NEET Chemistry

Alcohols Phenols And Ethers

Question 1. The general molecular formula, which represents the homologous series of alkanols is

1. \(\mathrm{C}_n \mathrm{H}_{2 n+2} \mathrm{O}\)
2. \(\mathrm{C}_n \mathrm{H}_{2 n} \mathrm{O}_2\)
3. \(\mathrm{C}_n \mathrm{H}_{2 n} \mathrm{O}\)
4. \(\mathrm{C}_n \mathrm{H}_{2 n+1} \mathrm{O}\)

Answer: 1. \(\mathrm{C}_n \mathrm{H}_{2 n+2} \mathrm{O}\)

All alcohols follow the general formula \(\mathrm{C}_n \mathrm{H}_{2 \pi+2} \mathrm{O}\).

⇒ \(\begin{gathered}
\mathrm{CH}_3 \mathrm{OH}\left[\mathrm{CH}_{2+2} \mathrm{O}\right] ; \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\left[\mathrm{C}_2 \mathrm{H}_{(2 \times 2)+2} \mathrm{O}\right] \\
n=1, \quad n=2
\end{gathered}
\)

Question 2. Consider the following reaction and identify the product (P).

Answer: 3

Question 3. Which of the following will be most readily dehydrated under acidic conditions?

Answer: 4

Question 4. Given below are two statements:

Statement-1: The acidic strength of monosubstituted nitrophenol is higher than phenol because of the electron-withdrawing nitro group.

Statement 2: 0-nitrophenol, m-nitrophenol, and p-nitrophenol will have the same acidic strength as they have one nitro group attached to the phenolic ring. In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both statement-1 and statement-2 are correct.
2. Both statement-1 and statement-2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but Statement 2 is

Answer: 3. Statement 1 is correct but statement 2 is incorrect.

Read and Learn More NEET MCQs with Answers

Electron withdrawing groups (for example, -NO₂) stabilize the phenoxide ion more by dispersing the negative charge relative to phenol (i.e. release of proton becomes easy) and thus, increase the acidic strength of phenols.

The particular effect is more when the substituent is present on o- and p-positions than in re-position to the phenolic group.

Thus, the acidic strength of nitrophenols decreases in the order: p-nitrophenol > o-nitrophenol > z-nitrophenol > phenol

Question 5. Given below are two statements:

Statement 1: In the Lucas test, primary, secondary, and tertiary alcohols are distinguished on the basis of their reactivity with the cone. HCl + ZnCl₂, known as Lucas Reagent.

Statement 2: Primary alcohols are most reactive and immediately produced turbidity at room temperature on reaction with Lucas Reagent.

In the light of the above statements, choose the most appropriate answer from the options given below.

1. Both statement 1 and statement 2 are correct.
2. Both statement 1 and statement 2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but statement 2 is correct.

Answer: 3. Statement 1 is correct but statement 2 is incorrect.

Tertiary alcohols are most reactive and immediately produce turbidity at room temperature while primary alcohols do not react with Lucas reagent at room temperature.

Question 6. The reaction between acetone and methyl magnesium chloride followed by hydrolysis will give

1. isopropyl alcohol
2. sec-butyl alcohol
3. fert-butyl alcohol
4. iso-butyl alcohol.

Answer: 3. fert-butyl alcohol

Question 7. The structure of intermediate A in the following reaction is

Answer: 3

Question 8. When vapors of a secondary alcohol is passed over heated copper at 573 K, the product formed is

1. A carboxylic acid
2. An aldehyde
3. A ketone
4. An alkene.

Answer: 3. A ketone

Question 9. In the reaction,

the electrophile involved is

1. Dichloromethyl cation \(\left({ }_{\mathrm{C}}^{+} \mathrm{HCl}_2\right)\)
2. Formyl cation \((\stackrel{\rightharpoonup}{\mathrm{C}} \mathrm{HO})\)
3. Dichloromethyl anion \(\left(\overline{\mathrm{C}} \mathrm{HCl}_2\right)\)
4. Dichlorocarbene \((:\left.\mathrm{CCl}_2\right)\)

Answer: 4. Dichlorocarbene \((:\left.\mathrm{CCl}_2\right)\)

It is Reimer-Tiemann reaction. The electrophile formed is dichlorocarbene (CCl) which is formed according to the following mechanism

Question 10. Compound A, C₈H₁₀O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with a characteristic smell. A and Y are respectively

Answer: 3

As the compound is giving yellow precipitate with NaOI that shows it is undergoing a haloform reaction. Haloform reaction is shown by the compounds having

Question 11. Identify the major products P, Q, and R in the following sequence of reactions

Answer: 4

Question 12. Which one is the most acidic compound?

