the least amount of \(\mathrm{KMnO}_4\).

Question 46. The reaction of aqueous KMnO4 with TT2O2 in acidic conditions gives

  1. \(\mathrm{Mn}^{4+}\) and \(\mathrm{O}_2\)
  2. \(\mathrm{Mn}^{2+}\) and \(\mathrm{O}_2\)
  3. \(\mathrm{Mn}^{2+}\) and \(\mathrm{O}_3\)
  4. \(\mathrm{Mn}^{4+}\) and \(\mathrm{MnO}_2\)

Answer: 2. \(\mathrm{Mn}^{2+}\) and \(\mathrm{O}_2\)

Hydrogen peroxide is oxidised to H2O and O2

Hydrogen peroxide is oxidised to \(\mathrm{H}_2 \mathrm{O}\) and \(\mathrm{O}_2\)
\(2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2\)

or, \(2 \mathrm{MnO}_4^{-}+5 \mathrm{H}_2 \mathrm{O}_2+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2\)

Question 47. Which of the statements is not true?

  1. On passing H2S through an acidified K2Cr2O7 solution, a milky colour is observed.
  2. Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis.
  3. K2Cr2O7 solution in an acidic medium is orange.
  4. K2Cr2O7 solution becomes yellow MB on increasing the pH beyond 7.

Solution: 2. Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis.

Potassium dichromate is preferred over sodium dichromate in volumetric analysis, primarily because the latter is hygroscopic in nature and therefore, accurate weighing is not possible in a normal atmosphere.

Question 48. The acidified K2Cr2O7 solution turns green when Na2SO3 is added to it. This is due to the formation of

  1. \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\)
  2. \(\mathrm{CrO}_4^{2-}\)
  3. \(\mathrm{Cr}_2\left(\mathrm{SO}_3\right)_3\)
  4. \(\mathrm{CrSO}_4\)

Answer: 1. \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\)

The acidified K2Cr2O7 solution turns green when Na2SO3 is added to it.

d And f Block Elements Acidic Conditions

Question 49. The number of moles of KMnO4 reduced by one mole of KI in an alkaline medium is

  1. One
  2. Two
  3. Five
  4. One Fifth.

Answer: 2. Two

d And f Block Elements Alkaline Medium

Question 50. K2Cr2O7 on heating with aqueous NaOH gives

  1. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\)
  2. \(\mathrm{Cr}(\mathrm{OH})_2\)
  3. \(\mathrm{CrO}_4^{2-}\)
  4. \(\mathrm{Cr}(\mathrm{OH})_3\)

Answer: 3. \(\mathrm{CrO}_4^{2-}\)

⇒ \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{NaOH} \rightarrow \mathrm{K}_2 \mathrm{CrO}_4+\mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{H}_2 \mathrm{O}\) or \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}\)

Question 51. KMnO4 reacts with oxalic acid according to the equation \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4{ }^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\), Here 20 mL of 0.1 M KMnO4 is equivalent to

  1. 50 mL of 0.5 M C2H2O4
  2. 20 mL of 0.1 M C2H2O4
  3. 20 mL of 0.5 M C2H2O4
  4. 50 mL of 0.1 M C2H2O4

Answer: 4. 50 mL of 0.1 M C2H2O4

⇒ \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

∴ 2 moles of MnO4 = 5 moles of C2O2-4

20 mL of 0.1 M KMnO4 = 2 mmol of KMnO4

Also, 50 mL of 0.1 M C2H2O4 = 5 mmol of C2O2-4

Therefore, these are equivalent

Question 52. The oxidation state of Cr in K2Cr2O7 is

  1. +5
  2. +3
  3. +6
  4. +7

Answer: 3. +6

Let, the oxidation state of Cr in K2Cr2O7 is x. Then

2+2x-14=0

⇒ 2x=12

‍∴ x=+6

Question 53. Gadolinium has a low value of third ionisation, enthalpy because of

  1. Small size
  2. High exchange enthalpy
  3. High electronegativity
  4. High basic character.

Answer: 2. High exchange enthalpy

Due to high exchange enthalpy Gd3+ (4f7) acquires extra stability and has low third ionisation enthalpy.

Question 54. Which one of the following statements related to lanthanons is incorrect?

  1. Europium shows a +2 oxidation state.
  2. The basicity decreases as the ionic radius decreases from Pr to Lu.
  3. All the lanthanoids are much more reactive than aluminium.
  4. Ce(+4) solutions are widely used as (H) oxidizing agents in volumetric analysis.

Answer: 3. All lanthanons are much more reactive than aluminium.

The first few members of the lanthanoid series are quite reactive, almost like calcium. However, with increasing atomic numbers, their behaviour becomes similar to that of aluminium.

Question 55. The electronic configurations of Eu (Atomic Number 63), Gd (Atomic Number 64) and Tb (Atomic Number 65) are

  1. \([\mathrm{Xe}] 4 f^6 5 d^1 6 s^2,[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2\) and \([\mathrm{Xe}] 4 f^8 5 d^1 6 s^2\)
  2. \([\mathrm{Xe}] 4 f^7 6 s^2,[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2\) and \([\mathrm{Xe}] 4 f^9 6 s^2\)
  3. \([\mathrm{Xe}] 4 f^7 6 s^2,[\mathrm{Xe}] 4 f^8 6 s^2\) and \([\mathrm{Xe}] 4 f^8 5 d^1 6 s^2\)
  4. \([\mathrm{Xe}] 4 f^6 5 d^1 6 s^2,[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2\) and \([\mathrm{Xe}] 4 f^9 6 s^2\)

Answer: 2. \([\mathrm{Xe}] 4 f^7 6 s^2,[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2\) and \([\mathrm{Xe}] 4 f^9 6 s^2\)

Question 56. Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?

  1. \([\mathrm{Xe}] 4 f^3 5 s^1\)
  2. \([\mathrm{Xe}] 4 f^7 5 d^1 6 s^2\)
  3. \([\mathrm{Xe}] 4 f^6 5 d^2 6 s^2\)
  4. \([\mathrm{Xe}] 4 f^8 6 d^2\)

Answer: 2. \([\mathrm{Xe}] 4 f^7 5 d^1 6 s^2\)

Question 57. Because of lanthanoid contraction, which of the following pairs of elements have nearly the same atomic radii? (Numbers in the parenthesis are atomic numbers)

  1. Zr(40) and Hf(72)
  2. Zr(40) and Ta(73)
  3. Ti(22) and Zr(40)
  4. Zr(40) and Nb(41)

Answer: 1. Zr(40) and Hf(72)

Zr and Hf have nearly the same radii due to lanthanoid contraction.

Question 58. The reason for lanthanoid contraction is

  1. Negligible screening effect of ‘f’ -orbitals
  2. Increasing nuclear charge
  3. Decreasing nuclear charge
  4. Decreasing screening effect.

Answer: 1. Negligible screening effect of ‘f’-orbitals

Due to the poor shielding effect of 4f orbitals nucleus will exert a strong attraction and the size of the atom or ion will decrease as moves in the series with an increase in atomic number.

Question 59. Which of the following lanthanoid ions is diamagnetic?

(Atomic numbers Ce = 58, Sm = 62, Eu = 63, Yb = 70)

  1. \(\mathrm{Eu}^{2+}\)
  2. \(\mathrm{Yb}^{2+}\)
  3. \(\mathrm{Ce}^{2+}\)
  4. \(\mathrm{Sm}^{2+}\)

Answer: 2. \(\mathrm{Yb}^{2+}\)

  1. \(\mathrm{Sm}^{2+}(Z=62):[\mathrm{Xe}] 4 f^6\)
  2. \(\mathrm{Eu}^{2+}(Z=63):[\mathrm{Xe}] 4 f^7\)
  3. \(\mathrm{Yb}^{2+}(Z=70):[\mathrm{Xe}] 4 f^{14}\)
  4. \(\mathrm{Ce}^{2+}(Z=58):[\mathrm{Xe}] 4 f^2\)

Only Yb2+ is diamagnetic

Question 60. Which of the following oxidation states is the most common among the lanthanoids?

  1. 4
  2. 2
  3. 5
  4. 3

Answer: 4. 3

The common stable oxidation state of all the lanthanoids is +3. The oxidation states of +2 and +4 are also exhibited by some of the elements. These oxidation states are only stable in those cases where stable 4f0, 4f7 or 4f14 configurations are achieved.

Question 61. Identify the incorrect statement among the following:

  1. Lanthanoid contraction is the accumulation of successive shrinkages.
  2. As a result of lanthanoid contraction, the properties of the 4d series of transition elements have no similarities with the 5d series of elements.
  3. The shielding power of 4f electrons is quite weak.
  4. There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.

Answer: 2. As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements.

In each vertical column of transition elements, the elements of the second and third transition series resemble each other more closely than the elements of the first and second transition series on account of lanthanide contraction. Hence the properties of elements of 4d series of transition elements resemble the properties of the elements of 5d series of transition elements.

Question 62. Lanthanoids are

  1. 14 elements in the sixth period (atomic number 90 to 103) that are filling 4f sublevel
  2. 14 elements in the seventh period (atomic number = 90 to 103) that are filling 5f sublevel
  3. 14 elements in the sixth period (atomic number = 58 to 71) that are filling the 4f sublevel
  4. 14 elements in the seventh period (atomic number = 58 to 71) that are filling 4f sublevel.

Answer: 3. 14 elements in the sixth period (atomic number = 58 to 71) that are filling the 4f sublevel

As the sixth period can accommodate only 18 elements in the table, 14 members of 4f series (atomic numbers 58 to 71) are separately accommodated in a horizontal row below the periodic table. These are called lanthanides.

Question 63. The correct order of ionic radii of Y3+, La3+, Eu3+ and Lu3+is (Atomic number Y = 39, La = 57, Eu = 63, Lu = 71)

  1. \(\mathrm{Y}^{3+}<\mathrm{La}^{3+}<\mathrm{Eu}^{3+}<\mathrm{Lu}^{3+}\)
  2. \(\mathrm{Y}^{3+}<\mathrm{Lu}^{3+}<\mathrm{Eu}^{3+}<\mathrm{La}^{3+}\)
  3. \(\mathrm{Lu}^{3+}<\mathrm{Eu}^{3+}<\mathrm{La}^{3+}<\mathrm{Y}^{3+}\)
  4. \(\mathrm{La}^{3+}<\mathrm{Eu}^{3+}<\mathrm{Lu}^{3+}<\mathrm{Y}^{3+}\)

Answer: 2. \(\mathrm{Y}^{3+}<\mathrm{Lu}^{3+}<\mathrm{Eu}^{3+}<\mathrm{La}^{3+}\)

Ongoing from La3+ to Lu3+, the ionic radius shrinks from 1.15 Å to 0.93 Å (lanthanide contraction) The radius of La3+ is also larger than that of Y3+ ion which lies immediately above it in the periodic table.

Question 64. The general electronic configuration of lanthanides is

  1. \((n-2) f^{1-14}(n-1) s^2 p^6 d^{0-1} n s^2\)
  2. \((n-2) f^{10-14}(n-1) d^{0-1} n s^2\)
  3. \((n-2) f^{0-14}(n-1) d^{10} n s^2\)
  4. \((n-2) d^{0-1}(n-1) f^{-14} n s^2\)

Answer: 1. \((n-2) f^{1-14}(n-1) s^2 p^6 d^{0-1} n s^2\)

The general electronic structure of lanthanides is: \((n-2) f^{1-14}(n-1) s^2 p^6 d^{0-1} n s^2 \text {. }\)

Question 65. Which of the following statements is not correct?

  1. La(OH)3 is less basic than Lu(OH)3.
  2. In the lanthanide series, the ionic radius of Ln3+ ion decreases.
  3. La is actually an element of transition series rather than lanthanides.
  4. Atomic radii of Zr and Hf are the same because of lanthanide contraction.

Answer: 1. La(OH)3 is less basic than Lu(OH)3

La(OH)3 is more basic than Lu(OH)3 In lanthanides the basic character of hydroxides decreases as the ionic radius decreases.

Question 66. The lanthanide contraction is responsible for the fact that

  1. Zr and Hf have about the same radius
  2. Zr and Zn have the same oxidation state
  3. Zr and Y have about the same radius
  4. Zr and Nb have similar oxidation states.

Answer: 1. Zr and Hf have about the same radius

Due to lanthanide contraction, the elements of the second and third transition series i.e., Zr and Hf resemble more with each other tiran the elements of the first and second transition series.

Question 67. Which of the following statements concerning lanthanide elements is false?

  1. All lanthanides are highly dense metals.
  2. The more characteristic oxidation state of lanthanide elements is +3.
  3. Lanthanides are separated from one another by the ion exchange method.
  4. Ionic radii of trivalent lanthanides steadily increase with the increase in the atomic number

Answer: 4. Ionic radii of trivalent lanthanides steadily increase with increase in the atomic number

Ionic radii of trivalent lanthanides decrease with an increase in atomic number.

Question 68. The incorrect statement among the following is

  1. Actinoids are highly reactive metals, especially when finely divided
  2. Actinoid contraction is greater for element to element-than lanthanoid contraction
  3. Most of the trivalent lanthanoid ions are colorless in the solid state
  4. Lanthanoids are good conductors of heat and electricity.

Answer: 3. Most of the trivalent lanthanoid ions are colourless in the solid state

Question 69. The reason for the greater range of oxidation states in actinoids is attributed to

  1. Actinoid contraction
  2. 5F 6d and 7s levels having comparable energies
  3. 4F and 5d levels are close in energies
  4. The radioactive nature of actinoids.

Answer: 2. 5F 6d and 7s levels having comparable energies

Actinoids have a greater range of oxidation states due to comparable energies if 5f, 6d and 7s orbitals. Hence all their electrons can take part in bond formation.

Question 70. Which of the following exhibits only +3 oxidation state?

  1. U
  2. Th
  3. Ac
  4. Pa

Answer: 3. Ac

U exhibits + 3,+ 4,+ 5, +6

Th exhibits + 3, + 4 ; Ac exhibits + 3 only

Pa exhibits + 3, + 4, + 5

Question 71. More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is

  1. The more active nature of the actinoids
  2. More energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
  3. Lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
  4. Greater metallic character of the lanthanoids than that of the corresponding actinoids.

Answer: 3. More active nature of the actinoids

The 5f-orbitals extend into space beyond the 6s and 6p-orbitals and participate in bonding. This is in direct contrast to the lanthanides where the 4f orbitals are buried deep inside the atom, totally shielded by outer orbitals and thus unable to take part in bonding.

Question 72. Which one of the following elements shows a maximum number of different oxidation states in its compounds?

  1. Gd
  2. La
  3. Eu
  4. Am

Answer: 4. Am

La forms compounds in which the oxidation number is +3.

‘Eu’ and ‘Gd’ exhibit +2 as well as +3 oxidation states and are not higher than that, due to stable (f) configuration. Whereas All exhibit the oxidation states +3, +4, +5, +6′ etc’ due to extremely large size and low ionisation energy.

Question 73. Match the catalyst with the process

d And f Block Elements Match The Catalyst And Process

Which of the following is the correct option?

  1. 1-C, 2-D, 3-A, 4-B
  2. 1-A, 2-B, 3-C, 4-D
  3. 1-A, 2-C, 3-B, 4-D
  4. 1-C, 2-A, 3-D, 4-B

Answer: 1. 1-C, 2-D, 3-A, 4-B

Question 74. HgCl2 and I2 both when dissolved in water containing L ions, the pair of species formed is

  1. \(\mathrm{HgI}_2, \mathrm{I}^{-}\)
  2. \(\mathrm{HgI}_4^{2-}, \mathrm{I}_3^{-}\)
  3. \(\mathrm{Hg}_2 \mathrm{I}_2, \mathrm{I}^{-}\)
  4. \(\mathrm{HgI}_2, \mathrm{I}_3^{-}\)

Answer: 2. \(\mathrm{HgI}_4^{2-}, \mathrm{I}_3^{-}\)

⇒ \(\mathrm{HgCl}_{2(\mathrm{mq})}+4 \mathrm{I}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{HgI}_4^{2-}(\mathrm{aq})+2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)

⇒ \(\mathrm{I}_{2(\mathrm{~s})}+\mathrm{I}_{(\mathrm{aq})} \longrightarrow{\mathrm{imq}} \mathrm{I}_{3(\mathrm{aq})}^{-}\)

Question 75. Which of the following elements is responsible for the oxidation of water to O2 in biological processes?

  1. Cu
  2. Mo
  3. Fe
  4. Mm

Answer: 3. Fe

Question 76. When calomel reacts with NH4OH, we get

  1. \(\mathrm{Hg}_2 \mathrm{O}\)
  2. \(\mathrm{HgO}\)
  3. \(\mathrm{HgNH}_2 \mathrm{Cl}\)
  4. \(\mathrm{NH}_2-\mathrm{Hg}-\mathrm{Hg}-\mathrm{Cl}\)

Answer: 3. \(\mathrm{HgNH}_2 \mathrm{Cl}\)

When calomel reacts with NH4OH it turns black due to the formation of a mixture of mercury and ammonium basic mercury (2) chloride.

⇒ \(\underset{\text{ Calmole }}{\mathrm{Hg}_2 \mathrm{Cl}_2}+2 \mathrm{NH}_4 \mathrm{OH} \rightarrow \mathrm{NH}_4 \mathrm{Cl}+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Hg}+\mathrm{HgNH}_2 \mathrm{Cl}\)

Question 77. Photographic films and plates have essential ingredients of

  1. Silver nitrate
  2. Silver bromide
  3. Sodium chloride
  4. Oleic acid.

Asnwer: 2. Silver bromide

AgBr is highly photosensitive and is used as an ingredient for photographic films and plates.

MCQs on s-block Elements for NEET

NEET Chemistry For The s Block Elements Multiple Choice Questions

Question 1. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.

Reason R: The deep blue solution is due to the formation of amide.

In the light of the above statements, choose the correct answer from the options given below:

  1. A is true but R is false.
  2. A is false but R is true.
  3. Both A and R are true and R is the correct explanation of A.
  4. Both A and R are true but R is not the correct explanation of A.

Answer: 1. A is true but R is false.

Sodium metal dissolves in liquid ammonia giving a deep blue solution which is conducting in nature. \(\mathrm{Na}+(x+y) \mathrm{NH}_3 \rightarrow\left[\mathrm{Na}\left(\mathrm{NH}_3\right)_x\right]^{+}+\left[e\left(\mathrm{NH}_3\right)_y\right]^{-}\)

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus, imparts blue colour to the solution.

The solution is paramagnetic and on standing, slowly liberates hydrogen gas resulting in the formation of amide. \(\mathrm{Na}_{(a m)}^{+}+e^{-}+\mathrm{NH}_{3(0)} \rightarrow \mathrm{NaNH}_{2(a m)}+1 / 2 \mathrm{H}_{2(\mathrm{~g})}\)

(where ‘am’ denotes solution in ammonia.)

Question 2. Identify the incorrect statement from the following.

  1. Alkali metals react with water to form their hydroxides.
  2. The oxidation number of K in KO2 is +4.
  3. The ionisation enthalpy of alkali metals decreases from top to bottom in the group.
  4. Lithium is the strongest reducing agent among the alkali metals.

Answer: 2. The oxidation number of K in KO2 is +4.

The superoxide species (KO2) is represented as O2, since the compound is neutral, therefore the oxidation state of potassium is +1. Remaining all the given statements are correct

Question 3. Match List-1 with List-2.

The s Block Elements Match The Lists

Choose the correct answer from the options given below:

  1. (1) -(D), (2) – (A), (3) – (C), (4) – (B)
  2. (1) -(C), (2) – (D), (3) – (B), (4) – (A)
  3. (1) -(A), (2) – (C), (3) – (D), (4) – (B)
  4. (1) -(B), (2) – (C), (3) – (A), (4) – (D)

Answer: 4. (1) -(B), (2) – (C), (3) – (A), (4) – (D)

The s Block Elements

Question 4. Ionic mobility of which of the following alkali metal ions is lowest when the aqueous solution of their salts is put under an electric field?

  1. K
  2. Rb
  3. Li
  4. Na

Asnwer: 3. Li

The hydration enthalpy of alkali metal ions decreases with an increase in ionic sizes i.e., Li+>Na+>K>Rb+>Cs+

Hence, lithium having a maximum degree of hydration will be the least mobile.

The order of ionic mobility is \(\left.\left[\mathrm{Li}_{(a q)}\right]^{+}<\left[\mathrm{Na}_{(a q)}\right)\right]^{+}<\left[\mathrm{K}_{(a q)}\right]^{+}<\left[\mathrm{Rb}_{(a q)}\right]^{+}\)

Read and Learn More NEET MCQs with Answers

Question 5. Which one of the alkali metals, forms only, the normal oxide, 2O on heating in the air?

  1. Rb
  2. K
  3. Li
  4. Na

Answer: 3. Li

When alkali metals are heated in an atmosphere of oxygen, the alkali metals ignite and form oxides. On combustion Li forms Li2O; sodium gives the peroxide Na2O2 and potassium and rubidium give superoxide (MO2).

Question 6. The ease of adsorption of the hydrated alkali metal ions on an ion-exchange resins follows the order

  1. Li+ < K+ < Na+ < Rb+
  2. Rb+ < K+ < Na+ < Li+
  3. K+ < Na+ < Rb+ < Li+
  4. Na+ < Li+ < K+ < Rb+

Answer: 2. Rb+ < K+ < Na+ < Li+

The order of decreasing hydration enthalpy of alkali metal ions is: Li+ > Na+ > K+ > Rb+

Thus, the ease of adsorption of hydrated ions is in the order: Rb+< K+< Na+< Li+

Question 7. The sequence of ionic mobility in aqueous solution is

  1. Rb+ > K+ > Cs+ > Na+
  2. Na+ > K+ > Rb+ > Cs+
  3. K+ > Na+ > Rb+ > Cs+
  4. Cs+ > Rb+ > K+ > Na+

Answer: 4. Cs+ > Rb+ > K+ > Na+

The smaller the size of the cation, the higher the hydration and its reflective size will increase hence mobility in an aqueous solution will decrease. Hence, the correct sequence of ionic mobility in an aqueous solution of the given cations is Cs+>Rb+>K+>Na+.

Question 8. When a substance (A) reacts with water it produces a combustible gas (B) and a solution of substance (C) in water. When another substance (D) reacts with this solution of (C), it also produces the same gas (B) on warming but (D) can produce gas (B) on reaction with dilute sulphuric acid at room temperature. Substance (A) imparts a deep golden-yellow colour to the smokeless flame of the Bunsen burner. Then (A), (B), (C) and (D) respectively are

  1. \(\mathrm{Ca}, \mathrm{H}_2, \mathrm{Ca}(\mathrm{OH})_2, \mathrm{Sn}\)
  2. \(\mathrm{K}, \mathrm{H}_2, \mathrm{KOH}, \mathrm{Al}\)
  3. \(\mathrm{Na}, \mathrm{H}_2, \mathrm{NaOH}, \mathrm{Zn}\)
  4. \(\mathrm{CaC}_2, \mathrm{C}_2 \mathrm{H}_2, \mathrm{Ca}(\mathrm{OH})_2, \mathrm{Fe}\)

Answer: 3. \(\mathrm{Na}, \mathrm{H}_2, \mathrm{NaOH}, \mathrm{Zn}\)

When a substance (A) reacts with water it produces a combustible gas (B) and a solution of substance (C) in water. When another substance (D) reacts with this solution of (C), it also produces the same gas (B) on warming but (D) can produce gas (B) on reaction with dilute sulphuric acid at room temperature. Substance (A) imparts a deep golden-yellow colour to the smokeless flame of the Bunsen burner.

Only ‘Na’ imparts a golden colour to Bunsen flame, therefore, A = Na, B = H2, C = NaOH, D = Zn.

⇒ \(\underset{\text{A}}{2 \mathrm{Na}}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text{A}}{2 \mathrm{NaOH}}+\underset{\text{B}}{\mathrm{H}_2}\)

⇒ \(\underset{\text{D}}{\mathrm{Zn}}+\underset{\text{C}}{2 \mathrm{NaOH}} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\underset{\text{B}}{\mathrm{H}_2}\)

⇒ \(\underset{\text{D}}{\mathrm{Zn}}+\mathrm{H}_2 \mathrm{SO}_4 \text { (dil.) } \rightarrow \mathrm{ZnSO}_4+\underset{\text{B}}{\mathrm{H}_2}\)

Question 9. Which one of the following properties of alkali metals increases in magnitude as the atomic number rises?

  1. Ionic radius
  2. Melting point
  3. Electronegativity
  4. First ionization energy

Answer: 1. Ionic radius

In a group, the ionic radius increases with an increase in atomic number whereas the m.pt. decreases down in a group due to the weakening of metallic bonds. Similarly, electronegativity and ionization energy also decrease down the group.

Question 10. In the case of alkali metals, the covalent character decreases in the order

  1. MF > MCI > MBr > MI
  2. MF > MCI > Ml > MBr
  3. MI > MBr > MCI > MF
  4. MCI > MI > MBr > MF

Answer: 3. MI > MBr > MCI > MF

Alkali metals are highly electropositive and halogens are electronegative. Thus, for the halides of a given alkali metal, the covalent character decreases with an increase in the electronegativity of halogens.

∴ The order of the covalent character of halides is

MI> MBr > MCI> MF

Question 11. The alkali metals form salt-like hydrides by direct synthesis at elevated temperatures. The thermal stability of these hydrides decreases in which of the following orders?

  1. NaH > LiH > KH > RbH > CsH
  2. LiH > NaH > KH > RbH > CsH
  3. CsH > RbH > KH > NaH > LiH
  4. KH > NaH > LiH > CsH > RbH

Answer: 2. LiH > NaH > KH > RbH > CsH

The ionic character of the bonds in hydrides increases from LiH to CsH due to the weakening of the M-H bond so, the thermal stability of these hydrides decreases in the order of LiH > NaH > KH > RbH > CsH.

Question 12. Which compound will show the highest lattice energy?

  1. RbF
  2. CsF
  3. NaF
  4. KF

Answer: 3. NaF

With the same anion, the smaller the size of the cation, the higher is the lattice energy. Therefore, NaF will show the highest lattice energy among the given compounds.

Question 13. Which of the alkali metal chloride (MCI) forms its dihydrate salt (MCl.2H2O) easily?

  1. LiCl
  2. CsCl
  3. RbCl
  4. KCl

Answer: 1. LiCl

LiCl is deliquescent and crystallises from aqueous solution as hydrates, LiCl.2H2O.

Question 14. Crude sodium chloride obtained by crystallisation of brine solution does not contain

  1. \(\mathrm{MgSO}_4\)
  2. \(\mathrm{Na}_2 \mathrm{SO}_4\)
  3. \(\mathrm{MgCl}_2\)
  4. \(\mathrm{CaSO}_4\)

Answer: 1. \(\mathrm{MgSO}_4\)

Crude sodium chloride, generally obtained by the crystallisation of brine solution contains sodium sulphate (Na2SO4), calcium sulphate (CaSO4), calcium chloride (CaCl2) and magnesium chloride (MgCl2) as impurities. Crude sodium chloride does not contain MgSO4

Question 15. In the Castner-Kellner cell for the production of sodium hydroxide

  1. Brine is electrolyzed using graphite electrodes
  2. Molten sodium chloride is electrolysed
  3. Sodium amalgam is formed at a mercury cathode
  4. Brine is electrolyzed with pt electrodes.

Answer: 3. Sodium amalgam is formed at a mercury cathode

In the Castner-Kellner cell, sodium amalgam is formed at the mercury cathode.

A brine solution is electrolysed using a mercury cathode and a carbon anode.

Question 16. Which of the following statements is incorrect?

  1. Pure sodium metal dissolves in liquid ammonia to give a blue solution.
  2. NaOH reacts with glass to give sodium silicate.
  3. Aluminium reacts with excess NaOH to give Al(OH)3.
  4. NaHCO3 on heating gives Na2CO3.

Answer: 3. Aluminium reacts with excess NaOH to give Al(OH)3.

Al reacts with NaOH to give sodium aluminate

Question 17. In which of the following processes, fused sodium hydroxide is electrolysed at a 330 °C temperature for extraction of sodium?

  1. Castner’s process
  2. Down’s process
  3. Cyanide process
  4. Both (2) and (3).

Answer: 1. Castner’s process

In Castner’s process, for the production of sodium metal, sodium hydroxide (NaOH) is electroplated at a temperature of 330°C.

Question 18. Which of the following is known as a fusion mixture?

  1. Mixture of Na2CO3 + NaHCO3
  2. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}\)
  3. Mixture of \(\mathrm{K}_2 \mathrm{CO}_3+\mathrm{Na}_2 \mathrm{CO}_3\)
  4. \(\mathrm{NaHCO}_3\)

Answer: 3. Mixture of \(\mathrm{K}_2 \mathrm{CO}_3+\mathrm{Na}_2 \mathrm{CO}_3\)

K2CO3 and Na2CO3 mixture is called a fusion mixture.

Question 19. Washing soda has a formula

  1. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 7 \mathrm{H}_2 \mathrm{O}\)
  2. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}\)
  3. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 3 \mathrm{H}_2 \mathrm{O}\)
  4. \(\mathrm{Na}_2 \mathrm{CO}_3\)

Answer: 2. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}\)

Na2CO3.10H2O is washing soda.

Question 20. The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.

  1. Iron
  2. Copper
  3. Calcium
  4. Potassium

Answer: 4. Potassium

Potassium ions are then the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP and, with solider, are responsible for the transmission of nerve signals.

Question 21. The function of the sodium pump is a biological process operating in each and every cell of all animals. Which of the following biologically important ions is also a constituent of this pump?

  1. K+
  2. Fe2+
  3. Ca2+
  4. Mg2+

Answer: 1. K+

Question 22. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s² 2s² 2p³, the simplest formula for this compound is

  1. \(\mathrm{Mg}_2 X_3\)
  2. \(\mathrm{Mg} X_2\)
  3. \(\mathrm{Mg}_2 X\)
  4. \(\mathrm{Mg}_3 X_2\)

Answer: 4. \(\mathrm{Mg}_3 X_2\)

Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s² 2s² 2p³

The electronic configuration of X is 1s², 2s² 2p³.

So, the valency of X will be 3.

The s Block Elements Electronic Configuration Of X

Magnesium ion = Mg2+

Formula: Mg3X2

Question 23. The electronic configuration of calcium atoms may be written as

  1. \([\mathrm{Ne}] 4 p^2\)
  2. \([\mathrm{Ar}] 4 s^2\)
  3. \([\mathrm{Ne}] 4 s^2\)
  4. \([\mathrm{Ar}] 4 p^2\)

Answer: 2. \([\mathrm{Ar}] 4 s^2\)

⇒ \({ }_{20} \mathrm{Ca} \longrightarrow 1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2\)

⇒ \({ }_{18} \mathrm{Ar} \longrightarrow 1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6\)

Hence, \({ }_{20} \mathrm{Ca} \longrightarrow[\mathrm{Ar}] 4 s^2\)

Question 24. Compared with the alkaline earth metals, the alkali metals exhibit

  1. Smaller ionic radii
  2. Highest boiling points
  3. Greater hardness
  4. Lower ionization energies.

Answer: 4. Lower ionization energies.

The alkali metals are larger in size and have smaller nuclear charge thus they have lower ionization energy in comparison to alkaline earth metals.

Question 25. Which of the following atoms will have the smallest size?

  1. Mg
  2. Na
  3. Be
  4. Li

Answer: 3. Be

The atomic size decreases within a period from left to right, therefore Li > Be and Na > Mg. The size increases in a group from top to bottom. Hence, the size of Na is greater than Li. Overall order Na > Mg > Li > Be. Thus, Be has the smallest size.

Question 26. The structures of beryllium chloride in solid state and vapour phase are

  1. Chain in both
  2. Chain and dimer, respectively
  3. Linear in both
  4. Dimer and linear, respectively.

Answer: 2. Chain and dimer, respectively

In the vapour phase, BeCl2 is found in dimer form.

The s Block Elements Dimer Form

While in a solid state, it is found as a polymer (chain structure)

The s Block Elements Solid State Is Found A Polymer

Question 27. Among the following alkaline earth metal halides one which is covalent and soluble in organic solvents is

  1. Beryllium chloride
  2. Calcium chloride
  3. Strontium chloride
  4. Magnesium chloride

Answer: 1. Beryllium chloride

Except for beryllium halides, all other halides of alkaline earth metals are ionic in nature. Beryllium halides are essentially covalent and soluble in organic solvents.

Question 28. HCl was passed through a solution of CaCl2, MgCl2 and NaCl. Which of the following compound(s) crystallise(s)?

  1. Both MgCl2 and CaCl2
  2. Only NaCl
  3. Only MgCl2
  4. NaCl, MgCl2 and CaCl2

Answer: 2. Only NaCl

CaCl2 and MgCI2 are more soluble than NaCl. Thus, when HCl was passed through a solution containing CaCl2, MgCI2, and NaCl, only NaCl crystallised.

Question 29. Which of the following is an amphoteric hydroxide?

  1. Be(OH)2
  2. Sr(OH)2
  3. Ca(OH)2
  4. Mg(OH)2

Answer: 1. Be(OH)2

Be(OH)2 is amphoteric in nature as it reacts with acid and alkali.

⇒ \(\mathrm{Be}(\mathrm{OH})_2+2 \mathrm{OH}^{-} \rightarrow\left[\mathrm{Be}(\mathrm{OH})_4\right]^{2-}\)

⇒ \(\mathrm{Be}(\mathrm{OH})_2+2 \mathrm{HCl}+2 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{Be}(\mathrm{OH})_4\right] \mathrm{Cl}_2\)

Question 30. Among CaH2, BeH2, BaH2, the order of ionic character is

  1. BeH2 < CaH2 < BaH2
  2. CaH2 < BeH2 < BaH2
  3. BeH2 < BaH2 < CaH2
  4. BaH2 < B2 < CaH2

Answer: 1. BeH2 < CaH2 < BaH2

BeH2 < CaH2< BaH2

On moving down the group, the metallic character of metals increases. So, the ionic character of metal hydrides increases.

Hence, BeH2 will be the least ionic

Question 31. On heating which of the following releases CO2 most easily?

  1. Na2CO3
  2. MgCO3
  3. CaCO3
  4. K2CO3

Answer: 2. MgCO3

The stability of carbonates increases down the group with an increase in the size of the metal ions. Also, the alkali metal carbonates are more stable than alkaline earth metal carbonates. Hence, MgCO3 is the least stable and it releases CO2 most easily.

⇒ \(\mathrm{MgCO}_3\)\(\underrightarrow{\Delta}\)\(\mathrm{MgO}+\mathrm{CO}_2\)

Question 32. Solubility of the alkaline earth metal sulphates in water decreases in the sequence

  1. Sr > Ca > Mg > Ba
  2. Ba > Mg > Sr > Ca
  3. Mg > Ca > Sr > Ba
  4. Ca > Sr > Ba > Mg

Answer: 3. Mg > Ca > Sr > Ba

The solubility of alkaline earth metal sulphates decreases down the group because hydration energy decreases.

Question 33. Which of the following compounds has the lowest melting point?

  1. CaCl2
  2. CaBr2
  3. Cal2
  4. CaF2

Answer: 3. Cal2

As the covalent character in the compound increases and the ionic character decreases, the melting point of the compound decreases. So, CaI2 has the highest covalent character and lowest melting point.

Question 34. Which of the following alkaline earth metal sulphates has a hydration enthalpy higher than the lattice enthalpy?

  1. CaSO4
  2. BeSO4
  3. BaSO4
  4. SrSO4

Answer: 2. BeSO4

The hydration enthalpy of BeSO4 is higher than its lattice energy. Within group 2, the hydration energy decreases down the group while lattice energy is almost the same.

Question 35. Which one of the following compounds is a peroxide?

  1. KO2
  2. BaO2
  3. MnO2
  4. NO2

Answer: 2. BaO2

BaO2 has peroxide linkage

Question 36. Property of the alkaline earth metals that increases with their atomic number

  1. Solubility of their hydroxides in water
  2. Solubility of their sulphates in water
  3. Ionization energy
  4. Electronegativity

Answer: 1. Solubility of their hydroxides in water

The solubility of an ionic compound depends on two factors:

  1. Lattice energy, and
  2. Hydration energy

In the case of alkaline earth metal hydroxides, the lattice energy decreases as we move down the group. This decrease is more than the decrease in the hydration energy down the group.

Question 37. Which of the following oxides is not expected to react with sodium hydroxide?

  1. CaO
  2. SiO2
  3. BeO
  4. B2O3

Answer: 1. CaO

CaO being a basic oxide does not react with NaOH, however, SiO2 (acidic oxide), BeO (amphoteric oxide) and B2O3 (acidic oxide) react with NaOH

Question 38. The correct order of increasing thermal stability of K2CO3, MgCO3, CaCO3 and BeCO3 is

  1. BeCO3 < MgCO3 < CaCO3 < K2CO3
  2. MgCO3 < BeCO3 < CaCO3 < K2CO3
  3. K2CO3 < MgCO3 < CaCO3 < BeCO3
  4. BeCO3 < MgCO3 < K2CO3 < CaCO3

Answer: 1. BeCO3 < MgCO3 < CaCO3 < K2CO3

In all cases, for a particular set of group 1 or group 2 compounds, the thermal stability increases down the group as the ionic radius of the cation increases, and its polarising power decreases.

Group 1 compounds tend to be more thermally stable than group 2 compounds because group 1 cation has a smaller charge and a larger ionic radius, and so, a lower polarising power, particularly when adjacent metals on the same period are compared.

Hence, the order of increasing thermal stability of K2CO3, MgCO3, CaCO3 and BeCO3 is BeCO3 < MgCO3 < CaCO3 < K2CO3

Question 39. In which of the following the hydration energy is higher than the lattice energy?

  1. MgSO4
  2. RaSO4
  3. SrSO4
  4. BaSO4

Answer: 1. MgSO4

When hydration energy exceeds lattice energy, the compound becomes soluble in water. The solubility of alkaline earth metal sulphates decreases in the order:

The s Block Elements Hydration Energy

The solubilities of BeSO4 and MgSO4 are due to the high energy of solvation of smaller Be2+ and Mg2+ ions.

Question 40. The solubility in water of sulphate down the Be group is Be > Mg > Ca > Sr > Ba. This is due to

  1. Decreasing lattice energy
  2. The high heat of solvation for smaller ions like Be2+
  3. Increase in melting points
  4. Increasing molecular weight.

Answer: 2. High heat of solvation for smaller ions like Be2+

As we move down the group from BeSO4 to BaSO4 the enthalpy of hydration of the positive ion becomes smaller due to increasing in ionic size. Salts of heavier metal ions are 1ess soluble than those of lighter ions.

Question 41. All the following substances react with water. The pair that gives the same gaseous product is

  1. K and KO2
  2. Na and Na2O2
  3. Ca and CaH2
  4. Ba and BaO2

Answer: 3. Ca and CaH2

The pair which gives the same gaseous product is Ca and CaH2.

⇒ \(\mathrm{Ca}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2\)

⇒ \(\mathrm{CaH}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{H}_2\)

Whereas, \(\mathrm{K}\) gives \(\mathrm{H}_2\) while \(\mathrm{KO}_2\) gives \(\mathrm{O}_2\) and \(\mathrm{H}_2 \mathrm{O}_2\).

⇒ \(2 \mathrm{~K}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_2\)

⇒ \(2 \mathrm{KO}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{O}_2+\mathrm{H}_2 \mathrm{O}_2\)

Similarly, \(\mathrm{Na}\) gives \(\mathrm{H}_2\), while \(\mathrm{Na}_2 \mathrm{O}_2\) gives \(\mathrm{H}_2 \mathrm{O}_2\).

⇒ \(2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2\)

⇒ \(\mathrm{Na}_2 \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O}_2\)

Likewise Ba gives \(\mathrm{H}_2\) while \(\mathrm{BaO}_2\) gives \(\mathrm{H}_2 \mathrm{O}_2\).

⇒ \(\mathrm{Ba}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ba}(\mathrm{OH})_2+\mathrm{H}_2\)

⇒ \(\mathrm{BaO}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ba}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{O}_2\)

Question 42. Which of the following statements is false?

  1. Strontium decomposes water more readily than beryllium.
  2. Barium carbonate melts at a higher temperature than calcium carbonate.
  3. Barium hydroxide is more soluble in water than magnesium hydroxide.
  4. Beryllium hydroxide is more basic than barium hydroxide.

Answer: 4. Beryllium hydroxide is more basic than barium hydroxide.

Beryllium hydroxide although amphoteric, is however less basic than barium hydroxide.

Question 43. In context with beryllium, which one of the following statements is incorrect?

  1. It is rendered passive by nitric acid.
  2. It forms Be2C.
  3. Its salts rarely hydrolyse.
  4. Its hydride is electron-deficient and polymeric.

Answer: 3. Its salts rarely hydrolyse.

Due to the very small size of Be2+, beryllium salts are readily hydrolysed because of high hydration energy. \(\mathrm{BeCl}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Be}(\mathrm{OH})_2+2 \mathrm{HCl}\)

Question 44. The suspension of slaked lime in water is known as

  1. Lime water
  2. Quicklime
  3. Milk of lime
  4. Aqueous solution of slaked lime.

Answer: 3. Milk of lime

⇒ \(\underset{\text { Quick lime }}{\mathrm{CaO}}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Slaked lime }}{\mathrm{Ca}(\mathrm{OH})_2+\text { Heat }}\)

This process is known as slaking of lime. The paste of lime in water (i.e., suspension) is called milk of lime while the filtered and clear solution is known as lime water

Question 45. The product obtained as a result of a reaction of nitrogen with CaC2 is

  1. CaCN3
  2. Ca2CN
  3. Ca(CN)2
  4. CaCN

Answer: 3. Ca(CN)2

Read Ca(CN)2 as CaCN2

⇒ \(\mathrm{CaC}_2+\mathrm{N}_2 \longrightarrow \underset{\text{Slaked lime}}{\mathrm{CaCN}_2}+\mathrm{C}\)

Question 46. Which one of the following is present as an active ingredient in bleaching powder for bleaching action?

  1. CaOCl2
  2. Ca(OCl)2
  3. CaO2Cl
  4. CaCl2

Answer: 2. Ca(OCl)2

The active ingredient in bleaching powder for bleaching action is Ca(OCl)2.

Question 47. Match List-1 with List-2 for the compositions of substances and select the correct answer using the code given:

The s Block Elements List The Substances And Composition

  1. (1)-(C), (2)-(D), (3)-(A), (4)-(B)
  2. (1)-(B), (2)-(C), (3)-(D), (4)-(A)
  3. (1)-(A), (2)-(B), (3)-(C), (4)-(E)
  4. (1)-(D), (2)-(C), (3)-(B), (4)-(A)

Answer: 2. (1)-(B), (2)-(C), (3)-(D), (4)-(A)

The s Block Elements List The Substances And Composition

Question 48. The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of CO2 is bubbled through an aqueous solution of B, and C is formed which is recovered in the solid form. Solid C on gentle heating gives back A. The compound is

  1. CaCO3
  2. Na2CO3
  3. K2CO3
  4. CaSO4•2H2O

Answer: 1. CaCO3

The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of CO2 is bubbled through an aqueous solution of B, and C is formed which is recovered in the solid form. Solid C on gentle heating gives back A.

The reactions can be summarised as follows:

The s Block Elements Compound A On Heating Gives A Colorless Gas And Residue

Question 49. Which of the following represents calcium chlorite?

  1. Ca(ClO3)2
  2. Ca(ClO2)2
  3. CaClO2
  4. Ca(ClO4)2

Answer: 2. Ca(ClO2)2

Since the valency of calcium is 2 and a chlorite ion is CIO2, therefore calcium chlorite is Ca(ClO2)2.

Question 50. Identify the correct statement.

  1. Plaster of Paris can be obtained by hydration of gypsum.
  2. Plaster of Paris is obtained by partial oxidation of gypsum.
  3. Gypsum contains a lower percentage of calcium than Plaster of Paris.
  4. Gypsum is obtained by heating Plaster of Paris.

Answer: 3. Gypsum contains a lower percentage of calcium than Plaster of Paris.

Gypsum is CaSO4.2H2O and plaster of Paris is (CaSO4)2.H2O. Therefore, gypsum contains a lower percentage of calcium than Plaster of Paris.

Question 51. Bleaching powder is obtained by the action of chlorine gas and

  1. Dilute solution of Ca(OH)2
  2. Concentrated solution of Ca(OH)2
  3. Dry CaO
  4. Dry-slaked lime.

Answer: 4. Dry slaked lime.

Cl2 gas reacts with dry slaked lime, Ca(OH)2 to give bleaching powder.

⇒ \(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{Cl}_2\) \(\underrightarrow{\Delta}\) \(\mathrm{CaOCl}_2+\mathrm{H}_2 \mathrm{O}\)

Question 52. Which one of the following statements is correct?

  1. The bone in the human body is an inert and unchanging substance.
  2. Mg plays roles in neuromuscular function and interneuronal transmission.
  3. The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g.
  4. All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor.

Answer: 3. The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g.

Bones are not the inert and unchanging substances. Calcium plays an important role in neuromuscular function and interneuronal transmission.

All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor.

Question 53. Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is

  1. Sr
  2. Be
  3. Mg
  4. Ca

Answer: 3. Mg

All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor

Question 54. Which of the following statements is false?

  1. Ca2+ ions are not important in maintaining the regular beating of the heart.
  2. Mg2+ ions are important in the green parts of the plants.
  3. Mg2+ ions form a complex with ATP.
  4. Ca2+ ions are important in blood clotting.

Answer: 1. Ca2+ ions are not important in maintaining the regular beating of the heart.

Ca2+ ions are required to trigger the contraction of muscles and to maintain the regular beating of the heart

Question 55. Which of the following metal ions play an important role in muscle contraction?

  1. K+
  2. Na+
  3. Mg2+
  4. Ca2+

Answer: 4. Ca2+

Calcium is an essential element for the contraction of muscles.

MCQs on p-Block Elements for NEET

NEET Chemistry For The p Block Elements Multiple Choice Questions

Question 1. Taking stability as the factor, which one of the following represents the correct relationship?

  1. AlCl > AlCl3
  2. TlI > TlI3
  3. TICI3 > TlCl
  4. InI3 > InI

Answer: 2. TlI > TlI3

As we move down the group, the stability of the +3 oxidation state decreases while that of the +1 oxidation state increases. Thus, the stability order for the +1 oxidation state is Al < Ga < In< Tl and the +3 oxidation state is Al > Ga > In > TI.

Hence, the stability order is, AlC3 > AICI, TII > TlI3, TlCl > TlCl3 and InI > InI3

Question 2. BF3 is a planar and electron-deficient compound. Hybridization and the number of electrons around the central atom, respectively are

  1. sp² and 8
  2. sp³ and 4
  3. sp³ and 6
  4. sp² and 6

Answer: 4. sp² and 6

BF3 is a planar and electron-deficient compound. Hybridization and the number of electrons around the central atom

BF3 is a sp²-hybridised planar molecule. It forms 3σ-bonds with 3F-atoms, hence has six electrons around it.

The p Block Elements Palanar Molecule

Question 3. The correct order of atomic radii in group 13 elements is

  1. B < Al < In < Ga < Tl
  2. B < Al < Ga < In < l
  3. B < Ga < Al < Tl < In
  4. B < Ga < Al < In < Tl

Answer: 4. B < Ga < Al < In < Tl

Question 4. AlF3 is soluble in HF only in the presence of KF. It is due to the formation of

  1. \(\mathrm{K}_3\left[\mathrm{AlF}_3 \mathrm{H}_3\right]\)
  2. \(\mathrm{K}_3\left[\mathrm{AlF}_6\right]\)
  3. \(\mathrm{AlH}_3\)
  4. \(\mathrm{K}\left[\mathrm{AlF}_3 \mathrm{H}\right]\)

Answer: 2. \(\mathrm{K}_3\left[\mathrm{AlF}_6\right]\)

AlF3 is insoluble in anhydrous HF because the F ions are not available in hydrogen-bonded HF molecules but, it becomes soluble in the presence of a small amount of KF due to the formation of complex, K3[AlF3].

⇒ \(\mathrm{AlF}_3+3 \mathrm{KF} \rightarrow \mathrm{K}_3\left[\mathrm{AlF}_6\right]\)

Read and Learn More NEET MCQs with Answers

Question 5. The stability of the +1 oxidation state among Al, Ga, In, and Tl increases in the sequence

  1. Al < Ga < In < Tl
  2. Tl < In < Ga < Al
  3. In < Tl < Ga < Al
  4. Ga < In < Al < Tl

Answer: 1. Al < Ga < In < Tl

In group 13 elements, the stability of the +3 oxidation state decreases down the group while that of the +1 oxidation state increases due to the inert pair effect. Hence, the stability of the +1 oxidation state increases in the sequence: Al <Ga<In<Tl.

Question 6. Aluminum (3) chloride forms a dimer because aluminum

  1. Belongs to 3rd group
  2. Can have a higher coordination number
  3. Cannot form a trimer
  4. Has high ionization energy.

Answer: 2. Can have a higher coordination number

AlCl3 forms a dimer, as Al due to the presence of 3d-orbitals can expand its covalency from four to six. Also, dimerization enables Al atoms to complete their octets.

The p Block Elements Dimer

Question 7. Which one of the following elements is unable to form \(\mathrm{MF}_6^{3-} \text {. }\) ion?

  1. Ga
  2. Al
  3. B
  4. In

Answer: 3. B

Boron does not have vacant d-orbitals in its valence shell, so it cannot expand its covalency beyond 4 i.e., ‘B’ cannot form the ions like \(\mathrm{MF}_6^{3-} \text {. }\)

Question 8. The tendency of BF3, BCl3, and BBr3 to behave as Lewis acid decreases in the sequence

  1. BCl3 > BF3 > BBr3
  2. BBr3 > BCl3 > BF3
  3. BBr3 > BF3 > BCl3
  4. BF3 > BCl3 > BBr3

Answer: 2. BBr3 > BCl3 > BF3

The relative Lewis acid character of boron trihalides is found to follow the following order, BBr3 > BCl3 > BF3, but the expected order on the basis of electronegativity of the halogens (electronegativity of halogens decreases from F to I) should be, BF3, > BCl3 > BBr3

This anomaly is explained on the basis of the relative tendency. of the halogen atom to back donate its unutilized electrons to the vacant p-orbital of the boron atom. In BF3, boron has a vacant 2p orbital and each fluorine has fully filled unutilized 2p-orbitals.

Fluorine transfers two electrons to the vacant 2p-orbital of boron, thus forming pπ – pπ bond

The p Block Elements 2p Orbital Of Boron

This type of bond has some double-bond character and is known as dative or back bonding. All the three bond lengths are the same. It is possible when a double bond is delocalized. The delocalization may be represented as :

The p Block Elements Double Bond Is Delocalised

The tendency to back donate decreases from F to I as the energy level difference between B and halogen atoms increases from F to L So, the order of Lewis acid strength is BF3 < BCl3 < BBr3

Question 9. Boron compounds behave as Lewis acids, because of their

  1. Ionisation property
  2. Electron deficient nature
  3. Acidic nature
  4. Covalent nature.

Answer: 2. Electron-deficient nature

Lewis acids are those substances that can accept a pair of electrons and boron compounds usually are deficient in electrons.

Question 10. Which of the following statements is not correct about diborane?

  1. There are two 3-center-2-electron bonds.
  2. The four terminal B – H bonds are two center two-electron bonds.
  3. The four terminal hydrogen atoms and the two boron atoms lie in one plane.
  4. Both the boron atoms are sp² hybridized.

Answer: 4. Both the boron atoms are sp² hybridized.

In diborane (B2H6), each boron (B) atom uses sp³ hybrid orbitals for bonding.

Question 11. Boric acid is an acid because its molecule

  1. Contains replaceable H+ ion
  2. Gives up a proton
  3. Accepts OH from water releasing a proton
  4. Combines with a proton from the water molecule.

Answer: 3. Accepts OH from water releasing a proton

Boric acid behaves as a Lewis acid, by accepting a pair of electrons from the OH ion of water thereby releasing a proton.

Question 12. Which of the following structures is similar to graphite

  1. B4C
  2. B2H6
  3. BN
  4. B

Answer: 3. BN

BN is known as inorganic graphite and has a structure similar to graphite.

Question 13. The type of hybridisation of boron in diborane is

  1. sp³-hybridisation
  2. sp²-hybridisation
  3. sp-hybridization
  4. sp³d²-hybridisation.

Answer: 1. sp³-hybridisation

Each ‘B’ atom in diborane (B2H6) is sp³-hybridised’ Of the 4-hybrid orbitals, three have one electron each, while the 4th is empty.

Two orbitals of each form o bond with two ‘H’- atoms, with one of the remaining hybrid orbital (either filled or empty), ls orbital of ‘H’ atom and one of the hybrid orbitals of other ‘B’ atom overlap to form three centered two-electron bond. So there exist two such types of three-centered bonds.

Question 14. Which of the following statements about H3BO3 is not correct?

  1. It has a layer structure in which planar BO3 units are joined by hydrogen bonds.
  2. It does not act as a proton donor but acts as a Lewis acid by accepting hydroxyl ions.
  3. It is a strong tribasic acid.
  4. It is prepared by acidifying an aqueous solution of borax.

Answer: 3. It is a strong tribasic acid.

H3BO3 is a weak monobasic acid. \(\mathrm{B}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{O} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+}\)

Question 15. Which of the following is an incorrect statement?

  1. SnF4 is ionic in nature.
  2. PbF4 is covalent in nature.
  3. SiCl4 is easily hydrolyzed.
  4. GeX4(X = F, Cl, Br, I) is more stable than GeX2.

Answer: 2. PbF4 is covalent in nature.

Generally, the halides of group-4 elements are covalent in nature. PbF4 and SnF4 are exceptions which are ionic in nature.

Question 16. Which of the following species is not stable?

  1. [SiCl6]2-
  2. [SiF6]2-
  3. [GeCl6]2-
  4. [Sn(OH)6]2-

Answer: 1. [SiCl6]2-

[SiCl6]2- is not stable due to steric hindrance by large-sized Cl atoms.

Question 17. It is because of the inability of ns² electrons of the valence shell to participate in bonding that

  1. Sn2+ is oxidizing while Pb4+ is reducing
  2. Sn2++ and Pb2+ are both oxidizing and reducing
  3. Sn4+ is reducing while Pb4+ is oxidizing
  4. Sn2+ is reducing while Pb4+ is oxidizing.

Answer: 4. Sn2+ is reducing while Pb4+ is oxidizing.

The inertness of s-subshell electrons towards bond formation is known as the inert pair effect. This effect increases down the group thus, for Sn, the +4 oxidation state is more stable whereas, for Pb, the +2 oxidation state is more stable, i.e., Sn2+ is reducing while Pb4+ is oxidizing.

Question 18. Which of the following oxidation states is the most characteristic of lead and tin respectively?

  1. +2, +4
  2. +4, +4
  3. +2, +2
  4. +4, +2

Answer: 1. +2, +4

When ns² electrons of the outermost shell do not participate in bonding then these ns² electrons are called inert pair and the effect is called the inert pair effect.  Due to this inert pair effect, Ge, Sn, and Pb of group 14 have a tendency to form Loth +a and +2 ions.

Now the inert pair effect increases down the group, hence the stability of M2+ ions increases, and M4+ ions decrease down the group. For this reason, Pb2+ is more stable than Pb4+ and Sn4+ is more stable than Sn2+

Question 19. Carbon and silicon belong to the (4) group. The maximum coordination number of carbon in commonly occurring compounds is 4, whereas that of silicon is 6. This is due to

  1. Availability of low lying d-orbitals in silicon
  2. The large size of the silicon
  3. More electropositive nature of silicon
  4. Both (2) and (3).