Answer: 3

Electron withdrawing groups increase the acidity while electron donating groups decrease the acidity of phenol.

Question 13. The reaction of phenol with chloroform in the presence of dilute sodium hydroxide finally introduces which one of the following functional groups?

1. -COOH
2. -CHCl₂
3. -CHO
4. -CH₂Cl

Answer: 3. -CHO

This is Reimer-Tiemann reaction.

Question 14. Which of the following reaction(s) can be used for the preparation of alkyl halides?

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HCl}\) \(\underrightarrow{\mathrm{Anh} . \mathrm{ZnCl} l_2}\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HCl} \longrightarrow\)
3. \(\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{HCl} \longrightarrow\)
4. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH}+\mathrm{HCl}\) \(\underrightarrow{\text { Anh. } \mathrm{ZnCl}_2}\)

1. 1 and 2 only
2. 4 only
3. 3 and 4 only
4. 1, 3 and 4 only

Answer: 4. 1, 3 and 4 only

1° and 2° alcohols react with HCI in the presence of an anhydrous ZICI₂ as a catalyst while in the case of 3° alcohols, ZnCI₂ is not required.

Quetsion 15. Which of the following will not be soluble in sodium hydrogen carbonate?

1. 2,4,6-Trinitrophenol
2. Benzoic acid
3. o-Nitropbenol
4. Benzene sulphonic acid

Answer: 3. o-Nitropbenol

The reaction is as follows: Acid + \(\mathrm{NaHCO}_3 \rightarrow\) Sodium salt of acid+ \(\mathrm{H}_2 \mathrm{CO}_3\)

Among all the given compounds, o-nitrophenol is a weaker acid than \(\mathrm{HCO}_3^{-}\). Hence, it does not react with \(\mathrm{NaHCO}_3\).

Question 16. The number of isomeric alcohols of molecular formula C₆H₁₄O which give positive iodoform test is

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

Answer: 2. Four

The iodoform test is positive for alcohols with the formula R – CHOH – CH₃.

Among C₆H₁₄O isomers, the ones with positive iodoform tests are:

Question 17. In the following sequence of reactions, the end product (C) is

1. Acetone
2. Methane
3. Acetaldehyde
4. Ethyl alcohol.

Answer: 4. Ethyl alcohol.

Quetsion 18. In the following reactions, the major products (A) and (C) are respectively

Answer: 2

Quetsion 19. Given are cyclohexanol(1), acetic acid(2), 2,4,6-trinitrophenol (3), and phenol (4). In these, the order of decreasing acidic character will be

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

Answer: 1. 3>2>4>1

Since phenols and carboxylic acids are more acidic than aliphatic alcohols, we find that cyclohexanol (1) is the least acidic. Out of the two given phenols, 3 is more acidic than 4.

This is because of the presence of three highly electron-withdrawing -NO₂ groups on the benzene ring which makes the O-H bond extremely polarized. This facilitates the release of H as H⁺.

Thus, 3 and 4.

ln acetic acid, the electron-withdrawing in the -COOH group polarises the O-H bond and increases the acidic strength. Acetic acid is therefore more acidic than phenol or cyclohexanol.

∴ The order of acidic character is 3 > 2 > 4 > 1

Question 20. Which of the following compounds has the most acidic nature?

Answer: 2

Phenol is the most acidic of all the given compounds. In phenol, the electron-withdrawing phenyl ring polarizes the O-H bond, thereby facilitating the release of H as H⁺ and hence, phenol is most acidic.

The electron-withdrawing effect of the phenyl ring is somewhat diminished by the – CH₂ group and it is therefore, less acidic than phenol. In (3) and (4), -OH group is attached to alkyl groups which, due to their +1 effect reduces the polarity of -OH bond and so, the acidic strength is low.

Question 21. Among the following four compounds

1. Phenol
2. Methyl phenol
3. Mefa-nitrophenol
4. Para-nitrophenol

The acidity order is

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

Answer: 1. 4>3>1>2

In phenols, the presence of electron-releasing groups decreases the acidity, whereas the presence of electron-withdrawing groups increases the acidity, compared to phenol.

Among the meta and para-nitrophenols, the latter is more acidic as the presence of – NO₂ in the group at the para position stabilizes the phenoxide ion to a greater extent than when it is present at the meta position.