Answer: 1. Availability of low-lying d-orbitals in silicon

Carbon and silicon belong to the (4) group. The maximum coordination number of carbon in commonly occurring compounds is 4, whereas that of silicon is 6.

Carbon has no d-orbitals, while silicon contains d-orbitals in its valence shell which can be used for bonding purposes.

Question 20. Match List-1 with List-2

The p Block Elements Allotropes Of Carbon

Choose the correct answer from the options given below:

  1. (1) – (C), (2) – (1), (3) – (D), (4) – (B)
  2. (1) – (C), (2) – (D), (3) – (A), (4) – (B)
  3. (1) – (B), (2) – (D), (3) – (A), (4) – (C)
  4. (1) – (D), (2) – (A), (3) – (B), (4) – (C)

Answer: 1. (1) – (C), (2) – (1), (3) – (D), (4) – (B)

Question 21. Choose the correct statement.

  1. Diamond and graphite have a two-dimensional network.
  2. Diamond is covalent and graphite is ionic.
  3. Diamond is sp³ hybridized and graphite is sp² hybridized.
  4. Both diamond and graphite are used as lubricants.

Answer: 3. Diamond is sp³ hybridized and graphite is sp² hybridized.

In diamond, each carbon atom undergoes sp³ hybridization and is linked to four other carbon atoms by using hybridized orbitals in a tetrahedral fashion. In graphite, each carbon atom in a hexagonal ring undergoes sp² hybridization and makes three sigma bonds with three neighboring carbon atoms. The fourth electron forms a π-bond.

Question 22. Which of the following does not show electrical conduction?

  1. Diamond
  2. Graphite
  3. Potassium
  4. Sodium

Answer: 1. Diamond

Except for diamonds other three conduct electricity. Potassium and sodium are metallic conductors, while graphite is a non-metallic conductor.

Question 23. The percentage of lead in lead pencils is

  1. 80
  2. 20
  3. Zero
  4. 70

Answer: 3. Zero

The lead pencil contains graphite and clay. It does not contain lead.

Question 24. In graphite, electrons are

  1. Localized on each C-atom
  2. Localized on every third C-atom
  3. Spread out between the structure
  4. Present in antibonding orbital.

Answer: 3. Spread out between the structure

In graphite, each carbon atom undergoes sp²-hybridisation and is covalently bonded to three other .urbor. atoms by single bonds. The fourth electron forms π-bond. The electrons are delocalized over the whole sheet.

i.e., electrons are spread out between the structure.

Question 25. Which of the following types of forces bind together the carbon atoms in a diamond?

  1. Ionic
  2. Covalent
  3. Dipolar
  4. VanderWaals

Answer: 2. Covalent

In diamond, each carbon atom is sp³ hybridized and thus, forms covalent bonds with four other carbon atoms lying at the corners of a regular tetrahedron.

Question 26. Which of the following is an insulator?

  1. Graphite
  2. Aluminium
  3. Diamond
  4. Silicon

Answer: 3. Diamond

All the above are conductors except diamond. Diamond is an insulator.

Question 27. Identify the correct statements from the following:

  1. CO2(g) is used as a refrigerant for ice cream and frozen food.
  2. The structure of C60 contains twelve six-carbon rings and twenty-five carbon rings.
  3. ZSM-5, a type of zeolite, is used to convert alcohol into gasoline.
  4. CO is a colorless and odorless gas.
  1. (1), (2), and (3) only
  2. (1) and (3) only
  3. (2) and (3) only
  4. (3) and (4) only

Answer: 4. (3) and (4) only

  1. Solid CO2 (dry ice) is used as a refrigerant for ice cream and frozen food.
  2. The structure of C60 contains twenty-six-membered rings and twelve five-membered rings.
  3. 3 and 4 are correct statements

Question 28. Which of the following compounds is used in cosmetic surgery?

  1. Silica
  2. Silicates
  3. Silicones
  4. Zeolites

Answer: 3. Silicones

Silicones being biocompatible are used in surgical and cosmetic Plants.

Question 29. Which of these is not a monomer for a high molecular mass silicone polymer?

  1. Me3SiCl
  2. PhSiCl3
  3. MeSiCl3
  4. Me2SiCl3

Answer: 1. Me3SiCl

It can form only dimer.

Question 30. The basic structural unit of silicates is

  1. \(\mathrm{SiO}_3^{2-}\)
  2. \(\mathrm{SiO}_4^{2-}\)
  3. \(\mathrm{SiO}^{-}\)
  4. \(\mathrm{SiO}_4^{4-}\)

Answer: 4. \(\mathrm{SiO}_4^{4-}\)

SiO4-4 orthosilicate is the basic unit of silicates.

Question 31. Which statement is wrong?

  1. Beryl is an example of cyclic silicate.
  2. Mg2SiO4 is orthosilicate.
  3. The basic structural unit in silicates is the SiO4 tetrahedron.
  4. Feldspars are not aluminosilicates.

Answer: 4. Feldspars are not aluminosilicates.

Feldspars are three-dimensional aluminosilicates.

Question 32. Name the two types of the structure of silicate in which one oxygen atom of [SiO4]4- is shared.

  1. Linear chain silicate
  2. Sheet silicate
  3. Borosilicate
  4. Three dimensional

Answer: 3. Pyrosilicate

Borosilicate contains two units of SiO44- joined along a corner containing oxygen-atom

The p Block Elements Pyrosiliate

Question 33. The straight-chain polymer is formed by

  1. Hydrolysis of CH3SiCl3 followed by condensation polymerization
  2. Hydrolysis of (CH3)4Si by addition polymerisation
  3. Hydrolysis of (CH3)2SiCl2 followed by condensation polymerisation
  4. Hydrolysis of (CH3)3SiCl followed

Answer: 3. Hydrolysis of (CH3)2SiCl2 followed by condensation polymerization

Hydrolysis of substituted chlorosilanes yields corresponding silanols which undergo polymerization. Out of the given chlorosilanes, only (CH3)2SiCl2 will give linear polymer on hydrolysis followed by polymerisation.

The p Block Elements Hydrolysis Of Substitued Chlorosilanes

Question 34. Which of the following anions is present in the chain structure of silicates?

  1. \(\left(\mathrm{Si}_2 \mathrm{O}_5^{2-}\right)_n\)
  2. \(\left(\mathrm{SiO}_3^{2-}\right)_n\)
  3. \(\mathrm{SiO}_4^{4-}\)
  4. \(\mathrm{Si}_2 \mathrm{O}_7^{6-}\)

Answer: 2. \(\left(\mathrm{SiO}_3^{2-}\right)_n\)

Chain silicates are formed by sharing two oxygen atoms by each tetrahedron. Anions of chain silicate have two general formula

  1. \(\left(\mathrm{SiO}_3\right)_n^{2 n-}\)
  2. \(\left(\mathrm{Si}_4 \mathrm{O}_{11}\right)_n^{6 n-}\)

Question 35. Which one of the following statements about the zeolite is false?

  1. They are used as cation exchangers.
  2. They have an open structure which enables them to take up small molecules.
  3. Zeolites are aluminosilicates having a three-dimensional network.
  4. Some of the SiO44- units are replaced by AlO45- and AlO69- ions in zeolites.

Answer: 4. Some of the SiO44- units are replaced by AlO45- and AlO69- ions in zeolites.

In zeolites, some of the Si4+ ions may be replaced by Al3+ ions. This results in an unbalanced anionic charge To maintain electrical neutrality, positive ions must be introduced.

Question 36. The substance used as a smoke screen in warfare is

  1. SiCl4
  2. PH3
  3. PCl5
  4. Acetylene.

Answer: 1. SiCl4

SiCl4 gets hydrolysed in moist air and gives white fumes which are used as a smoke screen in warfare.

MCQs on Hydrocarbons for NEET

NEET Chemistry For Hydrocarbons Multiple Choice Questions

Question 1. The dihedral angle of the least stable conformer of ethane is

  1. 120°
  2. 180°
  3. 60°

Answer: 1. 0°

The dihedral angle of the least stable conformer of ethane is 0°. The magnitude of torsional strain depends upon the angle of rotation about the C – C bond. This angle is also called dihedral angle or torsional angle. Of all the conformations of ethane, the staggered form has the least torsional strain, and the eclipsed form has the maximum torsional strain.

Question 2. Hydrocarbons Reagent Chemical

Consider the above reaction and identify the missing reagent/chemical.

  1. DIBAL-H
  2. B2H6
  3. Red Phosphorus
  4. CaO

Answer: 4. CaO

Hydrocarbons Missing Reagent Reaction

Question 3. Which of the following alkanes cannot be made in good yield by the Wurtz reaction?

  1. n-Hexane
  2. 2, 3-Dimethylbutane
  3. n-Heptane
  4. n-Butane

Answer: 3. n-Heptane

Wurtz reaction is used for the preparation of higher alkanes containing an even number of C-atoms. Thus this reaction cannot be used for the preparation of n-heptane.

Read and Learn More NEET MCQs with Answers

Question 4. The alkane that gives only one monochloride product on chlorination with Cl2 in the presence of diffused sunlight is

  1. 2,2-dimethylbutane
  2. Neopentane
  3. n-pentane
  4. Isopentane.

Answer: 2. Neopentane

In the chlorination of alkanes, hydrogen is replaced by chlorine. In neo-pentane, only one type of hydrogen is present, hence the replacement of any hydrogen atom will give the same product.

Hydrocarbons Chlorination Of Alkanes

Question 5. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is

  1. CH ≡ CH
  2. CH2 = CH2
  3. CH3 — CH3
  4. CH4

Answer: 4. CH4

Hydrocarbon Wurtz Reaction

Question 6. With respect to the conformers of ethane, which of the following statements is true?

  1. The bond angle changes but the bond length remains the same.
  2. Both bond angle and bond length change.
  3. Both bond angle and bond length remain the same.
  4. Bond angle remains the same but the bond length changes.

Answer: 3. Both bond angle and bond length remain the same.

Conformers of ethane have different dihedral angles but the same bond angle and bond lengths.

Question 7. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is

  1. The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
  2. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain
  3. The staggered conformation of ethane is less stable than eclipsed conformation because staggered conformation has torsional strain
  4. The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain.

Answer: 2. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain

Conformers of ethane have different dihedral angles but the same bond angle and bond lengths.

Hydrocarbons Newmans Projections Of Ethane

The magnitude of torsional strain depends upon the angle of rotation of the C-C bond. The staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. So, the staggered conformation of ethane is more stable than the eclipsed conformation.

Question 8. In the following the most stable conformation of n-butane is

Hydrocarbons n Butane

Answer: 2

The anti-conformation is the most stable conformation of n-butane as in this, the bulky methyl groups are as far apart as possible thereby keeping steric repulsion at a minimum

Question 9. Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by

  1. Oxidation
  2. Cracking
  3. Distillation under reduced pressure
  4. Hydrolysis.

Answer: 2. Cracking

The process of cracking converts higher alkanes into smaller alkanes and alkenes. This process can be used for the production of natural gas.

Question 10. Which of the following conformers for ethylene glycol is most stable?

Hydrocarbons Ethylene Glycol

Answer: 4

The conformation (4) is most stable because of intermolecular H-bonding.

Question 11. The dihedral angle in the staggered form of ethane is

  1. 120°
  2. 60°
  3. 180°

Answer: 3. 60°

The staggered form of ethane has the following structure and the dihedral angle is 60°, which means ‘H’ atoms are at an angle of 60° to each other.

Hydrocarbons Dihedral Angle

Question 12. Which of the following reactions is expected to readily give a hydrocarbon product in good yields?

Hydrocarbon Products Of Good Yields

Answer: 3.

When an aqueous solution of sodium or potassium salt of a carboxylic acid is electrolyzed, hydrocarbon is evolved at the anode.

Hydrocarbon Electrolysis Oxidation

At anode: \(2 \mathrm{RCOO}^{-}-2 e^{-} \rightarrow \underset{ Alkane }{R-R}+2 \mathrm{CO}_2\)

Question 13. In commercial gasoline, the type of hydrocarbons that are more desirable is

  1. Linear unsaturated hydrocarbon
  2. Toluene
  3. Branched hydrocarbon
  4. Straight-chain hydrocarbon.

Answer: 3. Branched hydrocarbon

The branching of the chain increases the octane number of a fuel. A high octane number means better fuel

Question 14. The most stable conformation of n-butane is

  1. Gauche
  2. Staggered
  3. Skew boat
  4. Eclipsed.

Answer: 2. Staggered

Newman projections for n-butane are

Hydrocarbons Newman Projections For n butane

The staggered conformation has minimum repulsion between the hydrogen atoms attached tetrahedrally to the two carbon atoms. Thus, it is the most stable conformation.

Question 15. Which of the following is used as an antiknocking material?

  1. Glyoxal
  2. Freon
  3. T.E.L.
  4. Ethyl alcohol

Answer: 3. T.E.L.

Tetraethyllead(C2H5)4Pb, is used as an antiknocking agent in gasoline used for running automobiles.

Question 16. The reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the order

  1. Tertiary > primary > secondary
  2. Primary > secondary > tertiary
  3. Both (1) and (2)
  4. Tertiary > secondary > primary.

Answer: 4. Tertiary > secondary > primary.

The reactivity of the H-atom depends upon the stability of free radicals, therefore reactivity of the H-atom follows the order: 3°>2°>1°.

Question 17. Compound X on reaction with O3 followed by Zn/H2O gives formaldehyde and 2-methylpropanal as products. The compound X is

  1. 3-methyl but-l-ene
  2. 2-methyl but-l-ene
  3. 2-methyl but-2-ene
  4. pent-2-ene.

Answer: 1

Hydrocarbons Formaldehyde And 2 Methylpropanal

So, X can be

Hydrocarbons 3 Methylbut 1 ene

Question 18. The major product of the following chemical reaction is

Hydrocarbons Major Products Of Chemical Reactions

Answer: 2

In the presence of peroxide, an addition reaction takes place.

Hydrocarbons 1 Bromo 3 Methylbutane

Question 19. An alkene on ozonolysis gives methanal as one of the products. Its structure is

Hydrocarbons Alkene On Ozonolysis Gives Methanal

Answer: 3

Hydrocarbons Alkene On Ozonolysis

Question 20. An alkene A on reaction with O3 and Zn—H2O gives propanone and ethanal in equimolar ratio. The addition of HCl to alkene A gives B as the major product. The structure of product B is

Hydrocarbons Propanone And Ethanal

Answer: 4

An alkene A on reaction with O3 and Zn—H2O gives propanone and ethanal in equimolar ratio. The addition of HCl to alkene A gives B as the major product.

Hydrocarbons HCl To An Alkene

Addition of HCI to an alkene (1) will take place according to Markownikoff’s rule.

Question 21. The most suitable reagent for the following conversion is

Hydrocarbons cis 2 butene

  1. Hg2+/H+,H2O
  2. Na/liquid NH2
  3. H2, Pd/C, quinoline
  4. Zn/HCl

Answer: 3. H2, Pd/C, quinoline

Hydrocarbons Quinoline And cis 2 Butene

Question 22. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?

Hydrocarbons Propane By reaction

Answer: 3

Hydrocarbons HBr Direct Eliminated

Question 23. The compound that will react most readily with gaseous bromine has the formula

  1. C3H6
  2. C2H2
  3. C4H10
  4. C2H4

Amswer: 1. C3H6

The rate of free radical substitution with Br2(g) depends upon the stability of free radicals. Propenyl free radical is allylic free radical which is more stable.

Question 24. In the reaction with HCl, an alkene reacts in accordance with the Markovnikovs rule to give a product 1-chloro-1-methyl-cyclohexane. The possible alkene is

Hydrocarbons Markovnikovs Rule

Answer: 3

Hydrocarbons Markovnikovs Addition With 1 Chloro 1 methylcyclohexane

Question 25. Which of the following is not the product of dehydration of Hydrocarbons Dehydration isomerism?

Hydrocarbons Product Of Dehydration

Answer: 1

Hydrocarbons Not A Product Of Dehydration

Question 26. In the following reaction

Hydrocarbons Major And Minor Products

Answer: 1

Hydrocarbons 2 and 3 Carbocations

Question 27. Which of the following compounds will exhibit cis-trans (geometrical) isomerism?

  1. Butanol
  2. 2-Butyne
  3. 2-Butenol
  4. 2-Butene

Answer: 4. 2-Butene

Cis-trans isomerism is exhibited by compounds having C – C, C – N, and N – N groups, due to restricted rotation around the double bond. Among the given options, only 2-butene exhibits geometrical isomerism.

Hydrocarbons Geometrical Isomerism

Question 28.

Hydrocarbons Predominantly

Answer: 4

Hydrocarbons 2 Bromo 2 Methylbutane

Question 29. Which of the compounds with molecular formula C5H10 yields acetone on ozonolysis?

  1. 3-Methyl-1-butene
  2. Cyclopentane
  3. 2-Methyl-1-butene
  4. 2-Methyl-2-butene

Answer: 4. 2-Methyl-2-butene

Hydrocarbons Acetone On Ozonolysis

Question 30. Which one of the following alkenes will react faster with H2 under catalytic hydrogenation conditions?

Hydrocarbons R Alkyl Substituent

Answer: 1

The relative rates of hydrogenation decrease with the increase of steric hindrance. Most stable alkene, slowly it undergoes hydrogenation to give the product. The least substituted alkene is less stable and more reactive.

Order of stability is:

Hydrocarbons Hydrogenation

Hence, alkene which will react faster with H, is that which is most unstable

Question 31. The reaction of HBr with propene in the presence of peroxide gives

  1. Isopropyl bromide
  2. 3-bromopropane
  3. Allyl bromide
  4. n-propyl bromide.

Answer: 4. n-propyl bromide.

The formation of n-propyl bromide in the presence of peroxide can be explained as follows:

Step 1: Peroxide undergoes fission to give free radicals \(\mathrm{R}-\mathrm{O}-\mathrm{O}-\mathrm{R} \longrightarrow 2 \mathrm{R}-\dot{\mathrm{O}}\)

Step 2: HBr combines with free radicals to form bromine free radicals. \(R-\dot{\mathrm{O}}+\mathrm{HBr} \longrightarrow \mathrm{R}-\mathrm{OH}+\dot{\mathrm{Br}}\)

Step 3: Br attacks the double bond of the alkene to form a more stable free radical

Hydrocarbons Br Attacks The Double Bond Of The Alkene

Step 4: More stable free radical attacks on HBr. \(\mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{HCH}_2 \mathrm{Br}+\mathrm{HBr} \longrightarrow \underset{n-\text { Propyl bromide }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\dot{\mathrm{Br}}}\)

Step 5: \(\dot{\mathrm{Br}}+\dot{\mathrm{Br}} \longrightarrow \mathrm{Br}_2\)

Question 32. The compound,  Hydrocarbons NaIO on reaction with NaIO4 in the presence of KMnO2 gives

  1. CH3COCH3
  2. CH3COCH3 + CH3COOH
  3. CH3COCH3 + CH3CHO
  4. CH3CHO + CO3

Answer: 2. CH3COCH3 + CH3COOH

Hydrocarbons Methyl Carbon Reaction

Question 33. Geometrical isomers differ in

  1. Position of functional group
  2. Position of atoms
  3. Spatial arrangement of atoms
  4. Length of the carbon chain.

Answer: 3. Spatial arrangement of atoms

Geometrical isomers are those isomers that possess the same molecular and structural formula but differ in the arrangement of atoms or groups in space due to hindered rotation around the double-bonded atoms.

Question 34. In preparation of alkene from alcohol using Al2O3 which is the effective factor?

  1. Porosity of Al2O3
  2. Temperature
  3. Concentration
  4. Surface area of Al2O3

Answer: 2. Temperature

Temperature is an effective factor because at different temperatures it forms different products

Hydrocarbon Temperature Effective Factor

This is intramolecular dehydration. At lower temperatures, intermolecular dehydration takes place between two molecules of alcohol and ether will be formed.

Question 35. Which reagent converts propane to 1-propanol?

  1. \(\mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{SO}_4\)
  2. \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}^{-}\)
  3. \(\mathrm{Hg}(\mathrm{OAc})_2, \mathrm{NaBH}_4 / \mathrm{H}_2 \mathrm{O}\)
  4. Aq. KOH

Answer: 2. \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}^{-}\)

Propene adds to diborane (B2H6) giving an additional product. The addition compound on oxidation gives l-propanol. Here the addition of water takes place according to anti-Markownikoff ‘s rule

Question 36. Which is maximum stable?

  1. 1-Butene
  2. cis-2-Butene
  3. trans-2-Butene
  4. All have the same stability.

Answer: 3. trans-2-Butene

This is the most stable as the repulsion between two methyl groups is the least.

Hydrocarbons Methyl Groups

Question 37. 2-Butene shows geometrical isomerism due to

  1. Restricted rotation about the double bond
  2. Free rotation about the double bond
  3. Free rotation about single bond
  4. Chiral carbon.

Answer: 1. Restricted rotation about the double bond

Due to restricted rotation about double bonds, 2-butene shows geometrical isomerism.

Hydrocarbons 2 Butene Shows Geometrical ISomerism

Question 38. 2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is

  1. trans-2-pentene
  2. 1-pentene
  3. 2-ethoxy pentane
  4. 2-ds-pentene.

Answer: 1. trans-2-pentene

Hydrocarbons 2 Bromo Pentane

Question 39. In a reaction,  Hydrocarbons Hypochlorous Acid  where, M = Molecule and R = Reagent. M and R are

  1. CH3CH2OH and HCl
  2. CH2 = CH2 and heat
  3. CH3CH2Cl and NaOH
  4. CH2Cl – CH2OH and aq. NaHCO3

Answer: 4. CH2Cl – CH2OH and aq. NaHCO3

Hydrocarbons Molecule And Reagent

Therefore, M = CH2CI-CH2OH and

R = ag.NaHCO3

Question 40. The reaction, CH2 = CH – CH2 + HBr → CH2CHBr – CH3 is

  1. Electrophilic substitution
  2. Free radical addition
  3. Nucleophilic addition
  4. Electrophilic addition.

Answer: 4. Electrophilic addition.

In this reaction, HBr undergoes heterolytic fission as HBr → H+ + Br

Hydrocarbon Heterolytic Fission

This is an example of an electrophilic addition reaction

Question 41. Which of the following has zero dipole moment?

  1. 1-Butene
  2. 2-Methyl-1-propene
  3. cis-2-Butene
  4. trans-2-Butene

Answer: 4. trans-2-Butene

Question 42. One of the following which does not observe the anti-Markownikoff addition of HBr, is

  1. Pent-2-ene
  2. Propene
  3. But-2-ene
  4. But-1-ene.

Answer: 3. But-2-ene

In the case of but-2-ene (CH3– CH-CH- CH2)both double-bonded carbons are identical. Therefore, it does not observe the anti-Markownikoff addition of HBr

Question 43. Reduction of 2-butyne with sodium in liquid ammonia gives predominantly

  1. cis-2-butene
  2. No reaction
  3. Trans-2-butene
  4. n-butane.

Answer: 3. Trans-2-butene

Reduction of non-terminal alkynes with Na in Liq. NH3 at 195 – 200 K gives trans-alkene.

Hydrocarbons Reduction Of Non Terminal Alkynes

Question 44. The restricted rotation about the carbon double bond in 2-butene is due to

  1. Overlap of one s and sp²-hybridized orbitals
  2. Overlap of two sp²-hybridized orbitals
  3. Overlap of one p and one sp²-hybridized orbitals
  4. Sideways overlap of two p-orbitals.

Answer: 4. Sideways overlap of two p-orbitals.

Hydrocarbons Acetylene Orbitals

Restricted rotation is due to sideways overlap of two p-orbitals.

Question 45. Which one of the following can exhibit cis-trans isomerism?

  1. CH3 – CHCl – COOH
  2. H – C ≡ C – Cl
  3. ClCH =CHCl
  4. ClCH2 – CH2Cl

Answer: 3. ClCH =CHCl

Hydrocarbons 1 2 Dichloroethene Exhibits cis trans ISomerism

1, 2-Dichloroethene exhibits cis-trans (geometrical) isomerism

Question 46. In the following reaction, Hydrocarbons Red Hot Iron Tube the number of sigma(σ) bonds present in the product A, is

  1. 21
  2. 9
  3. 24
  4. 18

Answer: 1. 21

Hydrocarbons Alkynes Sigma Bonds

There are 21 σ bonds.

Question 47. Which one is the correct order of acidity?

  1. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}\) \(>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3\)
  2. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{CH}_3\)
  3. \(\mathrm{CH}_3-\mathrm{CH}_3>\mathrm{CH}_2=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}\)
  4. \(\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}\)

Answer: 1. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}\) \(>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3\)

Alkanes, alkenes, and alkynes follow the following trend in their acidic behavior:

⇒ \(\stackrel{s p}{\mathrm{HC}} \equiv \stackrel{s p}{\mathrm{C}} \mathrm{H}>\stackrel{s p^2}{\mathrm{H}_2 \mathrm{C}}=\stackrel{s p^2}{\mathrm{CH}_2}>\stackrel{s p^3}{\mathrm{CH}_3}-\stackrel{s p^3}{\mathrm{CH}_3}\)

This is because sp-hybridised carbon is more electronegative than sp²-hybridized carbon which is further more electronegative than sp³-hybridised carbon.

Hence, in ethyne proton can be released more easily than ethene and ethane.

Among alkynes, the order of acidity is HC ≡ CH > CH3 -C ≡ CH > CH3 -C ≡ C- CH3 This is due to the +1 effect of – CH3 group.

Question 48. Predict the correct intermediate and product in the following reaction:

Hydrocarbons Alkene Products

Answer: 3

In the case of unsymmetrical alkynes addition of H2O occurs in accordance with Markownikoff’s rule.

Hydrocarbons Unsymmetrical Alkynes

Question 49. The pair of electrons in the given carbanion, CH3C≡ C, is present in which of the following orbitals?

  1. sp²
  2. sp
  3. 2p
  4. sp³

Answer: 2. sp²

⇒ \(\mathrm{CH}_3-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}}\)

Thus, a pair of electrons is present in the sp-hybridized orbital

Question 50. In the reaction H-C ≡ CH Hydrocarbon Butyne And HexyneX and Y are

  1. X = 2-butyne, Y = 2-hexyne
  2. X = 1-butyne, Y = 2-hexyne
  3. X = 1-butyne, Y = 3-hexyne
  4. X = 2-butyne, Y = 3-hexyne.

Answer: 3. X = 1-butyne, Y = 3-hexyne

Hydrocarbons 1 Butyne And 2 Hexyne

Question 51. Which of the following organic compounds has the same hybridization as its combustion product (CO2)?

  1. Ethane
  2. Ethyne
  3. Ethene
  4. Ethanol

Answer: 2. Ethyne

⇒ \(\mathrm{C}_2 \mathrm{H}_2+\frac{5}{2} \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Both ethyne and CO2 have sp-hybridization.

⇒ \(\mathrm{O}=\stackrel{s p}{\mathrm{C}}=\mathrm{O} \quad \stackrel{s p}{\mathrm{HC}} \equiv \stackrel{s p}{\mathrm{C}} \mathrm{H}\)

Question 52. In the following reaction: Hydrocarbon Victor Meyer Test Product ‘P’ will not give

  1. Tollens’ reagent test
  2. Bradys reagent test
  3. Victor Meyer test
  4. Iodoform test.

Answer: 3. Victor Meyer test

Hydrocarbons Acetaldehyde

Acetaldehyde does not give Victor Meyer a test.

Question 53. Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?

  1. NaNH2
  2. HCl
  3. O2
  4. Br2

Answer: 1. NaNH2

Terminal alkynes (l-butyne) react with NaNH2 to form sodium acetylide and evolve hydrogen but 2-butyne does not.

Question 54. Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear.

  1. CH3 — CH = CH — CH3
  2. CH3 — C ≡ C — CH3
  3. CH3=CH —CH3 —C ≡ CH
  4. CH3 — CH2 — CH3

Answer: 2. CH3 — C ≡ C — CH3

CH3 – C ≡ C-CH3

In the case of sp³ hybridized carbon, the bond angle is 109°28′; sp² hybridized carbon, the bond angle is 120° and sp hybridized carbon, the bond angle is 180°.

So, only Hydrocarbons Hybridised Carbon Is Linear  is linear.

Question 55. Base strength of

  1. \(\mathrm{H}_3 \mathrm{C} \stackrel{-}{\mathrm{C}} \mathrm{H}_2\),
  2. \(\mathrm{H}_2 \mathrm{C}=\stackrel{-}{\mathrm{C}} \mathrm{H}\)
  3. and \(\mathrm{H}-\mathrm{C} \equiv \overline{\mathrm{C}}\)

is in the order of

  1. (A) > (C) > (B)
  2. (A) > (B) > (C)
  3. (B) > (A) > (C)
  4. (C) > (B) > (A)

Answer: 2. (A) > (B) > (C)

⇒ \(\mathrm{H}-\underset{s p}{\mathrm{C}} \equiv \underset{s p}{\mathrm{C}}-\mathrm{H}>\underset{s p^2}{\mathrm{CH}_2}=\underset{s p^2}{\mathrm{CH}_2}>\underset{s p^3}{\mathrm{CH}_3} \mathrm{CH}_3\)

Conjugate base of the given acid: \(\stackrel{-}{\mathrm{C}} \equiv \mathrm{C}-\mathrm{H}<\stackrel{-}{\mathrm{C H}} =\mathrm{CH}_2<\stackrel{-}{\mathrm{C}} \mathrm{H}_2 \mathrm{CH}_3\)

The conjugate base of stronger acid is weaker and vice-versa.

Question 56. Predict the product C obtained in the following reaction of 1-butyne. 

Hydrocarbons Reaction Of 1 Butyne

Answer: 3

Hydrocarbons Hydrohalogenation To Unsymmetrical Alkene

According to Markownikoff’s rule, during hydrohalogenation to an unsymmetrical alkene, the negative part of the addendum adds to a less hydrogenated (i.e. more substituted) carbon atom.

Question 57. Products of the following reaction: Hydrocarbon Hydrolysis are

  1. CH3COOH + CO2
  2. CH3COOH + HOOCCH2CH3
  3. CH3CHO + CH3CH2CHO
  4. CH3COOH + CH3COCH3

Answer: 2. CH3COOH + HOOCCH2CH3

On ozonolysis, higher alkynes form diketones which are further oxidized to dicarboxylic acid.

Hydrocarbons Oxidised To Dicarboxylic Acid

Question 58. When CH3CH2CHCl2 is treated with NaNH2, the product formed is

Hydrocarbons Methylene

Answer: 2

Hydrocarbon Sodium Hydrogen

Question 59. When acetylene is passed through dil. H2SO4 in the presence of HgSO4, the compound formed is

  1. Acetic acid
  2. Ketone
  3. Ether
  4. Acetaldehyde.

Answer: 4. Acetaldehyde.

Hydrocarbons Acetylene

Question 60. The cylindrical shape of an alkyne is due to

  1. Two sigma C – C and one π C – C bonds
  2. One sigma C – C and two π C-C bonds
  3. Three sigma C – C bonds
  4. Three π C – C bonds.

Answer: 2. One sigma C – C and two π C-C bonds

In alkyne, two carbon atoms constituting the triple bond are sp-hybridised’Carbon undergoes sp-hybridization to form two sp-hybrid orbitals. The two 2p-orbitals remain unhybridized. Hybrid orbitals form one sigma bond while two π-bonds are formed by unhybridised orbitals.

Question 61. Hydrocarbons Reagent The reagent is

  1. Na
  2. HCl in H2O
  3. KOH in C2H5OH
  4. Zn in alcohol.

Answer: 3. KOH in C2H5OH

A powerful base is needed to carry out a second dehydrohalogenation reaction, for example, a hot alcoholic KOH solution or alkoxide ion.

Question 62. A compound is treated with NaNH2 to give sodium salt. Identify the compound.

  1. C2H2
  2. C6H6
  3. C2H6
  4. C2H4

Answer: 1. C2H2

A compound is treated with NaNH2 to give sodium salt.

Alkynes react with strong bases like NaNH2 to form sodium acetylide derivatives known as acrylics.

⇒ \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+\mathrm{NaNH}_2 \longrightarrow \mathrm{H}-\mathrm{C} \equiv \overline{\mathrm{C}} \mathrm{Na}^{+}+1 / 2 \mathrm{H}_2\)

Question 63. The shortest C-C bond distance is found in

  1. Diamond
  2. Ethane
  3. Benzene
  4. Acetylene.

Answer: 4. Acetylene

The shortest C- C distance (1.20 Å) is in acetylene.

Question 64. Acetylenic hydrogens are acidic because

  1. Sigma electron density of the C – H bond in acetylene is nearer to carbon, which has 50% s-character
  2. Acetylene has only open hydrogen in each carbon
  3. Acetylene contains the least number of hydrogens among the possible hydrocarbons having two carbons
  4. Acetylene belongs to the class of alkynes with the molecular formula, CnH2n-2.

Answer: 1. Sigma electron density of the C – H bond in acetylene is nearer to carbon, which has 50% s-character

Hydrocarbons Sideway Overlap Of Two p Orbitals

The formation of a C-H bond in acetylene involves a sp-hybridised carbon atom. Since s-electrons are closer to the nucleus than p-electrons, the electrons present in a bond having more s-characters will be closer to the nucleus. In alkynes, the character is 50%, and the electrons constituting this bond are more strongly bonded by the carbon nucleus.

Thus acetylenic C-atom becomes more electronegative in comparison to sp², sp³ and hence the hydrogen atom Present on the carbon atom (≡C-H) can be easily removed

Question 65. Which is the most suitable reagent among the following to distinguish compound (3) from the rest of the compounds?

  1. CH3-C ≡ C-CH3
  2. CH3 – CH2 – CH2 – CH3
  3. CH3-CH2C ≡ CH
  4. CH3-CH ≡ CH2
  1. Bromine in carbon tetrachloride
  2. Bromine in acetic acid
  3. Aik. KMnO4
  4. Ammoniacal silver nitrate

Answer: 4. Ammoniacal silver nitrate

Compound 3 possessing the terminal alkyne only reacts with ammoniacal AgNO3 and thus can be distinguished from 1, 2, and 4 compounds.

Question 66. Consider the following compound/species:

Hydrocarbons Compound Or Species

The number of compounds/species which obey Huckel’s rule is __________

  1. 2
  2. 5
  3. 4
  4. 6

Answer: 3. 4

Hydrocarbons Huckels Rule

These compounds obey Huckel’s rule as these contain (4n + 2)π electrons.

Question 67. Which compound amongst the following is not an aromatic compound?

Hydrocarbons Aromatic Compound

Answer: 4

Compound (4) is not an aromatic compound as reflected by the non-planarity of the methylene bridge (-CH2-) with respect to other atoms. However, the tropylium cation is aromatic due to planarity.

Question 68. Among the following the reaction that proceeds through an electrophilic substitution is

Hydrocarbons Electrophilic Substitution

Answer: 3

The attacking species in the reaction given in option (3) is an electrophile i.e., \(\begin{aligned}
& \delta+ \\
& \mathrm{Cl}
\end{aligned}\)

Therefore, it is an electrophilic substitution reaction.

Question 69. Which of the following can be used as the halide component for the Friedel-Crafts reaction?

  1. Chlorobenzene
  2. Bromobenzene
  3. Chloroethene
  4. Isopropyl chloride

Answer: 4. Isopropyl chloride

Friedel Crafts Reaction:

Hydrocarbons Friedel Craft Reaction

Chlorobenzene, bromobenzene, and chloroethene are not suitable halide components as the C-X bond acquires some double bond character due to the resonance of a lone pair of electrons with π-bond

Question 70. In which of the following molecules, all atoms are coplanar?

Hydrocarbons All Atoms Are Coplanar

Answer: 1

Biphenyl is coplanar as all C-atoms are sp² hybridized.

Question 71. In pyrrole, the electron density is maximum on

Hydrocarbons Pyrrole The Electron

  1. 2 and 3
  2. 3 and 4
  3. 2 and 4
  4. 2 and 5

Answer: 4. 2 and 5

Pyrrole has maximum electron density on 2 and 5. It generally interacts with electrophiles at the C-2 or C-5 due to the highest degree of stability of the protonated intermediate.

Attack at position 3 or 4 yields a carbonation that is a hybrid of structures (1) and (2). Attack at position 2 or 5  yields a carbonation that is a hybrid not only of structures (3) and (4) (analogous to 1 and 2) but also of structure (5).

The extra stabilization conferred by (5) makes this ion the more stable one. Also, attack at position 2 or 5 is laster because the developing positive charge is accommodated by three atoms of the ring instead of only two.

Question 72. In the given reaction, the product P isHydrocarbons Friedal Crafts Alkylation Product the product P is?

Hydrocarbons Friedal Crafts Alkylation

Answer: 3

Hydrocarbons Friedal Crafts Alkylation Reaction

Question 73. Consider the nitration of benzene using a mixed cone. H2SO4 and HNO3. If a large amount of KHSO4 is added to the mixture, the rate of nitration will be

  1. Unchanged
  2. Doubled
  3. Faster
  4. Slower.

Answer: 4. Slower.

Consider the nitration of benzene using a mixed cone. H2SO4 and HNO3. If a large amount of KHSO4 is added to the mixture,

Mechanism of nitration is: \(\mathrm{HNO}_3+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NO}_2^{+}+2 \mathrm{HSO}_4^{-}+\mathrm{H}_5 \mathrm{O}^{+}\)

If a large amount of KHSO4 is added then conc. of HSO4 ions increase and the reaction will be shifted in a backward direction hence, the rate of nilration will be slower.

Question 74. The oxidation of benzene by V2Os in the presence of air produces

  1. Maleic anhydride
  2. Benzoic acid
  3. Benzaldehyde
  4. Benzoic anhydride.

Answer: 1. Maleic anhydride

Hydrocarbons Maleic Anhydride

Question 75. What products are formed when the following compound is treated with Br2 in the presence of FeBr3?

Hydrocarbons Methane Products

Answer: 3

– CH3 group is o,p-directing. Because of crowding, no substitution occurs at the carbon atom between the two – CH3 groups in m-xylene, even though two – CH3 groups activate that position.

Hydrocarbons Bromo Benzene

Question 76. Some meta-directing substituents in aromatic substitution are given, Which one is most deactivating?

  1. — COOH
  2. — NO2
  3. — C ≡ N
  4. — SO3H

Answer: 2. — NO2

-NO2 is most deactivating due to – I and – M effect

Question 77. Which of the following compounds will not undergo Friedel-Crafts reaction easily?

  1. Nitrobenzene
  2. Toluene
  3. Cumene
  4. Xylene

Answer: 1. Nitrobenzene

Nitrobenzene is strongly deactivated, hence will not undergo Friedel-Crafts reaction.

Question 78. Which of the following chemical systems is nonaromatic?

Hydrocarbons Chemical System Is Non Aromatic

Answer: 4

The molecules that do not satisfy the Huckel rule or (4n + 2)π-electron rule are said to be non-aromatic. The compound (4) has a total of 4πe. It does not follow (4n + 2)π tule. So, it is a non-aromatic compound.

Question 79. Among the following compounds, the one that is most reactive toward electrophilic nitration is

  1. Benzoic acid
  2. Nitrobenzene
  3. Toluene
  4. Benzene.

Answer: 3. Toluene

As the +I effect increases reactivity towards electrophilic reactions decreases and as -I or -M elect increases, reactivity toward electrophilic reactions decreases thus, the order is

Hydrocarbons Electrophilic Reaction Decreases

Question 80. The reaction of toluene with Cl2 in the presence of FeCl3 gives X and the reaction in the presence of light gives Y. Thus, X and Y are

  1. X = benzal chloride, Y = o-chlorotoluene
  2. X = m-chlorotoluene, Y = p-chlorotoluene
  3. X = o- and p-chlorotoluene, Y = trichloromethyl benzene
  4. X = benzyl chloride, Y= m-chlorotoluene.

Answer: 3. X = o- and p-chlorotoluene, Y = trichloromethyl benzene

The reaction of Cl2 in the presence of FeCl3, with toluene yields a ring substitution product.

Hydrocarbons Toluene

In presence of sunlight, free radical reaction takes place.

Hydrocarbons Toluene Yields

Question 81. Benzene reacts with CH3Cl in the presence of anhydrous AlCl3 to form

  1. Chlorobenzene
  2. Benzyl chloride
  3. Xylene
  4. Toluene.

Answer: 4. Toluene.

This is Friedel-Crafts alkylation.

Hydrocarbons Friedel Craft Alkylation

Question 82. Nitrobenzene can be prepared from benzene by using a mixture of cones. HNO3 and cone. H2SO4. In the mixture, nitric acid acts as an

  1. Acid
  2. Base
  3. Catalyst
  4. Reducing agent.

Answer: 2. Base

Nitrobenzene can be prepared from benzene by using a mixture of cones. HNO3 and cone. H2SO4. In the mixture

Hydrocarbons Nitrobenzene

Question 83. Which one of the following is most reactive towards electrophilic attack?

Hydrocarbons Electropholic Attack

Answer: 1

Groups like, -CI and -NO, show -I effet -I group attached to the benzene ring decrease the electron density and hence less; prone to electrophilic attack. -OH not only shows the -I effect but also the +M effect which predominates the -I character and electron density is increased in the benzene ring which facilitates electrophilic attack.

Question 84. The order of decreasing reactivity towards an electrophilic reagent, for the following, would be

  1. Benzene
  2. Toluene
  3. Chlorobenzene
  4. Phenol
  1. (B) > (D) > (A) > (C)
  2. (D) > (C) > (B) > (A)
  3. (D) > (B) > (A) > (C)
  4. (A) > (B) > (C) > (D)

Answer: 3. (D) > (B) > (A) > (C)

Benzene having any activaling group i.e., – OH, – R undergoes electrophilic substitution easily as compared to benzene itself. Thus, toluene and phenol undergo electrophilic substitution easily. Chlorine due to -I-reflect deactivates the ring. So, it is difficult to Larry about the electrophilic substitution in chlorobenzene.

Hence, the order is C6H5OH > C6H5CH3 > C6H6 > C6H5Cl.

Question 85. Using anhydrous AlCl3 as a catalyst, which one of the following reactions produces ethylbenzene (PhEt)?

  1. H3C – CH2OH + C6H6
  2. CH3 – CH = CH2 + C6H6
  3. H2C = CH2 + C6H6
  4. H3C – CH3 + C6H6

Answer: 3. H2C = CH2 + C6H6

Question 86. Which one of the following is a free-radical substitution reaction?

Hydrocarbons Free Radical Substition

Answer: 1

Hydrocarbons Benzyl Free Radical

Question 87. The correct order of reactivity towards the electrophilic substitution of the compounds aniline (1), benzene (2), and nitrobenzene (3) is

  1. 3 > 2 > 1
  2. 2 > 3 > 1
  3. 1 < 2 > 3
  4. 1 > 2 > 3

Answer: 4. 1 > 2 > 3

– NH2 group is electron donating hence increasing electron density on ring. Benzene is also electron-rich due to the delocalization of electrons. -NO2 the group is electron withdrawing hence, decreases electron density on the ring. Thus, the correct order for electrophilic substitution is 1>2>3.

Question 88. Increasing the order of electrophilic substitution for the following compounds

Hydrocarbons Incresing Order Of Rlectrophilic Substition

  1. 4 < 1 < 2 < 3
  2. 3 < 2 < 1 < 4
  3. 1 < 4 < 3 < 2
  4. 2 < 3 < 1 < 4

Answer: 1. 4 < 1 < 2 < 3

Due to -I effect of F atom, -CF3 on the benzene ring, deactivates the ring and does not favor electrophilic substitution. While – CH3 and – OCH3 are an electron-donating group that favors electrophilic substitution in the benzene ring at the ‘ortho’ and ‘para’ positions. The +I effect of -OCH3 is more than – CH3, therefore the correct order for electrophilic substitution is

Hydrocarbons Electrophilic

Question 89. In Friedel-Crafts reaction, toluene can be prepared by

  1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)
  2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{CH}_4\)
  3. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_2 \mathrm{Cl}_2\)
  4. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)

In the Friedel-Crafts reaction, toluene is obtained by the action of CH3Cl on benzene in the presence of AlCl3

Hydrocarbons Friedal Crafts Reaction Toluence

Question 90. In Friedel-Crafts alkylation, besides AlCl3 the other reactants are

  1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)
  2. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_4\)
  3. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{NH}_3\)
  4. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)

In Friedel-Crafts reaction, Toluene an alkyl group is introduced into the benzene ring in the presence of a Lewis is acid (AlCl3) catalyst. The reaction is

Hydrocarbons Lewis Acid catalyst

Question 91. Which of the following compounds will be most easily attacked by an electrophile?

Hydrocarbons Most Easily Attack By Electrophile

Answer: 1

-OH, -Cl, and -CH3 groups in benzene are ortho, para directing, groups. But among these – OH group is strongly activating while – CH3 is weakly activating and – Cl is deactivating. Thus, phenol will be most easily attacked by an electrophile.

Question 92. Which one of these is not compatible with arenes?

  1. Electrophilic additions
  2. Delocalisation of π-electrons
  3. Greater stability
  4. Resonance

Answer: 1. Electrophilic additions

Arenes undergo electrophilic substitution reactions and are resistant to addition reactions, due to delocalization of m-electrons. These are also stabilized by resonance.

Question 93. Among the following compounds (1-3) the correct order of reaction with electrophiles is

Hydrocarbons Electron Benzene

  1. 1>2>3
  2. 1=2>3
  3. 2>3>1
  4. 3<1<2

Answer: 1. 1>2>3

In structure 3, the withdrawal of electrons by -NO2 causes a decrease in reaction rate while in structure I, there is an electron-releasing effect by the – OCH3 group which accelerates the reaction.

The order of reactivity towards electrophile is 1>2>3

Question 94. Electrophile in the case of chlorination of benzene in the presence of FeCl3 is

  1. Cl
  2. FeCl3
  3. Cl+
  4. Cl

Answer: 3. Cl+

⇒ \(\mathrm{Cl}_2+\mathrm{FeCl}_3 \longrightarrow \mathrm{FeCl}_4^{-}+\mathrm{Cl}^{+}\)

Question 95. The reactive species in the nitration of benzene is

  1. \(\mathrm{NO}_3\)
  2. \(\mathrm{HNO}_3\)
  3. \(\mathrm{NO}_2^{+}\)
  4. \(\mathrm{NO}_2^{-}\)

Answer: 3. \(\mathrm{NO}_2^{+}\)

Nitronium ion (NO+2) is an electrophile that actually attacks the benzene ring.

Question 96. Which is the correct symbol relating to the two Kekule structures of benzene?

  1. \(\rightleftharpoons\)
  2. \(\longrightarrow\)
  3. \(\equiv\)
  4. \(\longleftrightarrow\)

Answer: 4. \(\longleftrightarrow\)

Benzene shows Kekule structures which are resonating structures and these structures are separated by a double-headed arrow (\(\leftrightarrow\)).

Question 97. Select the true statement about benzene amongst the following

  1. Because of unsaturation benzene easily undergoes addition
  2. There are two types of C – C bonds in the benzene molecule
  3. There is cyclic delocalization of π-electrons in benzene
  4. Monosubstitution of benzene gives three isomeric products.

Answer: 3. There is cyclic delocalization of π-electrons in benzene

Due to resonance all the C – C bonds in the benzene possess the same nature and the resonating structures are obtained because of the delocalization of π-electrons.

MCQs on Electrochemistry for NEET

NEET Chemistry For Electrochemistry Multiple Choice Questions

Question 1. The standard electrode potential (E°) values of Al3+/Al, Ag+/Ag, K+/K, and Cr3+/Cr are -1.66 V, 0.80 V, -2.93 V and -0.74 V, respectively. The correct decreasing order of reducing power of the metal is

  1. Ag > Cr > Al > K
  2. K > Al > Cr > Ag
  3. K > Al > Ag > Cr
  4. Al > K > Ag > Cr

Answer: 2. K > Al > Cr > Ag

Given

The standard electrode potential (E°) values of Al3+/Al, Ag+/Ag, K+/K, and Cr3+/Cr are -1.66 V, 0.80 V, -2.93 V and -0.74 V, respectively.

The higher the value of E°red, the stronger is the oxidizing power. Thus, the decreasing order of reducing the power of the metals K>AI>Cr>Ag.

Question 2. A button cell used in watches function as following: \(\mathrm{Zn}_{(s)}+\mathrm{Ag}_2 \mathrm{O}_{(s)}+\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons 2 \mathrm{Ag}_{(s)}+\mathrm{Zn}_{(a q)}^{2+}+2 \mathrm{OH}_{(a q)}^{-}\) If half cell potentials are \(\mathrm{Zn}^{2+}(a q)+2 e^{-} \rightarrow \mathrm{Zn}_{(s)} ; E^{\circ}=-0.76 \mathrm{~V}\); \(\mathrm{Ag}_2 \mathrm{O}_{(s)}+\mathrm{H}_2 \mathrm{O}_{(l)}+2 e^{-} \rightarrow 2 \mathrm{Ag}_{(s)}+2 \mathrm{OH}_{(a q)}^{-} ; E^{\circ}=0.34 \mathrm{~V}\)

The cell potential will be

  1. 0.84 V
  2. 1.34 V
  3. 1.10 V
  4. 0.42 V

Answer: 3. 1.10 V

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {O.P. }}^{\circ}+E_{\text {R.P. }}^{\circ}=0.76+0.34=1.10 \mathrm{~V}\)

Question 3. The standard reduction potentials of the half-reactions are given below:

  • \(\mathrm{F}_{2(g)}+2 e^{-} \rightarrow 2 \mathrm{~F}_{(a q)}^{-} ; E^{\circ}=+2.85 \mathrm{~V}\)
  • \(\mathrm{Cl}_{2(g)}+2 e^{-} \rightarrow 2 \mathrm{Cl}_{(a q)}^{-} ; E^{\circ}=+1.36 \mathrm{~V}\)
  • \(\mathrm{Br}_{2(l)}+2 e^{-} \rightarrow 2 \mathrm{Br}_{(a q)}^{-} ; E^{\circ}=+1.06 \mathrm{~V}\)
  • \(\mathrm{I}_{2(s)}+2 e^{-} \rightarrow 2 \mathrm{I}_{(a q)}^{-} ; E^{\circ}=+0.53 \mathrm{~V}\)

The strongest oxidizing and reducing agents respectively are

  1. F2 and I
  2. Br2 and Cl
  3. Cl2 and Br
  4. Cl2 and I2

Answer: 1. F2 and I

Less lower the value of reduction potential, the stronger the reducing agent thus, I am the strongest reducing agent. More positive, the value of reduction potential shows good oxidizing properties thus, the strongest oxidizing agent is F2.

Read and Learn More NEET MCQs with Answers

Question 4. Standard electrode potentials of three metals X, Y, and Z are -1.2 V, + 0.5 V, and – 3.0 V respectively. The reducing power of these metals will be

  1. Y>Z>X
  2. Y> X> Z
  3. Z>X>Y
  4. X> Y> Z

Answer: 3. Z>X>Y

The more negative the value of reduction potential, the stronger will be the reducing agent.

So, Z (-3.0 V) > X (-1.2 V) > Y (+ 0.5 V)

Question 5. The standard electrode potential for Sn4++/ Sn2+ couple is +0.15 V and that for the Cr3+/Cr couple is -0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be

  1. + 1.19 V
  2. + 0.89 V
  3. + 0.18 V
  4. + 1.83 V

Answer: 2. + 0.89 V

Given

The standard electrode potential for Sn4++/ Sn2+ couple is +0.15 V and that for the Cr3+/Cr couple is -0.74 V. These two couples in their standard state are connected to make a cell.

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}\) = 0.15-(-0.74)=0.15+0.74=0.89 V

Question 6. A solution contains Fe2+, Fe3+, and I ions. This solution was treated with iodine at 35°C. E° for Fe3+/Fe2+ is + 0.77 V and E° for I2/2I = 0.536 V. The favorable redox reaction is

  1. I2 will be reduced to I
  2. There will be no redox reaction
  3. I will be oxidized to I2
  4. Fe2+ will be oxidized to Fe3+.

Answer: 3. I will be oxidized to I2

Given

A solution contains Fe2+, Fe3+, and I ions. This solution was treated with iodine at 35°C. E° for Fe3+/Fe2+ is + 0.77 V and E° for I2/2I = 0.536 V.

Since the reduction potential of \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) is greater than that of \(\mathrm{I}_{2} / \mathrm{I}^{-}\) will be reduced and I will be oxidised.

⇒ \(2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\)

Question 7. Consider the following relations for emf of an electrochemical cell

  1. EMF of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
  2. EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
  3. EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)
  4. EMF of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)

Which of the above relations is correct?

  1. (3) and (1)
  2. (1) and (2)
  3. (3) and (4)
  4. (2) and (4)

Answer: 4. (2) and (4)

EMF of a cell = Reduction potential of cathode – Reduction potential of anode

= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of the anode Oxidation potential of the cathode.

Question 8. On the basis of the following E° values, the strongest oxidizing agent is \({\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}+e^{-} ; E^{\circ}=-0.35 \mathrm{~V}}\);  \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+e^{-} ; E^{\circ}=-0.77 \mathrm{~V}\)

  1. \(\mathrm{Fe}^{3+}\)
  2. \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\)
  3. \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\)
  4. \(\mathrm{Fe}^{2+}\)

Answer: 1. \(\mathrm{Fe}^{3+}\)

⇒ \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}, E^{\circ}\) = \(+0.35 \mathrm{~V}\); \(\mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}^{2+} ; E^{\circ}=+0.77 \mathrm{~V}\)

The higher the +ve reduction potential stronger the oxidising agent. Oxidizing agent oxidizes other compounds and gets themselves reduced easily. Thus, Fe3+ is the strongest oxidizing agent.

Question 9. A hypothetical electrochemical cell is shown below: A/A+ (x M)||B+ (y M)| B The emf measured is + 0.20 V. The cell reaction is

  1. \(A+B^{+} \rightarrow A^{+}+B\)
  2. \(A^{+}+B \rightarrow A+B^{+}\)
  3. \(A^{+}+e^{-} \rightarrow A ; B^{+}+e^{-} \rightarrow B\)
  4. The cell reaction cannot be predicted.

Answer: 1. \(A+B^{+} \rightarrow A^{+}+B\)

From the given expression:

At anode A → A+ + e (oxidation)

At cathode B+ + e → B (reduction)

Overall reaction is A + B+ → A+ + B

Question 10. \(E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.441 \mathrm{~V} \text { and } E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.771 \mathrm{~V}\), the standard EMF of the reaction \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+}\) will be

  1. 0.111 V
  2. 0.330 V
  3. 1.653 V
  4. 1.212 V

Answer: 4. 1.212 V

⇒ \(\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; E^{\circ}=-0.441 \mathrm{~V}\)…..(1)

⇒ \(\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} ; E^{\circ}=0.771 \mathrm{~V}\)….(2)

⇒ \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \longrightarrow 3 \mathrm{Fe}^{2+} ; E^{\circ}=?\)

To get the above equation, (2) x 2 – (1)

Electrochemistry EMF Of The Reaction And Equation

Electrochemistry Electrodes Oxidation

Question 11. Standard electrode potentials are Fe2+/Fe; E° = -0.44 and Fe3+/Fe2+; E° = 0.77. Fe2+, Fe3+, and Fe blocks are kept together, then

  1. Fe3+ increases
  2. Fe3+ decreases
  3. Fe2+/Fe3+ remains unchanged
  4. Fe2+ decreases.

Answer: 2. Fe3+ decreases

⇒ \(E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V}\)

⇒ \(E_{\mathrm{Fe}^{3+}}^{\circ} / \mathrm{Fe}^{2+}=+0.77 \mathrm{~V}\)

If a cell is constructed by combining these two electrodes oxidation occurs at Fe2+ /Fe electrode.