Thus, correct order of acidity is: para -nitrophenol(4) > meta-nitrophenol(3) > phenol(1) > methyl phenol(2)

Quetsion 22. When glycerol is treated with excess of HI, it produces

1. 2-Iodopropane
2. Allyl Iodide
3. Propene
4. Glycerol Triiodide.

Answer: 1. 2-Iodopropane

Question 23. Consider the following reaction:

the product Z is

1. \(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3\)
2. \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}-\mathrm{SO}_3 \mathrm{H}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)
4. \(\mathrm{CH}_2=\mathrm{CH}_2\)

Answer: 3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)

Quetsion 24. HOCH₂CH₂OH on heating with periodic acid gives

Answer: 3

When 1,2-diol-like ethylene glycol is treated with HIO₂ each alcoholic group is oxidized to a carbonyl group by HIO₂. Since in glycol, both the -OH groups are terminal, so oxidation would yield two formaldehyde molecules.

Question 25. Consider the following reaction product Z is

1. Benzaldehyde
2. Benzoic acid
3. Benzene
4. Toluene.

Answer: 2. Benzoic acid

Quetsion 26. Ethylene oxide when treated with Grignard reagent yields

1. Primary alcohol
2. Secondary alcohol
3. Tertiary alcohol
4. Cyclopropyl alcohol.

Answer: 1. Primary alcohol

Quetsion 27. Which one of the following compounds is most acidic?

Answer: 3

Phenols are much more acidic than alcohols, due to the stabilization of phenoxide ion by resonance.

-NO₂ is the electron-withdrawing group and helps in stabilizing the negative charge on the oxygen hence equilibrium shifts in the forward direction, and more H⁺ are removed easily Hence, it is the most acidic.

CH₃ is the electron-donating group. Hence, electron density increases on the oxygen and destabilizes the product. Thus, equilibrium shifts in the backward direction.

Question 28. Which one of the following will not form a yellow precipitate on heating with an alkaline solution of iodine?

1. \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3\)
3. \(\mathrm{CH}_3 \mathrm{OH}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)

Answer: 3. \(\mathrm{CH}_3 \mathrm{OH}\)

The formation of a yellow precipitate on heating a compound with an alkaline solution of iodine is known as an iodoform reaction. Methyl alcohol does not respond to this test. Iodoform test is exhibited by ethyl alcohol, acetaldehyde, acetone, methyl ketones, and those alcohols that possess the CH₃CH(OH)- group.

Quetsion 29. n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?

1. PCl₅
2. Reduction
3. Oxidation with potassium dichromate
4. Ozonolysis

Answer: 3. Oxidation with potassium dichromate

n-Propyl alcohol on oxidation with potassium dichromate gives an aldehyde which on further oxidation gives an acid. Both aldehydes and acids contain the same number of C atoms as the original alcohol.

Isopropyl alcohol on oxidation gives a ketone with the same number of C atoms as the original alcohol.

Question 30. When phenol is treated with CHCl₃ and NaOH, the product formed is

1. Benzaldehyde
2. Salicylaldehyde
3. Salicylic acid
4. Benzoic acid.

Answer: 2. Salicylaldehyde

This reaction is called the Reimer-Tiemann reaction

Question 31. Which of the following is correct?

1. On reduction, any aldehyde gives secondary alcohol.
2. The reaction of vegetable oil with H₂SO₄ gives glycerine.
3. Alcoholic iodine with NaOH gives iodoform.
4. Sucrose in reaction with NaCl gives inverted sugar.

Answer: 3. Alcoholic iodine with NaOH gives iodoform.

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+4 \mathrm{I}_2+\mathrm{NaOH} \longrightarrow \mathrm{CHI}_3+\mathrm{NaI}\)

+ \(\mathrm{HCOONa}+\mathrm{H}_2 \mathrm{O}\)

Iodoform is a pale yellow solid which crystallizes in hexagonal plates.

Question 32. The correct acidic order of the following is

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

Answer: 2. 3>1>2

Phenol exists as a resonance hybrid of the following structures.

Thus, due to resonance, the oxygen atom of the – OH group acquires a positive charge and hence attracts the electron pair of the O-H bond leading to the release of a hydrogen atom as a proton.

Once the phenoxide ion is formed it stabilises itself by resonance which is more stable than the parent phenol as there is no charge separation.