If Fe2+, Fe3+, and Fe blocks are kept together then Fe3+ reacts with Fe to yield Fe2+ i.e., the concentration of Fe3+ is decreased and that of Fe2+ is increased

Question 12. Electrode potential for the following half-cell reactions are

  • \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e^{-}; E^{\circ}=+0.76 \mathrm{~V}\)
  • \(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 e^{-} ; E^{\circ}=+0.44 \mathrm{~V}\)

The EMF for the cell reaction \(\mathrm{Fe}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}\) will be

  1. -0.32 V
  2. + 1.20 V
  3. -1.20 V
  4. + 0.32 V

Answer: 4. + 0.32 V

⇒ \(E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{\circ}=+0.76 \mathrm{~V}\)

⇒ \(E_{\mathrm{Fe} / \mathrm{Fe}}{ }^{2+}=0.44 \mathrm{~V} \Rightarrow E_{\mathrm{Fe}}^{\circ}{ }^{2+} / \mathrm{Re}=-0.44 \mathrm{~V}\)

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {O.P. }}^{\circ}+E_{\text {R.P. }}^{\circ}=+0.76-0.44=+0.32 \mathrm{~V}\)

Question 13. An electrochemical cell is set up of: Pt;H2 (1 atm) | HCl(0.1M) || CH3COOH(0.1M) |H2( 1 atm); Pt. The e.m.f. of this cell will not be zero, because

  1. Acids used in two compartments are different
  2. e.m.f. depends on the molarities of acids used
  3. The temperature is constant
  4. pH of 0.1 M HCl and 0.1 M CH3COOH is not the same.

Answer: 4. pH of 0.1 M HCl and 0.1 M CH3COOH is not the same

Since it is a concentration cell and the concentration of H+ ions in two electrolyte solutions (HCl and CH33COOH) are different i.e., pH of 0.1 M HCl and 0.1 M CH3COOH is not the same, therefore e.m.f. of this cell will not be zero

Question 14. Standard reduction potentials at 25°C of Li+|Li, Ba2+|Ba, Na+|Na, and Mg2+|Mg are -3.05, -2.90, -2.71, and -2.37 volt respectively. Which one of the following is the strongest oxidizing agent?

  1. Ba2+
  2. Mg2+
  3. Na+
  4. Li+

Answer: 2. Mg2+

Given

Standard reduction potentials at 25°C of Li+|Li, Ba2+|Ba, Na+|Na, and Mg2+|Mg are -3.05, -2.90, -2.71, and -2.37 volt respectively.

The more positive or less negative the reduction potential value, the stronger is the oxidizing agent.

Question 15. A solution of potassium bromide is treated with each of the following. Which one would liberate bromine?

  1. Hydrogen iodide
  2. Sulfur dioxide
  3. Chlorine
  4. Iodine

Answer: 3. Chlorine

A stronger oxidizing agent (Cl2) displaces a weaker oxidizing agent (Br2) from its salt solution.

⇒ \(2 \mathrm{KBr}+\mathrm{Cl}_2 \rightarrow 2 \mathrm{KCl}+\mathrm{Br}_2\)

Question 16. Given below are two statements: One is labelled as Assertion A and the other is labeled as Reason R;

Assertion A: In equation ΔrG = -nFEcell, the value of ΔrG depends on n.

Reason R: cell is an intensive property and ArG is an extensive property.

In light of the above statements, choose the correct answer from the options given below.

  1. A is true but R is false.
  2. A is false but R is true.
  3. Both A and R are true and R is the correct explanation of A.
  4. Both A and R are true and R is not the correct explanation of A.

Answer: 4. Both A and R are true and R is not the correct explanation of A.

ΔrG =-nFEcell

Ecell, is an intensive parameter but ΔrG is an extensive thermodynamic property and the value of ΔrG depends on n.

Question 17. Given below are half-cell reactions: \(\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\); \(E_{\mathrm{Mn}^{2+} / \mathrm{MnO}_4^{-}}^{\circ}=-1.510 \mathrm{~V}\); \(\frac{1}{2} \mathrm{O}_2+2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}; E_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}=+1.223 \mathrm{~V}\) Will the permanganate ion, MnO4 water in the presence of an acid?

  1. Yes, because E°cell= +0.287 V
  2. No, because E°cell = -0.287 v
  3. Yes, because E°cell = +2.733 V
  4. No, because E°cell = -2.733 V

Answer: 1. Yes, because E°cell= +0.287 V

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=+1.510-1.223=+0.287 \mathrm{~V}\)

As \(E_{\text {cell }}^{\circ}\) is positive, hence the reaction is feasible.

Question 18. At 298 K the standard electrode potentials of Cu2+ / Cu, Zn2+/ Zn, Fe2+/ Fe, and Ag+/Ag are 0.34 V, -0.76 V, -0.44 V, and 0.80 V respectively. On the basis of standard electrode potential, predict which of the following reactions cannot occur.

  1. \(\mathrm{CuSO}_{4(a q)}+\mathrm{Zn}_{(s)} \longrightarrow \mathrm{ZnSO}_{4(a q)}+\mathrm{Cu}_{(s)}\)
  2. \(\mathrm{CuSO}_{4(a q)}+\mathrm{Fe}_{(s)} \longrightarrow \mathrm{FeSO}_{4(a q)}+\mathrm{Cu}_{(s)}\)
  3. \(\mathrm{FeSO}_{4(a q)}^{4(a q)}+\mathrm{Zn}_{(s)} \longrightarrow \mathrm{ZnSO}_{4(a q)}+\mathrm{Fe}_{(s)}\)
  4. \(2 \mathrm{CuSO}_{4(a q)}+2 \mathrm{Ag} g_{(s)} \longrightarrow 2 \mathrm{Cu}_{(s)}+\mathrm{Ag}_2 \mathrm{SO}_{4(a q)}\)

Answer: 4. \(2 \mathrm{CuSO}_{4(a q)}+2 \mathrm{Ag} g_{(s)} \longrightarrow 2 \mathrm{Cu}_{(s)}+\mathrm{Ag}_2 \mathrm{SO}_{4(a q)}\)

The values of the standard reduction potential of Cu and Ag suggest that Cu would undergo oxidation (lower reduction potential) and Ag would undergo reduction (higher reduction potential).

Hence, the cell reaction will be \(\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}\)

Question 19. Find the emf of the cell in which the following reaction takes place at \(298 K \mathrm{Ni}_{(s)}+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \longrightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}_{(s)}\)

Given that \(E_{\text {cell }}^{\circ}=10.5 \mathrm{~V}, \frac{2.303 R T}{F}=0.059 \text { at } 298 \mathrm{~K}\)

  1. 1.0385 V
  2. 1.385 V
  3. 0.9615 V
  4. 1.05 V

Answer: 3. 0.9615 V

According to the Nernst equation,

E = \(E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)

⇒ \(E_{\text {cell }}^{\circ}=1.05(\text { Given) }\)

E = \(1.05-\frac{0.059}{2} \log \frac{(0.001)}{(0.001)^2}\)

= \(1.05-\frac{0.059}{2} \log 10^3=1.05-\frac{0.059 \times 3}{2}\)

= \(1.05-0.0885=0.9615 \mathrm{~V}\)

Question 20. For the cell reaction: \(2 \mathrm{Fe}_{(a q)}^{3+}+2 \mathrm{I}_{(a q)}^{-} \rightarrow 2 \mathrm{Fe}_{(a q)}^{2+}+\mathrm{I}_{2(a q)}\) E°cell = 0.24 V at 298 K. The standard Gibbs’ energy (ΔrG°) of the cell reaction is [Given that Faraday constant, F = 96500 C mol-1]

  1. 23.16 kJ mol-1
  2. -46.32 kJ mol-1
  3. -23.16 kJ mol-1
  4. 46.32 kJ mol-1

Answer: 2. -46.32 kJ mol-1

The standard Gibbs energy, \(\left(\Delta G^{\circ}\right)=-n F E_{\text {cell }}^{\circ}\) Value of n=2

⇒ \(\Delta G^{\circ}=-2 \times 96500 \times 0.24=-46320 \mathrm{~J}\)

= \(-46.32 \mathrm{~kJ} / \mathrm{mol}\)

Question 21. For a cell involving one electron, E°cell= 0.59 V at 298 K, the equilibrium constant for the cell reaction is [Given that \(\frac{2.303 R T}{F}\)= 0.059 V at T = 298 K]

  1. 1.0 x 1030
  2. 1.0 x 102
  3. 1.0 x 105
  4. 1.0 x 1010

Answer: 4. 1.0 x 1010

According to Nernst equation, \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log Q_c\)

At equilibrium \(E_{\text {cell }}=0\), \(Q_c=K_c\)

⇒ \(E_{\text {cell }}^{\mathrm{a}}=\frac{0.059}{n} \log K_c \Rightarrow 0.59=\frac{0.059}{1} \log K_c\)

⇒ \(K_c=\text { antilog } 10 \Rightarrow K_c=1 \times 10^{10}\)

Question 22. In the electrochemical cell: Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu, the emf of this Daniell cell is E1 When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E1 From the followings, which one is the relationship between E1 and E2? (Given, RT/F = 0.059)

  1. E1 < E2
  2. E1 > E2
  3. E2 = 0 ≠ E1
  4. E1 = E2

Answer: 2. E1 > E2

Given

In the electrochemical cell: Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu, the emf of this Daniell cell is E1 When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E1

⇒ \(E_{\text {cell }}=E_{\text {cell }}^o-\frac{0.059}{n} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)

⇒ \(E_1=E^{\circ}-\frac{0.059}{2} \log \frac{0.01}{1}\)

⇒ \(E_1=E^{\circ}-\frac{0.059}{2}(-2)=E^{\circ}+0.059\)

⇒ \(E_2=E^{\circ}-\frac{0.059}{2} \log \frac{1}{0.01}=E^{\circ}-0.059\)

Hence, \(E_1>E_2\).

Question 23. If the E°cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG° and Keq?

  1. \(\Delta G^{\circ}>0 ; K_{\text {eq }}<1\)
  2. \(\Delta G^{\circ}>0 ; K_{\text {eq }}>1\)
  3. \(\Delta G^0<0 ; K_{\text {eq }}>1\)
  4. \(\Delta G^{\circ}<0 ; K_{\text {eq }}<1\)

Answer: 1. \(\Delta G^{\circ}>0 ; K_{\text {eq }}<1\)

⇒ \(\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}\)

If \(E_{\text {cell }}^{\circ}=- ve then \Delta G^{\circ}=+ ve\)

i.e; \(\Delta G^{\circ}>0\).

⇒ \(\Delta G^{\mathrm{a}}=-n R T \log K_{\mathrm{eq}}\)

For \(\Delta G^{\circ}=+ \text { ve, } K_{\text {eq }}=- \text { ve i.e., } K_{e q}<1 \text {. }\)

Question 24. The pressure of H2 required to make the potential of H2 electrode zero in pure water at 298 K is

  1. 10-10 atm
  2. 10-4 atm
  3. 10-14 atm
  4. 10-12 atm.

Answer: 3. 10-14 atm

pH = 7 for water.

⇒ \(-\log \left[\mathrm{H}^{+}\right]=7 \Rightarrow\left[\mathrm{H}^{+}\right]=10^{-7}\)

⇒ \(2 \mathrm{H}^{+}{ }_{(a q)}^{+}+2 e^{-} \longrightarrow \mathrm{H}_{2(g)}\)

⇒ \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{P_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\)

0 = \(0-\frac{0.0591}{2} \log \frac{P_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}\)

⇒ \(\log \frac{P_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}=0 \Rightarrow \frac{P_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}=1\) (because log 1=0)

⇒ \(p_{\mathrm{H}_2}=10^{-14} \mathrm{~atm}\)

Question 25. A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH =10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of the electrode would be

  1. 0.118 V
  2. 1.18 V
  3. 0.059 V
  4. 0.59 V

Answer: 4. 0.59 V

Given

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH =10 and by passing hydrogen gas around the platinum wire at one atm pressure.

⇒ \(\underset{1 \mathrm{~atm}}{\mathrm{H}_2} \longrightarrow \underset{10^{-10}}{2 \mathrm{H}^{+}}+2 e^{-}\)

⇒ \(E_{\mathrm{H}_2 / \mathrm{H}^{+}}=0-\frac{0.059}{2} \log \frac{\left(10^{-10}\right)^2}{1}\)

⇒ \(E_{\mathrm{H}_2 / \mathrm{H}^{+}}=+0.59 \mathrm{~V}\)

Question 26. Consider the half-cell reduction reaction \(\mathrm{Mn}^{2+}+2 e^{-} \rightarrow \mathrm{Mn}, E^{\circ}=-1.18 \mathrm{~V}\); \(\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+}+e^{-}, E^{\circ}=-1.51 \mathrm{~V}\) The \(E^{\circ}\) for the reaction, \(3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^0+2 \mathrm{Mn}^{3+}\) and possibility of the forward reaction are respectively

  1. -4.18 V and yes
  2. + 0.33 V and yes
  3. + 2.69 V and no
  4. – 2.69 V and no.

Answer: 4. – 2.69 V and no

⇒ \(\mathrm{Mn}^{2+}+2 e^{-} \longrightarrow \mathrm{Mn} ; E^{\circ}=-1.18 \mathrm{~V}\)

⇒ \(2 \mathrm{Mn}^{2+} \longrightarrow 2 \mathrm{Mn}^{3+}+2 e^{-} ; E^{\circ}=-1.51 \mathrm{~V}\)

By adding equations (1) and (2), we get an equation for the cell, \(3 \mathrm{Mn}^{2+} \longrightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}; E^{\circ}=-2.69 \mathrm{~V}\)

Since the E value is negative, so the process is non-spontaneous as ΔG° is positive.

Question 27. The Gibbs’ energy for the decomposition of Al2O3 at 500 °C is as follows \(\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_{\mathrm{r}} G=+960 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The potential difference needed for the electrolytic reduction of aluminum oxide (Al2O3) at 500 °C is at least

  1. 4.5 V
  2. 3.0 V
  3. 2.5 V
  4. 5.0 V

Answer: 3. 2.5 V

The Gibbs’ energy for the decomposition of Al2O3 at 500 °C is as follows \(\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_{\mathrm{r}} G=+960 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\Delta G^{\circ}=-n F E^{\circ}\)

F = \(96500, \Delta G^{\circ}=+960 \times 10^3 \mathrm{~J} / \mathrm{mol}\)

⇒ \(\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2\)

Total number of \(\mathrm{Al}\) atoms in \(\mathrm{Al}_2 \mathrm{O}_3=\frac{2}{3} \times 2=\frac{4}{3}\)

⇒ \(\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}\)

As \(3 e^{-}\) change occur for each Al-atom

total n = \(\frac{4}{3} \times 3=4\)

⇒ \(E^{\circ}=-\frac{\Delta G^{\circ}}{n F}=-\frac{960 \times 1000}{4 \times 96500} \Rightarrow E^{\circ}=-2.48 \approx-2.5 \mathrm{~V}\)

Question 28. The electrode potentials for, \(\mathrm{Cu}^{2+}{ }_{(a q)}+e^{-} \rightarrow \mathrm{Cu}_{(a q)}^{+} \text {and } \mathrm{Cu}^{+}{ }_{(a q)}+e^{-} \rightarrow \mathrm{Cu}_{(s)}\) are + 0.15 V and + 0.50 V respectively. The value of \(E^{\circ}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}\) will be

  1. 0.500 V
  2. 0.325 V
  3. 0.650 V
  4. 0.150 V

Answer: 2. 0.325 V

Given

The electrode potentials for, \(\mathrm{Cu}^{2+}{ }_{(a q)}+e^{-} \rightarrow \mathrm{Cu}_{(a q)}^{+} \text {and } \mathrm{Cu}^{+}{ }_{(a q)}+e^{-} \rightarrow \mathrm{Cu}_{(s)}\) are + 0.15 V and + 0.50 V respectively.

⇒ \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+e^{–} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{+} ; E_1^0=0.15 \mathrm{~V}\)

⇒ \(\mathrm{Cu}^{+}{ }_{(\mathrm{aq})}+e^{-} \longrightarrow \mathrm{Cu}_{(s)} ; E_2{ }^{\circ}=0.50 \mathrm{~V}\)

⇒ \(\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E^{\circ}=?\)

Now, \(\Delta G^a=\Delta G_1{ }^{\circ}+\Delta G_2{ }^{\circ}\)

or, \(-n F E^{\circ}=-n_1 F E_1{ }^{\circ}-n_2 F E_2{ }^{\circ}\)

or, \(E^{\circ}=\frac{n_1 E_1^{\circ}+n_2 E_2^{\circ}}{n}=\frac{1 \times 0.15+1 \times 0.50}{2}=0.325 \mathrm{~V}\)

Question 29. For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25 °C. The value of standard Gibbs energy, ΔG° will be (F = 96500 C mol-1)

  1. – 89.0 kJ
  2. – 89.0 J
  3. -44.5 kJ
  4. – 98.0 kJ

Answer: 1. – 89.0 kJ

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25 °C.

The cell reaction can be written as \(\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}\)

We know, \(\Delta G^{\circ}=-n F E^{\circ}{ }_{\text {cell }}\)

= – 2 x 96500 x 0.46 = – 88780 J

= – 88.78 kJ ≈ -89 kJ

Question 30. Given:

  1. \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}, E^{\circ}=0.337 \mathrm{~V}\)
  2. \(\mathrm{Cu}^{2+}+e^{-} \rightarrow \mathrm{Cu}^{+}, E^{\circ}=0.153 \mathrm{~V}\)

Electrode potential, E° for the reaction, \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}\), will be

  1. 0.90 V
  2. 0.30 V
  3. 0.38 V
  4. 0.52 V

Answer: 4. 0.52 V

Given, \(\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E_1^0=0.337 \mathrm{~V}\); \(\mathrm{Cu}^{2+}+e^{-} \longrightarrow \mathrm{Cu}^{+} ; E_2^{\circ}=0.153 \mathrm{~V}\)

The required reaction is \(\mathrm{Cu}^{+}+e^{-} \longrightarrow \mathrm{Cu} ; E_3^{\circ}=?\)

Applying, \(\Delta G^{\circ}=-n F E^{\circ}, \Delta G_3^{\circ}=\Delta G_1^{\circ}-\Delta G_2^{\circ}\)

⇒ \(-\left(n_3 F E_3^{\circ}\right)=-\left(n_1 F E_1^{\circ}\right)-\left(-n_2 F E_2^{\circ}\right)\)

or \(E_3^{\circ}=2 \times E_1^{\circ}-E_2^{\circ}\)

or \(E_3^{\circ}=(2 \times 0.337)-0.153=0.52 \mathrm{~V}\)

Question 31. Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4, and -8.2 for H2O(l), CO2(g), and pentane(g), respectively. The value of £°cell for the pentane-oxygen fuel cell is

  1. 1.0968 V
  2. 0.0968 V
  3. 1.968 V
  4. 2.0968 V

Answer: 1. 1.0968 V

Given

Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4, and -8.2 for H2O(l), CO2(g), and pentane(g), respectively.

⇒ \(\mathrm{C}_5 \mathrm{H}_{12(g)}+8 \mathrm{O}_{2(g)} \longrightarrow 5 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(l)}\)

⇒ \(\Delta G^{\circ}=[(-394.4 \times 5)+(-237.2 \times 6)]-[(-8.2)+(8 \times 0)]\) =-3387 kJ

Note that the standard free energy change of elementary substances is taken as zero.

For the fuel cell, the complete cell reaction is: \(\mathrm{C}_5 \mathrm{H}_{12(\mathrm{~g})}+8 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 5 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(0)}\)

which is the combination of the following two half-reactions: \(\mathrm{C}_5 \mathrm{H}_{12(g)}+10 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \longrightarrow 5 \mathrm{CO}_{2(g)}+32 \mathrm{H}^{+}+32 e^{-}\)

and \(8 \mathrm{O}_{2(\mathrm{~g})}+32 \mathrm{H}^{+}+32 e^{-} \longrightarrow 16 \mathrm{H}_2 \mathrm{O}_{(l)}\)

Therefore, the number of electrons exchanged is 32 here, i.e., n=32.

⇒ \(\Delta G^{\circ}=-n F E^{\circ}=-3387 \times 10^3 \mathrm{~J}\)

= \(-32 \times 96500 \mathrm{~J} / \text { Volt } \times E^0\)

Thus, \(E^{\circ}=1.0968 \mathrm{~V}\)

Question 32. The equilibrium constant of the reaction: \(\mathrm{Cu}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \rightarrow \mathrm{Cu}^{2+}{ }_{(a q)}+2 \mathrm{Ag}_{(s)}\); \(E^{\circ}=0.46 \mathrm{~V}\) at \(298 \mathrm{~K}\) is

  1. \(2.0 \times 10^{10}\)
  2. \(4.0 \times 10^{10}\)
  3. \(4.0 \times 10^{15}\)
  4. \(2.4 \times 10^{10}\)

Answer: 3. \(4.0 \times 10^{15}\)

For a cell reaction in equilibrium at 298 K, \(E_{\text {cell }}^{\circ}=\frac{0.0591}{n} \log K_c\)

where, \(K_c=\) equilibrium constant, n= number of electrons involved in the electrochemical cell reaction.

Given, \(E_{\text {cell }}^0=0.46 \mathrm{~V}, n=2\)

0.46 = \(\frac{0.0591}{2} \times \log K_c or, \log K_c=\frac{2 \times 0.46}{0.0591}=15.57\)

or, \(K_c=3.7 \times 10^{15} \approx 4 \times 10^{15}\)

Question 33. The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C. The equilibrium constant of the reaction would be

  1. 2.0 x 1011
  2. 4.0 x 1012
  3. 1.0 x 102
  4. 1.0 x 1010

(Given F = 96500 C mol-1, R = 8.314 J K-1 mol-1)

Answer: 4. 1.0 x 1010

E = \(E^{\circ}-\frac{0,0591}{n} \log _{10} \mathrm{Q}\) at \(25^{\circ} \mathrm{C}\)

At equilibrium, E=0, Q=K

0 = \(E^{\circ}-\frac{0.0591}{n} \log _{10} K\)

or, \(K=\mathrm{antilog}\left[\frac{n E^{\circ}}{0.0591}\right]\)

K = \(\mathrm{antilog}\left[\frac{2 \times 0.295}{0.0591}\right]=\mathrm{antilog}\left[\frac{0.590}{0.0591}\right]\)

= \(\mathrm{antilog} 10=1 \times 10^{10}\)

Question 34. On the basis of the information available from the reaction, \(4 / 3 \mathrm{Al}+\mathrm{O}_2 \rightarrow 2 / 3 \mathrm{Al}_2 \mathrm{O}_3, \Delta G=-827 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { of } \mathrm{O}_2\), the minimum e.m.f. required to carry out an electrolysis of Al2O3 is (F = 96500 C mol-1)

  1. 2.14 V
  2. 4.28 V
  3. 6.42 V
  4. 8.56 V

Answer: 1. 2.14 V

⇒ \(\Delta G^{\circ}=-n F E^{\circ}\)

⇒ \(E^{\circ}=\frac{\Delta G^{\circ}}{-n F}=\frac{-827000}{-4 \times 96500}=2.14 \mathrm{~V}\)

⇒ \(\left(1 \mathrm{Al} \equiv 3 e^{-}, \frac{4}{3} \mathrm{Al}=\frac{4}{3} \times 3 e^{-}=4 e^{-}\right)\)

Question 35. For the disproportionation of copper \(2 \mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}, E^{\circ}\) is (Given : \(E^{\circ}\) for \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) is \(0.34 \mathrm{~V}\) and \(E^{\circ}\) for \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\) is \(0.15 \mathrm{~V}\)

  1. 0.49 V
  2. -0.19 V
  3. 0.38 V
  4. -0.38 V

Answer: 3. 0.38 V

For the reaction, \(2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}\) the cathode is \(\mathrm{Cu}^{+} / \mathrm{Cu}\) and anode is \(\mathrm{Cu}^{+} / \mathrm{Cu}^{2+}\).

Given, \(\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}_2 E_1^{\circ}=0.34 \mathrm{~V}\)…..(1)

\(\mathrm{Cu}^{2+}+e^{-} \longrightarrow \mathrm{Cu}^{+} ; E_2^{\circ}=0.15 \mathrm{~V}\)….(2)

\(\mathrm{Cu}^{+}+e^{-} \longrightarrow \mathrm{Cu} ; E_3^{\circ}=?\)….(3)

Now, \(\Delta G_1^{\circ}=-n F E_1^{\circ}=-2 \times 0.34 \times F=-0.68 F\)

⇒ \(\Delta G_2^{\circ}=-1 \times 0.15 \times F, \Delta G_3^{\circ}=-1 \times E_3^{\mathrm{a}} \times F\)

Again \(\Delta G_1^{\circ}=\Delta G_2^{\circ}+\Delta G_3^{\circ}\)

-0.68 F = \(-0.15 F-E_3^0 \times F\)

⇒ \(E_3^{\circ}=0.68-0.15=0.53 \mathrm{~V}\)

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {cathode }\left(\mathrm{Cu}^{+} / \mathrm{Cu}\right)}^{\circ}-E_{\text {anode }\left(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\right)}^{\circ}\)

= \(0.53-0.15=0.38 \mathrm{~V}\)

Question 36. E° for the cell, \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}{ }_{(a q)}\right|\left|\mathrm{Cu}^{2+}{ }_{(a q)}\right| \mathrm{Cu}\) is \(1.10 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Zn}+\mathrm{Cu}_{(\text {aq) }}^{2+} \rightarrow \mathrm{Cu}+\mathrm{Zn}_{(\text {aq })}^{2+}\) is of the order

  1. \(10^{+18}\)
  2. \(10^{+17}\)
  3. \(10^{-28}\)
  4. \(10^{+37}\)

Answer: 4. \(10^{+37}\)

Nernst equation is E = \(E^{\circ}-\frac{0.059}{2} \log K\)

⇒ \(E^{\circ}=\frac{0.059}{2} \log K\) (E=0 at equilibrium condition)

⇒ 1.1 = \(\frac{0.059}{2} \log K \Rightarrow K=1.9 \times 10^{+37}\)

Question 37. The conductivity of a centriolar solution of KCI at 25°C is 0.0210 ohm-1 cm-1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of the cell constant is

  1. 1.26 cm-1
  2. 3.34 cm-1
  3. 1.34 cm-1
  4. 3.28 cm-1

Answer: 1. 1.26 cm-1

The conductivity of a centriolar solution of KCI at 25°C is 0.0210 ohm-1 cm-1 and the resistance of the cell containing the solution at 25°C is 60 ohm.

Given, \(\mathrm{K}=0.0210 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}, R=60 \mathrm{ohm}\)

k = \(\frac{1}{R} \times \frac{l}{a}\) (where \(\frac{l}{a}=\) cell constant) or \(0.0210=\frac{1}{60} \times \frac{l}{a}\)

or \(\frac{l}{a}=60 \times 0.0210=1.26 \mathrm{~cm}^{-1}\)

Question 38. The molar conductance of NaCl, HCl, and CH3COONa at infinite dilution are 126.45, 426.16, and 91.0 S cm2 mol-1 respectively. The molar conductance of CH3COOH at infinite dilution is.

Choose the right option for your answer.

  1. 540.48 S cm2 mol-1
  2. 201.28 S cm2 mol-1
  3. 390.71 S cm2 mol-1
  4. 698.28 S cm2 mol-1

Answer: 3. 390.71 S cm2 mol-1

The molar conductance of NaCl, HCl, and CH3COONa at infinite dilution are 126.45, 426.16, and 91.0 S cm2 mol-1 respectively.

⇒ \(\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}=\Lambda_{\mathrm{CH}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{NaCl}}^{\circ}\)

= \(91.0+426.16-126.45\)

= \(390.71 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)

Question 39. The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol-1. What is the dissociation constant of acetic acid? Choose the correct option.

⇒ \(\left[\begin{array}{c}
\Lambda_{\mathrm{H}^{+}}^{\circ}=350 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
\Lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}=50 \mathrm{Scm}^2 \mathrm{~mol}^{-1}
\end{array}\right]\)

  1. \(2.50 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
  2. \(1.75 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
  3. \(2.50 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
  4. \(1.75 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

Answer: 4. \(1.75 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol-1.

⇒ \(\Lambda_{m\left(\mathrm{CH}_3 \mathrm{COOH}\right)}^{\circ}=\lambda_{\mathrm{H}^{+}}^{\circ}+\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}\)

= \(350+50=400 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Degree of dissociation, \(\alpha=\frac{\Lambda_m^c}{\Lambda_m^{\circ}}=\frac{20}{400}=0.05\)

So, dissociation constant, \(K_a=c \alpha^2\) (for weak electrolytes)

= \(0.007(0.05)^2=1.75 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

 

MCQs of Redox Reactions for NEET

NEET Chemistry For Redox Reactions Multiple Choice Questions

Question 1. Which of the following is a redox reaction?

  1. Evaporation of H2O
  2. Both oxidation and reduction
  3. H2SO4 with NaOH
  4. In the atmosphere O3 from O2 by lightning

Answer: 2. Both oxidation and reduction

Redox reactions are those chemical reactions which involve both oxidation and reduction simultaneously.

Question 2. Without losing its concentration, ZnCl2 solution cannot be kept in contact with

  1. Pb
  2. Al
  3. Au
  4. Ag

Answer: 2. Al

Only ‘AI’ lies above ‘Zn’ in the electrochemical series, which can displace Zn from the ZnCl2 solution. Therefore, conc. of ZnCl2 will decrease when kept in an ‘Al’ container.

⇒ \(2 \mathrm{Al}+3 \mathrm{ZnCl}_2 \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{Zn}\)

Question 3. On balancing the given redox reaction, \(a \mathrm{Cr}_2 \mathrm{O}_7^{2-}+b \mathrm{SO}_{3(\mathrm{mq})}^{2-}+c \mathrm{H}_{(\mathrm{mq})}^{+} \rightarrow 2 a \mathrm{Cr}_{[(a q)}^{3+}+b \mathrm{SO}_{4(\mathrm{aq})}^{2-}+\frac{c}{2} \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\) the coefficient a, b and c are found to be respectively.

  1. 1,8,3
  2. 8,1,3
  3. 1,3,8
  4. 3,8, 1

Answer: 3. 1,3,8

⇒ \(\mathrm{Cr}_2 \mathrm{O}_{7(a q)}^{2-}+3 \mathrm{SO}_{3(a q)}^{2-}+8 \mathrm{H}_{(a q)}^{+} \longrightarrow 2 \mathrm{Cr}_{(a q)}^{3+}+3 \mathrm{SO}_{4(a q)}^{2-}+4 \mathrm{H}_2 \mathrm{O}_{(l)}\)

Question 4. Which of the following reactions is the metal displacement reaction? Choose the right option.

  1. \(2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow 2 \mathrm{PbO}+4 \mathrm{NO}_2+\mathrm{O}_2 \uparrow\)
  2. \(2 \mathrm{KClO}_3\)\(\longrightarrow{\Delta}\)\(2 \mathrm{KCl}+3 \mathrm{O}_2\)
  3. \(\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al}\)\(\longrightarrow{\Delta}\)\(\mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\)
  4. \(\mathrm{Fe}+2 \mathrm{HCl} \longrightarrow \mathrm{FeCl}_2+\mathrm{H}_2 \uparrow\)

Answer: 3. \(\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al}\)\(\longrightarrow{\Delta}\)\(\mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\)

When a metal from the electrochemical series is mixed with the ions of a metal lower down in the electrochemical series, then more active metal displaces the less active one, this is known as metal displacement.

Hence, the correct reaction is \(\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al}\)\(\longrightarrow{\Delta}\)\(\mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\)

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Question 5. What is the change in the oxidation number of carbon in the following reaction? \(\mathrm{CH}_{4(\mathrm{~g})}+4 \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{CCl}_{4(l)}+4 \mathrm{HCl}_{(\mathrm{g})}\)

  1. +4 to+4
  2. 0 to+4
  3. -4 to+4
  4. 0 to-4

Answer: 3. -4 to+4

In CH4 the oxidation number of carbon is -4 while in CCl4, the oxidation number of carbon is +4. Thus, the change in oxidation number of carbon in the given reaction is from -4 to +4.

Question 6. The correct structure of tribromooctaoxide is Which of the following reactions are disproportionation reactions?

Redox Reactions Tribromooctaoxide

Answer: 2

Question 7. Which of the following reactions are disproportionation reactions?

  1. \(2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}^0\)
  2. \(3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}\)
  3. \(2 \mathrm{KMnO}_4\) \(\longrightarrow{\Delta}\)\(\mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2\)
  4. \(2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}\)

Select the correct option from the following.

  1. (1) and (4) only
  2. (1) and (2) only
  3. (1), (2) and (3)
  4. (1), (3) and (4)

Answer: 2. (1) and (2) only

Disproportionation reactions are those in which the same element/compound gets oxidised and reduced simultaneously.

⇒ \(2 \mathrm{Cu}^{+}\)\(\longrightarrow\)\(\mathrm{Cu}^{2+}+\mathrm{Cu}^0\)

⇒ \(3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \stackrel{+7}{\mathrm{MnO}_4^{-}}+\stackrel{+4}{\mathrm{MnO}_2}+2 \mathrm{H}_2 \mathrm{O}\)

Question 8. The oxidation state of Cr in CrO5 is

  1. -6
  2. +12
  3. +6
  4. +4

Answer: 3. +6

CrO5 has a butterfly structure having two peroxo bonds.

Peroxo oxygen has -1 oxidation state.

Redox Reactions Butterfly structure

Let the oxidation state of Cr be ‘x’

CrO5: x+4(-1)+1(-2) =0 + x=+6

Question 9. The correct order of N-compounds in its decreasing order of oxidation states is

  1. \(\mathrm{HNO}_3, \mathrm{NO}, \mathrm{N}_2, \mathrm{NH}_4 \mathrm{Cl}\)
  2. \(\mathrm{HNO}_3, \mathrm{NO}, \mathrm{NH}_4 \mathrm{Cl}, \mathrm{N}_2\)
  3. \(\mathrm{HNO}_3, \mathrm{NH}_4 \mathrm{Cl}, \mathrm{NO}, \mathrm{N}_2\)
  4. \(\mathrm{NH}_4 \mathrm{Cl}, \mathrm{N}_2, \mathrm{NO}, \mathrm{HNO}_3\)

Answer: 1. \(\mathrm{HNO}_3, \mathrm{NO}, \mathrm{N}_2, \mathrm{NH}_4 \mathrm{Cl}\)

⇒ \(\stackrel{+5}{\mathrm{HNO}_3,} \stackrel{+2}{\mathrm{NO}}, \stackrel{0}{\mathrm{~N}}, \stackrel{-3}{\mathrm{NH}_4 \mathrm{Cl}}\)

Question 10. For the redox reaction, \(\mathrm{MnO}_4^{-}+\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\). The correct coefficients of the reactants for the balanced equation are

Redox Reactions Coefficient Of The Reactants For The Balanced Equation

Answer: 2

For the redox reaction, \(\mathrm{MnO}_4^{-}+\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\).

The correct balanced equation is \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Question 11. Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour?

  1. \(\mathrm{Cu}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)
  2. \(\mathrm{S}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 3 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)
  3. \(\mathrm{C}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CO}_2+2 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)
  4. \(\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}(\mathrm{NE}\)

Answer: 4. \(\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}(\mathrm{NE}\)

⇒ \(\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}\)

Here, the oxidation state of every atom remains the same so, it is not a redox reaction.

Question 12.

  1. \(\mathrm{H}_2 \mathrm{O}_2+\mathrm{O}_3 \longrightarrow \mathrm{H}_2 \mathrm{O}+2 \mathrm{O}_2\)
  2. \(\mathrm{H}_2 \mathrm{O}_2+\mathrm{Ag}_2 \mathrm{O} \longrightarrow 2 \mathrm{Ag}+\mathrm{H}_2 \mathrm{O}+\mathrm{O}_2\)

The role of hydrogen peroxide in the above reactions is respectively

  1. Oxidizing in (1) and reducing in (2)
  2. Reducing in (1) and oxidizing in (2)
  3. Reducing in (1) and (2)
  4. Oxidizing in (1) and (2)

Answer: 3. Reducing in (1) and (2)

Redox Reactions Reducting The Both Reactions

H2O2 acts as a reducing agent in both the reactions in which O2 is evolved

Question 13. The pair of compounds that can exist together is

  1. \(\mathrm{FeCl}_3, \mathrm{SnCl}_2\)
  2. \(\mathrm{HgCl}_2, \mathrm{SnCl}_2\)
  3. \(\mathrm{FeCl}_2, \mathrm{SnCl}_2\)
  4. \(\mathrm{FeCl}_3, \mathrm{KI}\)

Answer: 3. \(\mathrm{FeCl}_2, \mathrm{SnCl}_2\)

Both FeCI2 and SnCl2 are reducing agents with low oxidation numbers.

Question 14. A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number?

  1. S
  2. H
  3. Cl
  4. C

Answer: 3. Cl

A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated.

⇒ \(\stackrel{+1+5-2}{\mathrm{KClO}_3}+(\mathrm{COOH})_2+\mathrm{H}_2 \stackrel{+6}{\mathrm{~S}} \mathrm{O}_4 \longrightarrow\)\(\mathrm{K}_2^{+6} \mathrm{SO}_4+\mathrm{KCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 15. Oxidation numbers of P in PO43-, of S in SO42- and of Cr in Cr2O7 are respectively

  1. +3, +6 and +5
  2. +5, +3 and +6
  3. -3, +6 and +6
  4. +5, +6 and +6

Answer: 4. +5, +6 and +6

Let the oxidation number of P in \(\mathrm{PO}_4^{3-}\) be r.

∴ x+4(-2)=-3 = x=+5

Let the oxidation number of S in \(\mathrm{SO}_4^{2-}\) be y.

∴ y+4(-2)=-2 + y=+6

Let oxidation number of Cr in \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) be z.

∴ 2z+7(-2)=-2 + z=*6

Question 16. The number of moles of MnO4 required to oxidize one mole of ferrous oxalate completely in an acidic medium will be

  1. 7.5 moles
  2. 0.2 moles
  3. 0.6 moles
  4. 0.4 moles.

Answer: 4. 0.4 moles.

⇒ \(\left[5 e^{-}+\mathrm{MnO}_4^{-}+8 \mathrm{H}^* \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \ldots\right.(1)] \times 2\)

⇒ \(\left[\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 e^{-}+2 \mathrm{CO}_2 \ldots \text { (2) }\right] \times 5\)

Question 17. Which is the best description of the behaviour of bromine in the reaction given below? \(\mathrm{H}_2 \mathrm{O}+\mathrm{Br}_2 \rightarrow \mathrm{HOBr}+\mathrm{HBr}\)

  1. Proton acceptor only
  2. Both oxidised and reduced
  3. Oxidised only
  4. Reduced only

Answer: 2. Both oxidised and reduced

Question 18. The oxidation states of sulphur in the anions \(\mathrm{SO}_3^2-\), \(\mathrm{S}_2 \mathrm{O}_4^{2-}\) and \(\mathrm{S}_2 \mathrm{O}_6^{2-}\) follow the order

  1. \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)
  2. \(\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)
  3. \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}<\mathrm{SO}_3^{2-}\)
  4. \(\mathrm{S}_2 \mathrm{O}_6^{2-}<\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}\)

Answer: 1. \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)

⇒ \(\mathrm{SO}_3^{2-}: x+(-2) 3=-2\) or \(x-6=-2\) or \(x=+4\)

⇒ \(\mathrm{S}_2 \mathrm{O}_4^{2-}: 2 x+(-2) 4=-2\)

or \(2 x-8=-2\) or \(2 x=+6\)

∴ x = +3

⇒ \(\mathrm{S}_2 \mathrm{O}_6^{2-}: 2 x+(-2) 6=-2\)

or \(2 x-12=-2\) or \(2 x=+10\)

∴ x = +5

Oxidation states follow the order : \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)

Question 19. The oxidation state of Fe in Fe2O3 is

  1. 5/4
  2. 4/5
  3. 3/2
  4. 8/3

Answer: 4. 8/3

⇒ \(\mathrm{Fe}_3 \mathrm{O}_4: 3 x+4(-2)=0 \Rightarrow x=+\frac{8}{3}\)

Question 20. Reaction of sodium thiosulphate with iodine gives

  1. Tetrathionate ion
  2. Sulphide ion
  3. Sulphate ion
  4. Sulphite ion.

Answer: 1. Tetrathionate ion

⇒ \(2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \underset{\mathrm{Sodium tetrathionate}}{\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6}+2 \mathrm{NaI}\)

Question 21. The oxide, which cannot act as a reducing agent is

  1. CO2
  2. ClO2
  3. NO2
  4. SO2

Answer: 1. CO2

Since carbon is in its maximum oxidation state of +4, therefore, carbon dioxide (CO2) cannot act as a reducing agent.

Question 22. Which substance is serving as a reducing agent in the following reaction? \(14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+3 \mathrm{Ni} \rightarrow 7 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Cr}^{3+}+3 \mathrm{Ni}^{2+}\)

  1. H+
  2. Cr2O2-7
  3. H2O
  4. Ni

Answer: 4. Ni

Since the oxidation number of Ni increases from 0 to 2, therefore it acts as a reducing agent.

Question 23. The oxidation state of I in H4IO6 is

  1. +1
  2. – 1
  3. + 7
  4. + 5

Answer: 3. + 7

Let r = Oxidation state of l. Since oxidation state of H = +1 and oxidation state of O = – 2, therefore for H4IO6, we get (4 x 1) + x + (6x – 2) = -1 or x = +7

Question 24. Consider the change in the oxidation state of bromine corresponding to different emf values as shown in the given diagram:

Redox Reactions Oxidation State Of Bromine

Then the species undergoing disproportionation is

  1. BrO3
  2. BrO4
  3. Br2
  4. HBrO

Answer: 4. HBrO

For a reaction to be spontaneous, E°cell, should be positive as ΔG° = -nFE°cell

HBrO → Br2; E° = 1.595 V, SRP (cathode)

HBrO → BrO3; E° = -1.5 V, SOP (anode)

2HBrO → Br2 + BrO3

cell = SRP (cathode) – SRP (anode)

= 1.595 – 1.5 = 0.095 V

cell> 0 ⇒ ΔG° < 0 (spontaneous)

NEET Chemistry MCQs on Chemical Bonding And Molecular Structure

NEET Chemistry For Chemical Bonding And Molecular Structure Multiple Choice Questions

Question 1. Amongst the following, the total number of species not having eight electrons around the central atom in its outermost shell, is NH3, AlCl3, BeCl2, CCl4, PCl5

  1. 4
  2. 1
  3. 3
  4. 2

Answer: 3. 3

AICl3, BeCl2, and PCl5 do not have eight electrons around the central atom in their outermost shells.

Chemical Bonding And Molecular Structure Do Not have Eight Elctrons Around Central Atom

Question 2. In \(\mathrm{PO}_4^{3-}\) ion, the formal charge on each oxygen atom and P—O bond order respectively are

  1. -0.75, 1.25
  2. -0.75, 1.0
  3. -0.75,0.6
  4. -3, 1.25

Answer: 1. -0.75, 1.25

The total charge = -3

So, the average formal charge on each ‘O’ atom is -3/4= -0.75

Chemical Bonding And Molecular Structure Average Formal Charge On Each O Atoms

⇒ Average P-O bond order = \(\frac{\text { Total no. of bonds }}{\text { Total no. of resonating structures }}=\frac{5}{4}=1.25\)

Question 3. Among LiCl, BeCl2, BCl3 and CCl4, the covalent bond character follows the order

  1. \(\mathrm{BeCl}_2>\mathrm{BCl}_3>\mathrm{CCl}_4<\mathrm{LiCl}\)
  2. \(\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4<\mathrm{LiCl}\)
  3. \(\mathrm{LiCl}<\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4\)
  4. \(\mathrm{LiCl}>\mathrm{BeCl}_2>\mathrm{BCl}_3>\mathrm{CCl}_4\)

Answer: 3. \(\mathrm{LiCl}<\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4\)

Along the period, as we move from Li>Be>B>C, the electronegativity increases, and hence the EN difference between the element and Cl decreases and accordingly, the covalent character increases. Thus LiCI < BeCl2 < BCl3 < CCl4 is the correct order of covalent bond character.

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Question 4. Which one of the following formulae does not correctly represent the bonding capacities of the two atoms involved?

Chemical Bonding And Molecular Structure Bonding Capacities Of Two Atoms

Answer: 4

Chemical Bonding And Molecular Structure Asterisk Carbon Has A Valency Of 5

The asterisk (*) marked carbon has a valency of 5 and hence, this formula is not correct because carbon has a maximum valency of 4.

Question 5. Among the following, which compound will show the highest lattice energy?

  1. KF
  2. NaF
  3. CsF
  4. RbF

Answer: 2. NaF

For compounds containing ions of the same charge, lattice energy increases as the size of ions decreases. Thus, NaF has the highest lattice energy

Question 6. Which of the following molecules is non-polar in nature?

  1. NO2
  2. POCl3
  3. CH2O
  4. SbCl5

Answer: 4. SbCl5

Among the given molecules, SbCl5 is non-polar in nature

Question 7. Which of the following, set of molecules will have zero dipole moment?

  1. Ammonia, beryllium difluoride, water, 1, 4-dichlorobenzene
  2. Boron trifluoride, hydrogen fluoride, carbon dioxide, 1, 3-dichlorobenzene
  3. Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene
  4. Boron trifluoride, beryllium difluoride, carbon dioxide, 1, 4-dichlorobenzene

Answer: 4. Boron trifluoride, beryllium difluoride, carbon dioxide, 1, 4-dichlorobenzene

Chemical Bonding And Molecular Structure Set Of Molecules Hae Zero Dipole Moment

Question 8. Which of the following is the correct order of dipole moment?

  1. \(\mathrm{NH}_3<\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{H}_2 \mathrm{O}\)
  2. \(\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{NH}_3<\mathrm{H}_2 \mathrm{O}\)
  3. \(\mathrm{BF}_3<\mathrm{NH}_3<\mathrm{NF}_3<\mathrm{H}_2 \mathrm{O}\)
  4. \(\mathrm{H}_2 \mathrm{O}<\mathrm{NF}_3<\mathrm{NH}_3<\mathrm{BF}_3\)

Answer: 2. \(\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{NH}_3<\mathrm{H}_2 \mathrm{O}\)

Chemical Bonding And Molecular Structure Order Of Dipole Moment

Question 9. The species, having bond angles of 120° is

  1. \(\mathrm{ClF}_3\)
  2. \(\mathrm{NCl}_3\)
  3. \(\mathrm{BCl}_3\)
  4. \(\mathrm{PH}_3\)

Answer: 3. \(\mathrm{BCl}_3\)

BCl3-Trigonal planar sp²-hybridised, 120° angle.

Question 10. Consider the molecules CH4, NH3, and H2O. Which of the given statements is false?

  1. The H — O — H bond angle in H2O is smaller than the H — N — H bond angle in NH3.
  2. The H — C — H bond angle in CH4 is larger than the H — N — H bond angle in NH3.
  3. The H — C — H bond angle in CH4, the H — N — H bond angle in NH3, and the H — O — H bond angle in H2O are all greater than 90°.
  4. The H — O — H bond angle in H2O is larger than the H — C — H bond angle in CH4.

Answer: 4. The H — O — H bond angle in H2O is larger than the H — C — H bond angle in CH4.

Chemical Bonding And Molecular Structure Bod Angle Water Is Larger Than Metane

Question 11. Which of the following molecules has the maximum dipole moment?

  1. CO2
  2. CH4
  3. NH3
  4. NF3

Answer: 3. NH3

Chemical Bonding And Molecular Structure Ammonia Nitrogen

In NH3, H is less electronegative than N and hence dipole moment of each N-H bond is towards N and creates a high net dipole moment whereas in NF3, F is more electronegative than N, the dipole moment of each N-F bond is opposite to that of lone pair, hence reducing the net dipole moment.

Question 12. The correct order of increasing bond length of C – H, C – O, C – C and C = C is

  1. C-H<C=C<C-O<C-C
  2. C-C<C=C<C-O<C-H
  3. C-O<C-H<C-C<C = C
  4. C-H<C-O<C-C<C = C

Answer: 1. C-H<C=C<C-O<C-C

Increasing order of bond length is C-H<C-C<C-O<C-C

Question 13. Which of the following structures is the most preferred and hence of the lowest energy for SO3?

Chemical Bonding And Molecular Structure Lowest Energy Of SO3

Answer: 4

Chemical Bonding And Molecular Structure Maximum Number Of Covalent Bonds

It has a maximum number of covalent bonds involving pπ – dπ bonding also.

Question 14. The correct order of increasing bond angles in the following triatomic species is

  1. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2<\mathrm{NO}_2^{-}\)
  2. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}<\mathrm{NO}_2\)
  3. \(\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}<\mathrm{NO}_2\)
  4. \(\mathrm{NO}_2^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}\)

Answer: 4. \(\mathrm{NO}_2^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}\)

Structures of \(\mathrm{NO}_2^{-}, \mathrm{NO}_2\) and \(\mathrm{NO}_2^{+}\) is given as

Chemical Bonding And Molecular Structure Triatomic Species

The correct order of increasing bond angles in the following triatomic species is \(\mathrm{NO}_2^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}\)

Question 15. The correct order of C – O bond length among CO, CO2-3, CO2 is

  1. \(\mathrm{CO}<\mathrm{CO}_3^{2-}<\mathrm{CO}_2\)
  2. \(\mathrm{CO}_3^{2-}<\mathrm{CO}_2<\mathrm{CO}\)
  3. \(\mathrm{CO}<\mathrm{CO}_2<\mathrm{CO}_3^{2-}\)
  4. \(\mathrm{CO}_2<\mathrm{CO}_3^{2-}<\mathrm{CO}\)

Answer: 3. \(\mathrm{CO}<\mathrm{CO}_2<\mathrm{CO}_3^{2-}\)

The more single bond character in resonance hybrid more is the bond length. Hence, the increasing bond length is \(\mathrm{CO}<\mathrm{CO}_2<\mathrm{CO}_3^{2-}\)

Question 16. The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH3 (1.5 D) is larger than that of NF3 (0.2 D). This is because

  1. In NH3 the atomic dipole and bond dipole are in the opposite directions whereas in NF3 these are in the same direction
  2. In NH3 as well as in NF3 the atomic dipole and bond dipole are in the same direction
  3. In NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite directions
  4. In NH3 as well as in NF3 the atomic dipole and bond dipole are in opposite directions.

Answer: 3. In NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite directions

The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH3 (1.5 D) is larger than that of NF3 (0.2 D).

The dipole moment of NF3 is 0.24 D, and of NH3 is 1.48 D. The difference is due to the fact that the dipole moment due to N – F bonds in NF3 are in opposite directions to the direction of the dipole moment of the lone pair on N atom which partly cancels out.

The dipole moment of N – H bonds in NH3 are in the same direction as the dipole moment of the lone pair on the N atom which adds up as shown

Chemical Bonding And Molecular Structure Atomic Dipole And Bond Dipole

Question 17. The correct order in which the O-O bond length increases in the following is

  1. \(\mathrm{O}_2<\mathrm{H}_2 \mathrm{O}_2<\mathrm{O}_3\)
  2. \(\mathrm{O}_3<\mathrm{H}_2 \mathrm{O}_2<\mathrm{O}_2\)
  3. \(\mathrm{H}_2 \mathrm{O}_2<\mathrm{O}_2<\mathrm{O}_3\)
  4. \(\mathrm{O}_2<\mathrm{O}_3<\mathrm{H}_2 \mathrm{O}_2\)

Answer: 4. \(\mathrm{O}_2<\mathrm{O}_3<\mathrm{H}_2 \mathrm{O}_2\)

Bond lengths of \(\mathrm{O}-\mathrm{O}\) in \(\mathrm{O}_2\) is 1.21Å, in \(\mathrm{H}_2 \mathrm{O}_2\) is 1.48 Å and in \(\mathrm{O}_3\) is 1.28 Å. Therefore, correct order of the \(\mathrm{O}-\mathrm{O}\) bond length is \(\mathrm{H}_2 \mathrm{O}_2>\mathrm{O}_3>\mathrm{O}_2\).

Question 18. The correct sequence of increasing covalent character is represented by

  1. LiCl < NaCl < BeCl2
  2. BeCl2 < LiCl < NaCl
  3. NaCl < LiCl < BeCl2
  4. BeCl2 < NaCl < LiCl

Answer: 3. NaCl < LiCl < BeCl2

The covalent character in a compound is found by Fajan’s rule.

Fajan’s Rule: The smaller the size of the cation and the larger the size of the anion, the greater is the covalent character of an ionic bond. The greater the charge on the cation, the greater is the covalent character of the ionic bond.

Question 19. Which of the following would have a permanent dipole moment?

  1. SiF4
  2. SF4
  3. XeF4
  4. BF3

Answer: 2. SF4

For dipole moment, we have to know the hybridization and shape.

Chemical Bonding And Molecular Structure Dioole Moment Hybridisation And Shape

Question 20. H2O is dipolar, whereas BeF2 is not. It is because

  1. The electronegativity of F is greater than that of O
  2. H2O involves hydrogen bonding whereas BeF2 is a discrete molecule
  3. H2O is linear and BeF2 is angular
  4. H2O is angular and BeF2 is linear.

Answer: 4. H2O is angular and BeF2 is linear.

The overall value of the dipole moment of a polar molecule depends on its geometry and shape, i.e., vectorial addition of the dipole moment of the constituent bonds. Water has an angular structure with a bond angle of 105°, it has a dipole moment. However, BeF2 is a linear molecule thus, dipole moment summation of all the bonds present in the molecule cancels each other.

Chemical Bonding And Molecular Structure Dipole Moment Of A Polar Molecule

Question 21. Which of the following molecules does not possess a permanent dipole moment?

  1. \(\mathrm{CS}_2\)
  2. \(\mathrm{SO}_3\)
  3. \(\mathrm{H}_2 \mathrm{~S}\)
  4. \(\mathrm{SO}_2\)

Answer: 1. \(\mathrm{CS}_2\)

The structure of CS2 is linear and therefore it does not have a permanent dipole moment. It is represented as S = C = S.

Question 22. The table shown below gives the bond dissociation energies (Ediss) for single covalent bonds of carbon (C) atoms with elements A, B, C, and D. Which element has the smallest atoms?

Chemical Bonding And Molecular Structure Bond Dissociation Energies For Single Covalent Bonds Of Carbon Atoms

  1. C
  2. D
  3. A
  4. B

Answer: 2. D

The smaller the atom, the stronger the bond, and the greater the bond dissociation energy. Therefore, the bond C-D has the greatest energy or D has the smallest atoms.

Question 23. The strongest bond is in between

  1. CsF
  2. NaCl
  3. Both (1) and (2)
  4. None of the above.

Answer: 1. CsF

According to the Fajans rule, ionic character increases with an increase in the size of the cation, (Cs > Rb > K > Na) and with a decrease in the size of the anion (F > CI > Br > I). Thus, CsF has a higher ionic character than NaCl and hence, the bond in CsF is stronger than in NaCl.

Question 24. Which of the following bonds will be most polar?

  1. N – Cl
  2. O – F
  3. N – F
  4. N – N

Answer: 3. N – F

The polarity of the bond depends upon the electronegativity difference of the two atoms forming the bond. The greater the electronegativity difference, the more is the polarity of the bond.

⇒ \(\begin{array}{cccc}
\mathrm{N}-\mathrm{Cl} & \mathrm{O}-\mathrm{F} & \mathrm{N}-\mathrm{F} & \mathrm{N}-\mathrm{N} \\
3.04-3.16 & 3.5-4.0 & 3.04-4.0 & 3.04-3.04
\end{array}\)

Question 25. Amongst the following which one will have the maximum lone pair lone pair electron

  1. \(\mathrm{ClF}_3\)
  2. \(\mathrm{IF}_5\)
  3. \(\mathrm{SF}_4\)
  4. \(\mathrm{XeF}_2\)

Answer: 4. \(\mathrm{XeF}_2\)

ClF3, IF3, SF4, and Xe2, contain 2, 1, 1, and 3 lone pairs of electrons on the central atom respectively. Hence, XeF2 has maximum lone pair – lone pair repulsions.

Question 26. Match List-A with List-B

Chemical Bonding And Molecular Structure Match The Column

Choose the correct answer from the options given below.

  1. (1) – (D), (2) – (C), (3) – (B), (4) – (A)
  2. (1) – (D), (2) – (C), (3) – (A), (4) – (B)
  3. (1) – (B), (2) – (C), (3) – (D), (4) – (A)
  4. (1) – (C), (2) – (A), (3) – (A), (4) – (B)

Answer: 2. (1) – (D), (2) – (C), (3) – (A), (4) – (B)

PCl5: Trigonal bipyramidal

SF6: Octahedral

BrF5: Square pyramidal

BF3: Trigonal planar

Question 27. In the structure of ClF3, the number of lone pairs of electrons on the central atom ‘Cl’ is

  1. One
  2. Two
  3. Four
  4. Three

Answer: 2. Two

The structure of CIF3 is

Chemical Bonding And Molecular Structure Cl Has Two pair Of Electrons

Hence, Cl has 2 lone pairs of electrons

Question 28. Predict the correct order among the following:

  1. Bond pair – bond pair > lone pair-bond pair > lone pair – lone pair
  2. Lone pair – bond pair > bond pair-bond pair > lone pair – lone pair
  3. Lone pair – lone pair > lone pair-bond pair > bond pair-bond pair
  4. Lone pair – lone pair > bond pair-bond pair > lone pair-bond pair

Answer: 3. Lone pair – lone pair > lone pair-bond pair > bond pair-bond pair

According to VSEPR theory, the repulsive forces between lone pair and lone pair are greater than between lone pair and bond pair which are further greater than bond pair and bond pair.

Question 29. Which of the following species contains three bond pairs and one lone pair around the central atom?

  1. \(\mathrm{H}_2 \mathrm{O}\)
  2. \(\mathrm{BF}_3\)
  3. \(\mathrm{NH}_2^{-}\)
  4. \(\mathrm{PCl}_3\)

Answer: 4. \(\mathrm{PCl}_3\)

Chemical Bonding And Molecular Structure Three Bond Pairs And One Lone Pair Around Central Atom

Question 30. Which of the following is not a correct statement?

  1. Multiple bonds are always shorter than corresponding single bonds.
  2. The electron-deficient molecules can act as Lewis acids.
  3. The canonical structures have no real existence.
  4. Every AB5 molecule has have square pyramid structure.

Answer: 4. Every AB5 molecule does in fact have a square pyramid structure

For AB5 molecules, there are three possible geometries i.e. plantar pentagonal, square pyramidal, and trigonal bipyramidal.

Chemical Bonding And Molecular Structure Three Possible Geometric Shapes

Out of these three geometries, it is only a trigonal bipyramidal shape in which bond pair-bond pair repulsions are minimal and hence, this geometry is the most probable geometry of AB5 molecule.

Question 31. Which of the following is not isostructural with SiCl4?

  1. \(\mathrm{NH}_4^{+}\)
  2. \(\mathrm{SCl}_4\)
  3. \(\mathrm{SO}_4^{2-}\)
  4. \(\mathrm{PO}_4^{3-}\)

Answer: 2. \(\mathrm{SCl}_4\)

∴ \(\mathrm{SiCl}_4, \mathrm{NH}_4^{+}, \mathrm{SO}_4^{2-}\) and \(\mathrm{PO}_4^{3-}\) ions are the examples of molecules/ions which are of AB, type and have tetrahedral structures. SCl4 is AB4 (lone pair) type species.

Although the arrangement of five sp³d hybrid orbitals in space is trigonal bipyramidal, due to the presence of one lone pair of electrons in the basal hybrid orbital, the shape of AB5 (lone pair) species gets distorted and trichomes distorted tetrahedral or see-saw.

Question 32. In which of the following molecules all the bonds are not equal?

  1. \(\mathrm{NF}_3\)
  2. \(\mathrm{ClF}_3\)
  3. \(\mathrm{BF}_3\)
  4. \(\mathrm{AlF}_3\)

Answer: 2. \(\mathrm{ClF}_3\)

The Cl – F (Cl – Feq) bond length is equal to 1.60 Å while each of the two axial CI – F (Cl – Fa) bond lengths is equal to 1.70 Å.

Question 33. Which of the following molecules has trigonal planar geometry?

  1. \(\mathrm{BF}_3\)
  2. \(\mathrm{NH}_3\)
  3. \(\mathrm{PCl}_3\)
  4. \(\mathrm{IF}_3\)

Answer: 1

Chemical Bonding And Molecular Structure Molecules Has Trigonal Planar Geometry

Question 34. In a regular octahedral molecule, MX6 the number of X- M -A bonds at 180° is

  1. Three
  2. Two
  3. Six
  4. Four.

Answer: 1. Three

In an octahedral molecule, six hybrid orbitals direct towards the corners of a regular octahedron with a bond angle of 90°.

Chemical Bonding And Molecular Structure Octahedral Molecules

According to this geometry., the X number of X – M – X bonds at 180° must be three

Question 35. In BrF3 molecule, the lone pairs occupy equatorial positions to minimize

  1. Lone pair – bond pair repulsion only
  2. Bond pair – bond pair repulsion only
  3. Lone pair-lone pair repulsion and lone pair-bond pair repulsion
  4. Lone pair – lone pair repulsion only

Answer: 4. Lone pair – lone pair repulsion only

Chemical Bonding And Molecular Structure Bent T Shaped Geometry

Bent T-shaped geometry in which both lone pairs occupy the equatorial positions of the trigonal bipyramid. Here (lp – lp) repulsion = 0, (lp – bp) repulsion = 4, and (bp – bp) repulsion = 2.

Question 36. In \(\mathrm{NO}_3^{-}\) ion, the number of bond pair and lone pair of electrons on the nitrogen atom are

  1. 2,2
  2. 3, 1
  3. 1,3
  4. 4,0

Answer: 4. 4,0

Chemical Bonding And Molecular Structure Nitrogen Has Found Bond Pairs

In NO3 ion, nitrogen has 4 bond pairs of electrons and no lone pair of electrons.

Question 37. In which of the following bond angles is maximum?

  1. \(\mathrm{NH}_3\)
  2. \(\mathrm{NH}_4^{+}\)
  3. \(\mathrm{PCl}_3\)
  4. \(\mathrm{SCl}_2\)

Answer: 2. \(\mathrm{NH}_4^{+}\)

The bond angle is maximum in NH4+ tetrahedral molecules with a bond angle of 109º.

Question 38. BCl3 is a planar molecule whereas NCl3 is pyramidal because

  1. The nitrogen atom is smaller than the boron atom
  2. BCl3 has no lone pair but NCl3 has a lone pair of electrons
  3. B—Cl bond is more polar than N—Cl bond
  4. The N-Cl bond is more covalent than the B—Cl bond.

Answer: 2. BCl3 has no lone pair but NCl3 has a lone pair of electrons

There is no lone pair on boron in BCl3. hence, no repulsion takes place. There is a lone pair of nitrogen in NCl3, hence, repulsion takes place. Therefore, BCl3 is a planar molecule but NCl3 is a pyramidal molecule.

Question 39. In compound X, all the bond angles are exactly 109°28′, and X is

  1. Chloromethane
  2. Carbon tetrachloride
  3. Iodoform
  4. Chloroform.

Answer: 2. Carbon tetrachloride

As C – Cl bonds are directed towards the corner of a regular tetrahedron

Question 40. Which of the following species contains an equal number of σ and π bonds?

  1. \((\mathrm{CN})_2\)
  2. \(\mathrm{CH}_2(\mathrm{CN})_2\)
  3. \(\mathrm{HCO}_3^{-}\)
  4. \(\mathrm{XeO}_4\)

Answer: 4. \(\mathrm{XeO}_4\)

Chemical Bonding And Molecular Structure Species Contains Equal Number Of Bonds

Question 41. Which one of the following molecules contains no π bond?

  1. \(\mathrm{SO}_2\)
  2. \(\mathrm{NO}_2\)
  3. \(\mathrm{CO}_2\)
  4. \(\mathrm{H}_2 \mathrm{O}\)

Answer: 4. \(\mathrm{H}_2 \mathrm{O}\)

Chemical Bonding And Molecular Structure Molecule Contains No Pi Bonds

Question 42. Which one of the following statements is not correct for sigma- and pi- bonds formed between two carbon atoms?

  1. A sigma-bond is stronger than a pi-bond.
  2. Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively.
  3. Free rotation of atoms about a sigma bond is allowed but not in the case of a pi bond.
  4. Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard.

Answer: 2. Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively.

Sigma bond dissociation energy. = 347 kJ/mol

Pi-bond dissociation energy = 264kJ/mol

Question 43. The main axis of a diatomic molecule is z, molecular orbital px, and py overlap to form which of the following orbitals?

  1. π molecular orbital
  2. σ molecular orbital
  3. δ molecular orbital
  4. No bond will form

Answer: 1. π molecular orbital

For π overlap, the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

Chemical Bonding And Molecular Structure Perpendicular To The Line Joining The Nuclei

Hence, only sidewise overlapping takes place

Question 44. Which statement is not correct?

  1. A sigma bond is weaker than a pi bond.
  2. A sigma bond is stronger than a pi bond.
  3. A double bond is stronger than a single bond.
  4. A double bond is shorter than a single bond.

Answer: 1. A sigma bond is weaker than a π bond.

A σ bond is stronger than a t-bond

Question 45. Linear combination of two hybridized orbitals belonging to two atoms each having one electron leads to the formation of

  1. Sigma bond
  2. Double bond
  3. Co-ordinate covalent bond
  4. Pi bond.

Answer: 1. Sigma bond

Question 46. Which of the following does not apply to metallic bonds?

  1. Overlapping valence orbitals
  2. Mobile valence electrons
  3. Delocalized electrons
  4. Highly directed bonds

Answer: 4. Highly directed bonds

Metallic bonds have electrostatic attractions on all sides and hence, do not have directional characteristics.

Question 47. The angle between the overlapping of one s-orbital and one p-orbital is

  1. 180°
  2. 120°
  3. 109°28′
  4. 120°, 60°

Answer: 1. 180°

The type of overlapping between s- and p-orbitals occurs along the internuclear axis and hence, the angle is 180°.

Chemical Bonding And Molecular Structure Overlapping Between s And p Orbitals

Question 48. Which of the following pairs of compounds is isoelectronic and isostructural?

  1. \(\mathrm{TeI}_2, \mathrm{XeF}_2\)
  2. \(\mathrm{IBr}_2^{-}, \mathrm{XeF}_2\)
  3. \(\mathrm{IF}_3, \mathrm{XeF}_2\)
  4. \(\mathrm{BeCl}_2, \mathrm{XeF}_2\)

Answer: None

Chemical Bonding And Molecular Structure Isoelectronic Should Be Same Number Of Valence Electrons

In this question, in place of isoelectronic, there should be the same number of valence electrons.

Question 49. The hybridizations of atomic orbitals of nitrogen in \(\mathrm{NO}_2^{+}, \mathrm{NO}_3^{-} \text {and } \mathrm{NH}_4^{+}\) respectively are

  1. \(s p, s p^3\) and \(s p^2\)
  2. \(s p^2, s p^3\) and \(s p\)
  3. \(s p, s p^2\) and \(s p^3\)
  4. \(s p^2, s p\) and \(s p^3\)

Answer: 3. \(s p, s p^2\) and \(s p^3\)

X = \(\frac{1}{2}(V E+M A-c+a)\)

For \(\mathrm{NO}_2^{+}, X=\frac{1}{2}(5+0-1)=2\) i.e., sp hybridisation

For \(\mathrm{NO}_3^{-}, X=\frac{1}{2}(5+0+1)=3\) i.e., \(s p^2\) hybridisation

For \(\mathrm{NH}_4^{+}, X=\frac{1}{2}(5+4-1)=4\) i.e, \(s p^3\) hybridisation

Question 50. Which of the following pairs of ions is isoelectronic and isostructural?

  1. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)
  2. \(\mathrm{ClO}_3^{-}, \mathrm{CO}_3^{2-}\)
  3. \(\mathrm{SO}_3^{2-}, \mathrm{NO}_3^{-}\)
  4. \(\mathrm{ClO}_3^{-}, \mathrm{SO}_3^{2-}\)

Answer: 1. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\) and 4. \(\mathrm{ClO}_3^{-}, \mathrm{SO}_3^{2-}\)

1. \(\mathrm{CO}_3^{2-}: 6+24+2=32 ; s p^2\); trigonal planar \(\mathrm{NO}_3^{-}: 7+24+1=32 ; s p^2\); trigonal planar

Hence, these are isoelectronic as well as isostructural.

2. \(\mathrm{ClO}_3^{-}: 17+24+1=42 ; s p^3\), trigonal pyramidal \(\mathrm{CO}_3^{2-}: 6+24+2=32 ; s p^2\), trigonal planar

Hence, these are neither isoelectronic nor isostructural.

3. \(\mathrm{SO}_3^{2-}: 16+24+2=42 ; s p^3\), trigonal pyramidal \(\mathrm{NO}_3^{-}: 7+24+1=32 ; s p^2\), trigonal planar These are neither isoelectronic nor isostructural.

4. \(\mathrm{ClO}_3^{-}: 17+24+1=42; s p^3\), trigonal pyramidal \(\mathrm{SO}_3^{2-}: 16+24+2=42; s p^3\), trigonal pyramidal Hence, these are isoelectronic as well as isostructural.

Question 51. The correct geometry and hybridization for XeF4 are

  1. Octahedral, sp³d²
  2. Trigonal bipyramidal, sp³d
  3. Planar triangle, sp³d³
  4. Square planar, sp³d2.

Answer: 1. Octahedral, sp³d²

Chemical Bonding And Molecular Structure Octahedral Geometry Dquare Planar Shape

Question 52. In which of the following pairs, both the species are not isostructural?

  1. Diamond, Silicon carbide
  2. \(\mathrm{NH}_3, \mathrm{PH}_3\)
  3. \(\mathrm{XeF}_4, \mathrm{XeO}_4\)
  4. \(\mathrm{SiCl}_4, \mathrm{PCl}_4^{+}\)

Answer: 3. \(\mathrm{XeF}_4, \mathrm{XeO}_4\)

In diamond and silicon carbide, the central atom is sp³ hybridized and hence, both are isostructural.

NH3 and PH3, both are pyramidal and the central atom in both cases is sp³ hybridized. SiCl4 and PCl+4, both are tetrahedral and the central atom in both cases is sp³ hybridized.

Chemical Bonding And Molecular Structure Diamond And Silicon Carbide

In XeF4, Xe is sp³d² hybridized and the structure is square planar while in XeO4, Xe is sp³ hybridized and the structure is tetrahedra.

 

Question 53. Maximum bond angle at nitrogen is present in which of the following?

  1. \(\mathrm{NO}_2^{+}\)
  2. \(\mathrm{NO}_3^{-}\)
  3. \(\mathrm{NO}_2\)
  4. \(\mathrm{NO}_2^{-}\)

Answer: 1. \(\mathrm{NO}_2^{+}\)

Chemical Bonding And Molecular Structure Maximu Bond Angles

So, \(\mathrm{NO}_2^{+}\) has maximum bond angle.

Question 54. Which one of the following species has a planar triangular shape?

  1. \(\mathrm{N}_3\)
  2. \(\mathrm{NO}_3^{-}\)
  3. \(\mathrm{NO}_2^{-}\)
  4. \(\mathrm{CO}_2\)

Answer: 2. \(\mathrm{NO}_3^{-}\)

Chemical Bonding And Molecular Structure Trigonal Planar

Question 55. XeF2 is isostructural with

  1. \(\mathrm{SbCl}_3\)
  2. \(\mathrm{BaCl}_2\)
  3. \(\mathrm{TeF}_2\)
  4. \(\mathrm{ICl}_2^{-}\)

Answer: 4. \(\mathrm{ICl}_2^{-}\)

Chemical Bonding And Molecular Structure Iso Structural

Question 56. Which of the following is a polar molecule?

  1. \(\mathrm{SiF}_4\)
  2. \(\mathrm{XeF}_4\)
  3. \(\mathrm{BF}_3\)
  4. \(\mathrm{SF}_4\)

Answer: 4. \(\mathrm{SF}_4\)

SF4 has sp³d-hybridisation and see-saw shape with (4 bp + 1 lp)

Chemical Bonding And Molecular Structure Polar Molecule

 

Question 57. In which of the following pairs, the two species are isostructural?

  1. \(\mathrm{SO}_3^{2-}\) and \(\mathrm{NO}_3^{-}\)
  2. \(\mathrm{BF}_3\) and \(\mathrm{NF}_3\)
  3. \(\mathrm{BrO}_3^{-}\) and \(\mathrm{XeO}_3\)
  4. \(\mathrm{SF}_4\) and \(\mathrm{XeF}_4\)

Answer: 3. \(\mathrm{BrO}_3^{-}\) and \(\mathrm{XeO}_3\)

Hybridisation of Brin \(\mathrm{BrO}_3^{-}\): H = 1/2(7 + 0 – 0 + 1) = 4 i.e.sp³ hybridisation

Hybridisation of Xe in XeO3 : H = 1/2(8 +0 -0+0)= 4 i.e. sp³ hybridisation

In both BrO3 and XeO3, the central atom is sp³ hybridized and contains one lone pair of electrons, hence in both cases, the structure is trigonal pyramidal.

Chemical Bonding And Molecular Structure Trigonal Pyramidal

 

 

Question 58. Which of the following species has a linear shape?

  1. \(\mathrm{O}_3\)
  2. \(\mathrm{NO}_2^{-}\)
  3. \(\mathrm{SO}_2\)
  4. \(\mathrm{NO}_2^{+}\)

Answer: 4. \(\mathrm{NO}_2^{+}\)

∴ \(\mathrm{NO}_2^{-}\): Due to sp² hybridization of N-atom and the presence of one lone pair on it, \(\mathrm{NO}_2^{-}\) has an angular shape.

Chemical Bonding And Molecular Structure Angle Shape

Chemical Bonding And Molecular Structure Angular V Shaped Structure

SO2: Due to the presence of one lone pair of electrons in one of the three sp²-hybrid orbitals, the SO2 molecule has an angular (V-shaped) structure.

Chemical Bonding And Molecular Structure One Lone Pair Of Electrons

∴ \(\mathrm{NO}_2^{+}\): Due to sp hybridisation of \(\mathrm{N}^{+}, \mathrm{NO}_2^{+}\) ion has linear shape.

Chemical Bonding And Molecular Structure Linear Shape

Question 59. The correct order regarding the electronegativity of hybrid orbitals of carbon is

  1. \(s p<s p^2<s p^3\)
  2. \(s p>s p^2<s p^3\)
  3. \(s p>s p^2>s p^3\)
  4. \(s p<s p^2>s p^3\)

Answer: 3. \(s p>s p^2>s p^3\)

The electronegativity of carbon atom is not fixed. It varies with the state of hybridization. The electronegativity of carbon increases as the s-character of the hybrid orbital increases.

C(sp)>C(sp²)>c(sp³)

Question 60. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.

  1. \(\mathrm{NO}<\mathrm{O}_2^{-}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
  2. \(\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
  3. \(\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}\)
  4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Answer: 4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Chemical Bonding And Molecular Structure Diatomic Species

Thus, bond order increases as: \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Question 61. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas?

  1. Dipole-dipole interaction
  2. Covalent bonds
  3. London dispersion force
  4. Hydrogen bonding

Answer: 4. Hydrogen bonding

Methanol can undergo intermolecular association through H-bonding as the -OH group in alcohols is highly polarised.

Chemical Bonding And Molecular Structure Methanol

As a result, in order to convert Iiquid CH3OH to a gaseous state, the strong hydrogen bonds must be broken.

Question 62. Which of the following has pπ – dπ bonding?

  1. \(\mathrm{NO}_3^{-}\)
  2. \(\mathrm{SO}_3^{2-}\)
  3. \(\mathrm{BO}_3^{3-}\)
  4. \(\mathrm{CO}_3^{2-}\)

Answer: 2. \(\mathrm{SO}_3^{2-}\)

In sulfite ion, the central atom sulfur is sp³ hybridisedElectronic structure of S atom in an excited state

Chemical Bonding And Molecular Structure Electronic Structure Of S Atom

The three p electrons form o bonds with three oxygen atoms with one position (of the tetrahedron) being occupied by a lone pair. The d electron (excluded from hybridization) forms π bond with one oxygen atom. i.e. pπ – dπ bonding occurs.

MCQs on Thermodynamics for NEET

NEET Chemistry For Thermodynamics Multiple Choice Questions

Question 1. Which of the following are not state functions?

  1. q+w
  2. q
  3. w
  4. H-TS
  1. (1), (2) and (3)
  2. (2) and (3)
  3. (1) and (4)
  4. (2),(3) and (4)

Answer: 2. (2) and (3)

State functions or state variables are those which depend only on the state of the system and not on how the state was reached.

q + w = ΔE (internal energy)

H – Ts = G (free energy)

Path function depends on the path followed during a process. Work and heat are the path functions.

Question 2. In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true?

  1. ΔE= W≠0, q=0
  2. ΔE= W=q≠0
  3. ΔE =0, W = q≠0
  4. W=0, ΔE=q≠0

Answer: 1. ΔE= W≠0, q=0

In a closed insulated container a liquid is stirred with a paddle to increase the temperature

The mathematical form of the first law of thermodynamics q = ΔE + w

Since the system is closed and insulated, q = 0

Paddlework is done on the system.

∴ w ≠ 0.

Temperature and hence internal energy of the system increases.

∴ ΔE ≠ 0.

Read and Learn More NEET MCQs with Answers

MCQs on Thermodynamics for NEET

Question 3. Which of the following is the correct equation?

  1. ΔU= ΔW+ ΔQ
  2. ΔU= ΔQ – W
  3. ΔW = ΔU+ ΔQ
  4. None of these

Answer: 2. ΔU= ΔQ – W

This is the mathematical relation of the first law of thermodynamics. Here ΔU = change in internal energy; ΔQ = heat absorbed by the system and W = work done by the system.

Question 4. Which of the following options is the correct relation between change in enthalpy and change in internal energy?

  1. ΔH – ΔU= -ΔnRT
  2. ΔH + ΔU = ΔnR
  3. ΔH = ΔU – ΔngRT
  4. ΔH= ΔU+ ΔngRT

Answer: 4. ΔH= ΔU+ ΔngRT

Question 5. Which of the following P-V curve represents maximum work done?

Thermodynamics P V Curve

Answer: 2

Question 6. Which one among the following is the correct option for the right relationship between CP and CV for one mole of an ideal gas?

  1. CV=RCP
  2. CP+CV=R
  3. CP– CV=R
  4. CP = RCV

Answer: 3. CP – Cv=R

CP = CV+ nR

For one mole of an ideal gas, CP= CV + R or CP – CV = R

Question 7. The correct option for free expansion of ideal gas under adiabatic conditions is

  1. q = 0, ΔT = 0 and w = 0
  2. q = 0, ΔT < 0 and w > 0
  3. q < 0, ΔT = 0 and w = 0
  4. q > 0, ΔT > 0 and w > 0

Answer: 1. q = 0, ΔT = 0 and w = 0

For the free expansion of an ideal gas, Pex = 0,

w=-Pex ΔV=0

For adiabatic process, q = 0

According to the first law of thermodynamics, ΔU=q+w=0

As the internal energy of an ideal gas is a function of temperature, ΔU=0,

∴ T =0

Question 8. Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is [Given that 1 L bar = 100 J]

  1. 30 J
  2. -30 J
  3. 5 kJ
  4. 25J

Answer: 2. -30 J

Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar.

Expansion of a gas against a constant external pressure is an irreversible process. The work done in an irreversible process

= \(-P_{\text {ext }} \Delta \mathrm{V}=-P_{\text {ext }}\left(V_2-V_1\right)=-2(0.25-0.1)\)

= \(-2 \times 0.15 \mathrm{~L} \text { bar }=-0.30 \times 100 \mathrm{~J}=-30 \mathrm{~J}\)

Question 9. Reversible expansion of an ideal gas under isothermal and adiabatic conditions as shown in the

AB → Isothermal expansion

AC → Adiabatic expansion

Thermodynamics Reversible Expansion

Which of the following options is not correct?

  1. \(\Delta S_{\text {isothermal }}>\Delta S_{\text {adiabatic }}\)
  2. \(T_A=T_B\)
  3. \(W_{\text {isothermal }}>W_{\text {adiabatic }}\)
  4. \(T_C>T_A\)

Answer: 4. \(T_C>T_A\)

For an ideal gas, internal energy is a function of temperature. The final temperature i.e., TC for adiabatic process is less than its initial temperature i.e., TA

∴ TC < TA

Question 10. An ideal gas expands isothermally from 10-3 m³ to 10-2 m³ at 300 K against a constant pressure of 105 N m-2. The work done on the gas is

  1. +270 kJ
  2. -900 J
  3. +900 kJ
  4. -900 kJ

Answer: 2. -900 J

An ideal gas expands isothermally from 10-3 m³ to 10-2 m³ at 300 K against a constant pressure of 105 N m-2.

w = \(-P d V=-P\left(V_2-V_1\right)\)

= \(-10^5 \mathrm{~N} \mathrm{~m}^{-2}\left(10^{-2}-10^{-3}\right) \mathrm{m}^3=-10^5 \mathrm{~N} \mathrm{~m}^{-2}\left(9 \times 10^{-3}\right) \mathrm{m}^3\)

= \(-9 \times 10^2 \mathrm{~N} \mathrm{~m}=-900 \mathrm{~J}\)

Question 11. A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

  1. -500 J
  2. -505 J
  3. +505 J
  4. 1136.25 J

Answer: 2. -505 J

A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L.

w = \(-P_{{ext}} \Delta V=-2.5(4.50-2.50)\)

= \(-5 \mathrm{~L} \text { atm }=-5 \times 101.325 \mathrm{~J}=-506.625 \mathrm{~J}\)

ΔU=q+w

As, the container is insulated, thus q=0

Hence, \(\Delta U=w=-506.625 \mathrm{~J}\)

Question 12. Equal volumes of two monatomic gases, A and B at the same temperature and pressure are mixed. The ratio of specific heat (CP/CV) of the mixture will be

  1. 0.83
  2. 1.50
  3. 3.3
  4. 1.67

Answer: 4. 1.67

Equal volumes of two monatomic gases, A and B at the same temperature and pressure are mixed.

CP for a monoatomic gas mixture of the same volume

= \(\frac{5}{2} R, C_V=\frac{3}{2} R\)

∴ \(\frac{C_P}{C_V}=\frac{\frac{5}{2} R}{\frac{3}{2} R}=\frac{5}{3}=1.67\)

Question 13. Which of the following is the correct option for free expansion of an ideal gas under adiabatic conditions?

  1. q=0, ΔT≠0, w = 0
  2. q≠0, ΔT= 0, w= 0
  3. q=0, ΔT=0, w=0
  4. q=0, ΔT<0, w≠0

Answer: 3. q=0, ΔT=0, w=0

For free expansion of an ideal gas under adiabatic condition q = 0, ΔT = 0, w = 0.

For free expansion, w = 0, adiabatic process, q = 0

ΔU = q + w = 0

Internal energy remains constant means ΔT = 0.

Question 14. Three moles of an ideal gas expanded spontaneously into the vacuum. The work done will be

  1. Infinite
  2. 3 Joules
  3. 9 Joules
  4. Zero.

Answer: 4. Zero.

Three moles of an ideal gas expanded spontaneously into the vacuum.

Since the ideal gas expands spontaneously into the vacuum, Pext = 0, hence work done is also zero.

Question 15. Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE?

  1. \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}\)
  2. \(\mathrm{H}_{2(g)}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}\)
  3. \(\mathrm{C}_{(s)}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightarrow 2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}\)
  4. \(\mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(\mathrm{~g})}\)

Answer: 2. \(\mathrm{H}_{2(g)}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}\)

∴ \(\Delta H=\Delta E+\Delta n_g R T\)

For \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}\)

∴ \(\Delta n_g=2-(1+1)=0 \text {. i.e. } \Delta H=\Delta E\)

Question 16. The work done during the expansion of a gas from a volume of 4 dm³ to 6 dm³ against a constant external pressure of 3 atm is (1 L atm = 101.32 J)

  1. -6J
  2. -608J
  3. +304J
  4. -304J

Answer: 2. -608J

Work = -Pext x volume change = -3 x (6 – 4) x 101.32 = 6 x 101.32 = – 607.92 J = – 608 J

Question 17. For the reaction, \(\mathrm{C}_3 \mathrm{H}_{8(\mathrm{~g})}+5 \mathrm{O}_{2(g)} \rightarrow 3 \mathrm{CO}_{2(\mathrm{~g})}+4 \mathrm{H}_2 \mathrm{O}_{(l)}\) at constant temperature, ΔH – ΔE is

  1. +RT
  2. -3RT
  3. +3RT
  4. -RT

Answer: 2. -3RT

⇒ \(\mathrm{C}_3 \mathrm{H}_{g(g)}+5 \mathrm{O}_{2(g)} \rightarrow 3 \mathrm{CO}_{2(g)}+4 \mathrm{H}_2 \mathrm{O}_{(l)}\)

⇒ \(\Delta n_g=3-6=-3\)

∴ \(\Delta H=\Delta E+P \Delta V\) or \(\Delta H-\Delta E=P \Delta V\)

∴ \(\Delta H-\Delta E=\Delta n_g R T=-3 R T\)

Question 18. The molar heat capacity of water at constant pressure, C, is 75 J K-1 mol-1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is

  1. 1.2 K
  2. 2.4 K
  3. 4.8 K
  4. 6.6 K

Answer: 2. 2.4 K

The molar heat capacity of water at constant pressure, C, is 75 J K-1 mol-1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand

Molar heat capacity = 75J K-1 mol-1

18 g of water = 1 mole = 75J K-1 mol-1

1 g of water = \(\frac{75}{18}\) J K-1

100 g ofwater= \(\frac{75}{18}\) x 100 J K-1

Q = \(m \cdot C \cdot \Delta T \text { or } 1000=100 \times \frac{75}{18} \times \Delta T\)

⇒ \(\Delta T=\frac{10 \times 18}{75}=2.4 \mathrm{~K}\)

Question 19. When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct?

  1. q = w = 500 J, ΔE = 0
  2. q = ΔE = 500 J, w = 0
  3. q = w = 500 J, ΔE = 0
  4. ΔE = 0, q = w = -500 J

Answer: 2. q = ΔE = 500 J, w = 0

When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J.

ΔH = ΔE + PΔV

When ΔV = 0; w = 0.

ΔH = ΔE + 0 or ΔH = ΔE

As ΔE = q + w, ΔE = q

In the present problem, ΔH = 500 J

ΔH= ΔE = 500 J, q = 500 J, w = 0

Question 20. For the reaction, \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)}+3 \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}+3 \mathrm{H}_2 \mathrm{O}_{(l)}\) which one is true?

  1. ΔH = ΔE – RT
  2. ΔH = ΔE + RT
  3. ΔH = ΔE + 2RT
  4. ΔH = ΔE – 2RT

Answer: 1. ΔH = ΔE – RT

ΔH = ΔE + PΔV also PV = nRT (ideal gas equation)

or PΔV = ΔngRT

Δng = Change in the number of gaseous moles

∴ ΔH= ΔE+ ΔngRT

⇒ Δng = 2 – 3 = -1

⇒ ΔH = ΔE – RT

Question 21. In an endothermic reaction, the value of ΔH is

  1. Negative
  2. Positive
  3. Zero
  4. Constant.

Answer: 2. Positive

In endothermic reactions, the energy of reactants is less than the energy of products. Thus, ER < EP.

ΔH = EP – ER = +ve

Question 22. One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 litres. The ΔE for this process is (R = 2cal mol-1 K-1)

  1. 1381.1 cal
  2. Zero
  3. 163.7 cal
  4. 9 L atm

Answer: 2. Zero

One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 litres.

Change in internal energy depends upon temperature. At constant temperature, the internal energy of the gas remains constant, so ΔE = 0.

Question 23. During isothermal expansion of an ideal gas, its

  1. Internal energy increases
  2. Enthalpy decreases
  3. Enthalpy remains unaffected
  4. Enthalpy reduces to zero.

Answer: 3. Enthalpy remains unaffected

During isothermal expansion of an ideal gas, ΔT = 0, ΔE = 0

∴ ΔH = 0

H = E + PV

ΔH = ΔE + Δ(PV) = ΔE + Δ(nRT)

∴ ΔH= ΔE + nRΔT = 0 + 0 = 0

Change in enthalpy is zero, which means its enthalpy remains the same or unaffected.

Question 24. For the reaction, \(\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3, \Delta H\)=?

  1. ΔE + 2RT
  2. ΔE – 2RT
  3. ΔH = RT
  4. ΔE – RT

Answer: 2. ΔE – 2RT

Δng =2 – 4= -2, ΔH = ΔE -2RT

Question 25. If ΔH is the change in enthalpy and ΔT, the change in internal energy accompanying a gaseous reaction, then

  1. ΔH is always greater than ΔT
  2. ΔH < ΔE only if the number of moles of the products is greater than the number of moles of the reactants
  3. ΔH is always less than ΔE
  4. ΔH < ΔE only if the number of moles of products is less than the number of moles of the reactant

Answer: 4. ΔH < ΔE only if the number of moles of products is less than the number of moles of the reactant

lf np < nr; Δng= np – nr= -ve.

Hence, ΔH < ΔE.

Question 26. Three thermochemical equations are given below:

  1. \(\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(g)} ; \Delta_{,} H^{\circ}=x \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  2. \(\mathrm{C}_{\text {(graphite) }}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)} ; \Delta_r H^{\circ}=y \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  3. \(\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta_{\mathrm{r}} H^{\circ}=z \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Based on the above equations, find out which of the relationships given below is correct.

  1. z = x + y
  2. x = y + z
  3. y = 2z – x
  4. x = y – z

Answer: 2. x = y + z

According to Hesst’s law, equation (1) is equal to equations (2) + (3) i.e., x = y + z

Question 27. The standard enthalpy of vaporisation ΔvapH° for water at 100°C is 40.66 kJ mol-1. The internal energy of the vaporisation of water at 100°C (in kJ mol-1) is

  1. +37.56
  2. -43.76
  3. +43.76
  4. +40.66

(Assume water vapour to behave like an ideal gas)

Answer: 1. +37.56

The standard enthalpy of vaporisation ΔvapH° for water at 100°C is 40.66 kJ mol-1.

∴ \(\Delta_{\text {vap }} H^{\circ}=40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

T = \(100+273=373 \mathrm{~K}, \Delta E=?\)

∴ \(\Delta H=\Delta E+\Delta n_g R T \Rightarrow \Delta E=\Delta H-\Delta n_g R T\)

⇒ \(\Delta n_g\) = number of gaseous moles of products – number of gaseous moles of reactants

⇒ \(\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_2 \mathrm{O}_{(g)}\)

⇒ \(\Delta n_g=1-0=1\)

∴ \(\Delta E=\Delta H-R T\)

ΔE = (40.66 x 10³) – (8314 x 373)

= 37559 J/mol or 37.56 kJ/mol

Question 28. Consider the following processes:

⇒ \(\begin{array}{lc}
& \Delta \boldsymbol{H}(\mathbf{k J} / \mathbf{m o l}) \\
1 / 2 A \rightarrow B & +150 \\
3 B \rightarrow 2 C+D & -125 \\
E+A \rightarrow 2 D & +350
\end{array}\)

For \(B+D \rightarrow E+2 C, \Delta H\) will be

  1. \(525 \mathrm{~kJ} / \mathrm{mol}\)
  2. \(-175 \mathrm{~kJ} / \mathrm{mol}\)
  3. \(-325 \mathrm{~kJ} / \mathrm{mol}\)
  4. \(325 \mathrm{~kJ} / \mathrm{mol}\)

Answer: 2. \(-175 \mathrm{~kJ} / \mathrm{mol}\)

Adding all the equations, we get

⇒ \(\begin{array}{lc}
& \Delta H \\
A \rightarrow 2 B & 300 \mathrm{~kJ} / \mathrm{mol} \\
3 B \rightarrow 2 C+D & -125 \mathrm{~kJ} / \mathrm{mol} \\
2 D \rightarrow A+E & -350 \mathrm{~kJ} / \mathrm{mol} \\
\hline B+D \rightarrow E+2 C ; \Delta H=(300-125-350)=-175 \mathrm{~kJ} / \mathrm{mol}
\end{array}\)

Question 29. The following two reactions are known

  • \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(g)} ; \Delta H=-26.8 \mathrm{~kJ}\)
  • \(\mathrm{FeO}_{(s)}+\mathrm{CO}_{(g)} \rightarrow \mathrm{Fe}_{(s)}+\mathrm{CO}_{2(g)} ; \Delta H=-16.5 \mathrm{~kJ}\)

The value of \(\Delta H\) for the following reaction \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+\mathrm{CO}_{(g)} \rightarrow 2 \mathrm{FeO}_{(s)}+\mathrm{CO}_{2(g)}\) is

  1. +10.3 kJ
  2. -43.3 kJ
  3. -10.3 kJ
  4. +6.2kJ

Answer: 4. +6.2kJ

⇒ \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(g)}\); \(\Delta H=-26.8 \mathrm{~kJ}\)…..(1)

⇒ \(\mathrm{FeO}_{(s)}+\mathrm{CO}_{(g)} \rightarrow \mathrm{Fe}_{(s)}+\mathrm{CO}_{2(g)} ; \Delta H=-16.5 \mathrm{~kJ}\)….(2)

⇒ \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+\mathrm{CO}_{(g)} \rightarrow 2 \mathrm{FeO}_{(s)}+\mathrm{CO}_{2(g)} ; \Delta H=?
\)…(3)

Equation (3) can be obtained as: (1) -2(2) = \(-26.8-2(-16.5)=-26.8+33.0=+6.2 \mathrm{~kJ}\)

Question 30. For which one of the following equations is \(\Delta H^{\circ} \text { reaction }\) equal to \(\Delta H^{\circ}{ }_f\) Affreaction equal to ïH°f, for the product?

  1. \(\mathrm{N}_{2(g)}+\mathrm{O}_{3(g)} \rightarrow \mathrm{N}_2 \mathrm{O}_{3(g)}\)
  2. \(\mathrm{CH}_{4(g)}+2 \mathrm{Cl}_{2(g)} \rightarrow \mathrm{CH}_2 \mathrm{Cl}_{2(b)}+2 \mathrm{HCl}_{(g)}\)
  3. \(\mathrm{Xe}_{(g)}+2 \mathrm{~F}_{2(g)} \rightarrow \mathrm{XeF}_{4(g)}\)
  4. \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}\)

Answer: 3. \(\mathrm{Xe}_{(g)}+2 \mathrm{~F}_{2(g)} \rightarrow \mathrm{XeF}_{4(g)}\)

For (3), \(\Delta H_{\text {reaction }}^{\circ}\)

= \(\Delta H^{\circ}{ }_f\left(\mathrm{XeF}_4\right)-\left[\Delta H_f^{\circ}(\mathrm{Xe})+2 \Delta H^{\circ}{ }_f\left(\mathrm{~F}_2\right)\right]\)

Enthalpies of formation of elementary substances \(\mathrm{Xe}\) and \(\mathrm{F}_2\) are taken as zero.

Thus, \(\Delta H_{\text {reaction }}^{\circ}=\Delta H_f^{\circ}\left(\mathrm{XeF}_4\right)\)

Question 31. Heat of combustion \(\Delta H\) for \(\mathrm{C}_{(s)}, \mathrm{H}_{2(g)}\) and \(\mathrm{CH}_{4(\mathrm{~g})}\) are -94,-68 and -213 \(\mathrm{kcal} / \mathrm{mol}\), then \(\Delta H\) for \(\mathrm{C}_{(s)}+2 \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(g)}\) is

  1. -17 k cal
  2. -111 k cal
  3. -170 k cal
  4. -85 k cal

Answer: 1. -17 k cal

  1. \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H_{\mathrm{i}}=-94 \mathrm{kcal} / \mathrm{mole}\)
  2. \(2 \mathrm{H}_{2(\mathrm{gg})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{f})} ; \Delta H_{\mathrm{ii}}=-68 \times 2 \mathrm{kcal} / \mathrm{mole}\)
  3. \(\mathrm{CH}_{4(g)}+2 \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(l)} ; \Delta H_{\mathrm{iii}}=-213 \mathrm{kcal} / \mathrm{mole}\)
  4. \(\mathrm{C}_{(s)}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_{4(\mathrm{~g}} ; \Delta H_{\mathrm{iv}}=\)?

By applying Hess’s law, we can compute \(\Delta H_{\mathrm{iv}}\).

∴ \(\Delta H_{\mathrm{iv}}=\Delta H_{\mathrm{1}}+\Delta H_{\mathrm{1}}-\Delta H_{\mathrm{3}}\)

= \((-94-68 \times 2+213) \mathrm{kcal}=-17 \mathrm{kcal}\)

Question 32. Change in enthalpy for reaction, \(2 \mathrm{H}_2 \mathrm{O}_{2(l)} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(l)}+\mathrm{O}_{2(\mathrm{~g})}\) if the heat of formation of H2O2(l) and H2O(l) are -188 and -286 kJ/mol respectively, is

  1. -196 kJ/mol
  2. +196 kJ/mol
  3. +948 kJ/mol
  4. -948 kJ/mol

Answer: 1. -196 kJ/mol

∴ \(\Delta H_f^{\circ}=\Sigma H_{f \text { (products) }}^{\circ}-\Sigma H_{f \text { (reactants) }}^{\circ}\)

For the given reaction, \(2 \mathrm{H}_2 \mathrm{O}_{2(\mathrm{~J})} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{J})}+\mathrm{O}_{2(\mathrm{~g})}\)

∴ \(\Delta H_f^{\circ}=2 \times \Delta H_{f\left(\mathrm{H}_2 \mathrm{O}\right)}^{\mathrm{o})}-2 \times \Delta H_{f\left(\mathrm{H}_2 \mathrm{O}_2\right)}^{\mathrm{o}}\)

= \(2 \times-286 \mathrm{~kJ} \mathrm{~mol}^{-1}-2 \times(-188) \mathrm{kJ} \mathrm{mol}^{-1}\)

= \(-196 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 33. Enthalpy of \(\mathrm{CH}_4+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{OH}\) is negative. If the enthalpy of combustion of CH4 and CH3OH are x and y respectively, then which relation is correct?

  1. x>y
  2. x<y
  3. x=y
  4. x≥y

Answer: 1. x>y

Enthalpy of \(\mathrm{CH}_4+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{OH}\) is negative. If the enthalpy of combustion of CH4 and CH3OH are x and y respectively

⇒ \(\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}, \Delta H_1=-x\)…..(1)

⇒ \(\mathrm{CH}_3 \mathrm{OH}+\frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}, \Delta H_2=-y\)….(2)

Subtracting (2) from (1), we get \(\mathrm{CH}_4+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{OH}, \Delta H_3=- \text { ve }\)

i.e., \(-x-(-y)=-\mathrm{ve}\)

y – x = -ve. Hence, x>y.

Question 34. In the reaction \(\mathrm{S}+3 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3+2 x\) and \(\mathrm{SO}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3+y\) kcal the heat of formation of SO2 is

  1. (2x + y)
  2. (x – y)
  3. (x + y)
  4. (2x – y)

Answer: 4. (2x – y)

⇒ \(\mathrm{S}+\frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3+2 x \) kcal….(1)

⇒ \(\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3+y \) kcal….(2)

By subtracting equation (2) from (1) we get, \(\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2+(2 x-y)\) kcal

The heat of formation of \(\mathrm{SO}_2\) is \((2 x-y)\) kcal/mole.

Question 35. Given that \(\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2, \Delta H^{\circ}=-x \mathrm{~kJ}\); \(2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2, \Delta H^{\circ}=-y \mathrm{~kJ}\) The enthalpy of formation of carbon monoxide will be

  1. \(\frac{y-2 x}{2}\)
  2. \(2 x-y\)
  3. \(y-2 x\)
  4. \(\frac{2 x-y}{2}\)

Answer: 1. \(\frac{y-2 x}{2}\)

⇒ \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H^{\circ}=-x \mathrm{~kJ}\)….(1)

⇒ \(\mathrm{CO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H^{\circ}=\frac{-y}{2} \mathrm{~kJ}\)….(2)

By subtracting equation (2) from (1) we get, \(\mathrm{C}_{(s)}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(g)} ;\)

∴ \(\Delta H^{\circ}=-x-\left(-\frac{y}{2}\right)=\frac{y-2 x}{2} \mathrm{~kJ}\)

Question 36. If enthalpies of formation for C2H4(g), CO2(g) and H2O(l) at 25°C and 1 atm pressure are 52, -394 and -286 kJ/mol respectively, then enthalpy of combustion of C2H4(g) will be

  1. +141.2 kJ/mol
  2. +1412 kJ/mol
  3. -141.2 kJ/mol
  4. -1412 kJ/mol

Answer: 4. -1412 kJ/mol

⇒ \(\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

∴ \(\Delta H^{\circ}=\Delta H_{\text {products }}^o-\Delta H_{\text {reactants }}^{\circ}\)

= \(2 \times(-394)+2 \times(-286)-(52+0)=-1412 \mathrm{~kJ} / \mathrm{mol}\)

Question 37. The bond dissociation energies of X2, Y2 and XY are in the ratio of 1: 0.5: 1. ΔH for the formation of XY is -200 kJ mol-1. The bond dissociation energy of X2 will be

  1. 200 kJ mol-1
  2. 100 kJ mol-1
  3. 800 kJ mol-1
  4. 400 kJ mol-1

Answer: 3. 800 kJ mol-1

The bond dissociation energies of X2, Y2 and XY are in the ratio of 1: 0.5: 1. ΔH for the formation of XY is -200 kJ mol-1.

Let B.E. of \(X_2, Y_2\) and \(X Y\) are \(x \mathrm{~kJ} \mathrm{~mol}^{-1}, 0.5 x \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(x \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively.

⇒ \(\frac{1}{2} X_2+\frac{1}{2} Y_2 \rightarrow X Y ; \Delta H=-200 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\Delta H=\Sigma(B . E .)_{\text {Reactants }}-\Sigma(B . E .)_{\text {Products }}\)

∴ \(-200=\left[\frac{1}{2} \times(x)+\frac{1}{2} \times(0.5 x)\right]-[1 \times(x)]\)

B.E. of \(X_2=x=800 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 38. The heat of combustion of carbon to CO2 is -393.5 kJ/mol. The heat released upon the formation of 35.2 g of CO2 from carbon and oxygen gas is

  1. +315 kJ
  2. -630 kJ
  3. -3.15 kJ
  4. 315 kJ

Answer: None

Given: \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \longrightarrow \mathrm{CO}_{2(g)}, \Delta H=-393.5 \mathrm{~kJ} / \mathrm{mol}\)

Amount of heat released on formation of 44 g CO2 = 393.5kJ

∴ Amount of heat released on formation of 35.2 g of CO2

= \(\frac{393.5}{44} \times 35.2=314.8 \approx 315 \mathrm{~kJ}\)

-ve or +ve sign considering the reaction is exothermic or endothermic.

Question 39. When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0ºC and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in \(\mathrm{kJ}\left(\Delta H_{\text {comb }}\left(\mathrm{CH}_4\right)=890 \mathrm{~kJ} \mathrm{~mol}^{-1}\right.\), \(\left.\Delta H_{\text {comb }}\left(\mathrm{C}_3 \mathrm{H}_8\right)=2220 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) is\)

  1. 38
  2. 317
  3. 477
  4. 32

Answer: 2. 317

When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0ºC and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed.

⇒ \(\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \rightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}\)

Number of moles in gaseous mixture \(\mathrm{CH}_4+\mathrm{C}_3 \mathrm{H}_8=\frac{5}{22.4}=0.22 \text { moles }\)

Number of moles of \(\mathrm{O}_2=\frac{16}{22.4}=0.71 \mathrm{moles}\)

Let x moles of CH4 be there in a gaseous mixture so, a number of moles of C3H8 would be 0.22 – x.

Then moles of O2 consumed, 2x + (0.22 – x)5 = 0.71 or x = 0.13

The total amount of heat liberated = 0.13 x 890 + 0.09 x 22210 = 315.5 J

Question 40. Enthalpy change for the reaction, 4 \(\mathrm{H}_{(g)} \rightarrow 2 \mathrm{H}_{2(g)} \text { is }-869.6 \mathrm{~kJ}\) The-dissociation energy of H – H bond is

  1. -434.8 kJ
  2. -869.6 kJ
  3. +434.8 kJ
  4. +217.4 kJ

Answer: 3. +434.8 kJ

The dissociation energy of H-H bond is \(\frac{869.6}{2}=434.8 \mathrm{~kJ}\)

Question 41. From the following bond energies:

  1. H – H bond energy: 431.37 k) mol-1
  2. C: C bond energy: 606.10 kJ mol-1
  3. C – C bond energy: 336.49 kJ mol-1
  4. C – H bond energy: 410.50 kJ mol-1

Enthalpy for the reaction, will be

  1. -243.6 kJ mol-1
  2. -120.0 kJ mol-1
  3. 553.0 kJ mol-1
  4. 1523.6 kJ mol-1

Answer: 2. -120.0 kJ mol-1

For the given reaction, the enthalpy of the reaction can be calculated as

= \(\Sigma B \cdot E_{.}(\text {reactants })-\Sigma B \cdot E .(\text { products) }\)

= \(\left[B \cdot E_{(\mathrm{C}=\mathrm{C})}+B \cdot E_{(\mathrm{H}-\mathrm{H})}+4 \times B_{\cdot} E_{(\mathrm{C}-\mathrm{H})}\right]\)

∴ \(-\left[B \cdot E_{\cdot(\mathrm{C}-\mathrm{C})}+6 \times B \cdot E_{\cdot(\mathrm{C}-\mathrm{H})}\right]\)

= \([606.10+431.37+4 \times 410.50]-[336.49+6 \times 410.50]\)

= \(2679.47-2799.49=-120.02 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 42. Bond dissociation enthalpy of H2, Cl2, and HCI are 434,242 and 431 kJ mol-1 respectively. Enthalpy of the formation of HCl is

  1. -93 kJ mol-1
  2. 245 kJ mol-1
  3. 93 kJ mol-1
  4. -245 kJ mol-1

Answer: 1. -93 kJ mol-1

⇒ \(\mathrm{H}_2+\mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}\)

⇒ \(\Delta H_{\text {reaction }}=\Sigma(B \cdot E)_{\text {reactants }}-\Sigma(B \cdot E)_{\text {products }}\)

= \(\left[(B \cdot E)_{\mathrm{H}-\mathrm{H}}+(B \cdot E)_{\mathrm{Cl}-\mathrm{Cl}}\right]-\left[2 B \cdot E_{(\mathrm{H}-\mathrm{Cl})}\right]\)

= \(434+242-(431) \times 2\)

∴ \(\Delta H_{\text {reaction }}=-186 \mathrm{~kJ}\)

Question 43. Consider the following reactions:

  1. \(\mathrm{H}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-}=\mathrm{H}_2 \mathrm{O}_{(b)}, \Delta H=-X_1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  2. \(\mathrm{H}_{2(g)}+1 / 2 \mathrm{O}_{2(g)}=\mathrm{H}_2 \mathrm{O}_{(l)}, \Delta H=-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  3. \(\mathrm{CO}_{2(g)}+\mathrm{H}_{2(\mathrm{~g})}=\mathrm{CO}_{(g)}+\mathrm{H}_2 \mathrm{O}_{(l)}, \Delta H=-X_3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  4. \(\mathrm{C}_2 \mathrm{H}_{2(g)}+5 / 2 \mathrm{O}_{2(g)}=2 \mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(j)},\Delta H=+X_4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Enthalpy of the formation of H2O(l) is

  1. \(+X_3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  2. \(-X_4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  3. \(+X_1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  4. \(-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 4. \(-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The amount of heat absorbed or released when 1 mole of a substance is directly obtained from its constituent elements is called the heat of formation or enthalpy of formation.

Equation (1) represents the neutralisation reaction, (3) represents the hydrogenation reaction and (4) represents the combustion reaction.

Thus, enthalpy of formation of \(\mathrm{H}_2 \mathrm{O}_{(J)} \text {is }-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 44. Given that the bond energy of H – H and Cl – Cl are 430 kJ mol-1 and 240 kJ mol-1 respectively and ΔHf for HCl is -90 kJ mol-1, the bond enthalpy of HCl is

  1. 380 kJ mol-1
  2. 425 kJ mol-1
  3. 245 kJ mol-1
  4. 290 kJ mol-1

Answer: 2. 425 kJ mol-1

Given that the bond energy of H – H and Cl – Cl are 430 kJ mol-1 and 240 kJ mol-1 respectively and ΔHf for HCl is -90 kJ mol-1

∴ \(\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \rightarrow \mathrm{HCl}\)

ΔH = \(=\Sigma B \cdot E_{\text {(reactants) }}-\Sigma B \cdot E_{\text {(products) }}\)

= \(\frac{1}{2}\left[B \cdot E_{\left(\mathrm{H}_2\right)}+B \cdot E_{\cdot\left(\mathrm{Cl}_2\right)}\right]-B \cdot E_{(\mathrm{HCl})}=-90\)

∴ \(\frac{1}{2}(430+240)-B \cdot E_{\cdot(\mathrm{HCl})}=-90\)

∴ \(B \cdot E_{(\mathrm{HCl})}=\frac{1}{2}(430+240)+90=425 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 45. The absolute enthalpy of neutralisation of the reaction: \(]\mathrm{MgO}_{(s)}+2 \mathrm{HCl}_{(a q)} \rightarrow \mathrm{MgCl}_{2(a q)}+\mathrm{H}_2 \mathrm{O}_{(f)}\) will be

  1. -57.33 kJ mol-1
  2. Greater than -57.33 kJ mol-1
  3. Less than -57.33 kJ mol-1
  4. 57.33 kJ mol-1

Answer: 3. Less than -57.33 kJ mol-1

MgO is the oxide of a weak base and we know that the heat of neutralisation of 1 eq. of a strong acid with a strong base is -57.33 kJ/mol.

⇒ With a weak base, some heat is absorbed in the dissociation of the weak base.

⇒ The heat of neutralisation of weak base with strong acid will be less than -57.33kJ/mol