Effect of substituent: The presence of electron withdrawing groups (-NO₂, -X, -CN) increases the acidity of phenols while the presence of electron releasing groups (-NH₂ -CH₃) decreases the acidity of phenols.

This explains the following order of acidity: p-nitrophenol > phenol >p-cresol.

Question 33. The reaction of   with RMgX leads to the formation of

Answer: 1

Question 34. When 3,3-dimethyl-2-butanol is heated with H₂SO₄, the major product obtained is

1. 2,3-dimethyl-2-butene
2. cis and trans isomers of 2,3-dimethyl-2-butene
3. 2,3-dimethyl-1-butene
4. 3,3-dimethyl 1-butene

Answer: 1. 2,3-dimethyl-2-butene

Question 35. The alkene R – CH = CH₂ reacts readily with B₂H₆ and the product on oxidation with alkaline hydrogen peroxides produces

Answer: 4

Question 36. On heating glycerol with cone. H₂SO₄, a compound is obtained which has a bad odor. The compound is

1. Acrolein
2. Formic acid
3. Allyl alcohol
4. Glycerol sulfate.

Answer: 1. Acrolein

Question 37. Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether, due to the presence of

1. H-bonding in ethanol
2. H-bonding in dimethyl ether
3. CH₃ group in ethanol
4. CH₃ group in dimethyl ether.

Answer: 1. H-bonding in ethanol

Question 38. Increasing order of acid strength among p-methoxyphenyl,p-methyl phenol and p-nitrophenol is

1. p-nitrophenol, p-methoxyphenyl, p-methyl phenol
2. p-methyl phenol, p-methoxyphenyl, p-nitrophenol
3. p-nitrophenol, p-methyl phenol, p -methoxyphenyl
4. p-methoxyphenyl, p-methylphenol, p-nitrophenol.

Answer: 4. p-methoxyphenyl, p-methylphenol, p-nitrophenol.

– OCH₃, – CH₃ are electron donating groups and decrease the acidic character of phenols. -NO₂, is an electron-withdrawing group and tends to increase the acidic character. The electron-donating effect of -OCH₃ group (+R effect) is more than that of – CH₃ group (+1 effect). Thus, the order is p-methoxyphenol < p-methylphenol < p-nitrophenol.

Quetsion 39. Which one of the following on oxidation gives a ketone?

1. Primary alcohol
2. Secondary alcohol
3. Tertiary alcohol
4. All of these

Answer: 2. Secondary alcohol

2° alcohols on oxidation give ketones, 1° alcohols form aldehydes.

Question 40. What is formed when a primary alcohol undergoes catalytic dehydrogenation?

1. Aldehyde
2. Ketone
3. Alkene
4. Acid

Answer: 1. Aldehyde

Primary alcohol undergoes catall.tic dehydrogenation to give aldehyde.

Question 41. How many isomers of C₅H₁₁OH will be primary alcohols?

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

Answer: 2. 4

4-isomers are possible for C₅H₁₁OH.

Question 42. HBr reacts fastest with

1. 2-methyl propan-1-ol
2. methyl propan-2-ol
3. propan-2-ol
4. propan-l-ol.

Answer: 2. methylpropan-2-ol

Generates 3° carbocation which is a very stable intermediate, thus it will react more rapidly with HBr.

Question 43. When phenol is treated with excess bromine water. It gives

1. m-bromophenol
2. o- and p-bromophenols
3. 2,4-bromophenol
4. 2,4,6-tribromophenol.

Answer: 4. 2,4,6-tribromophenol.

Phenol in reaction with excess bromine water gives 2,4,6 – tribromophenol.

Question 44. The compound which reacts fastest with Lucas reagent at room temperature is

1. Butan-1-ol
2. Butan-2-ol
3. 2-Methylpropan-1-ol
4. 2-Methylpropan-2-ol.

Asnwer: 4. 2-Methylpropan-2-ol.

2-Methylpropan-2-ol reacts rapidly with Lucas reagent at room temperature.

Quetsion 45. Which one of the following compounds will be most readily attacked by an electrophile?

1. Chlorobenzene
2. Benzene
3. Phenol
4. Toluene

Answer: 3. Phenol

-OH group being an electron donor increases the electron density in phenol. Thus, the electron density in phenol is higher than that of toluene, benzene, and chlorobenzene.

Question 46. Propene, CH₃CH=CH₂ can be converted into 1-propanol by oxidation. Indicate which set of reagents amongst the following is ideal for the above conversion.

1. KMnO₄ (alkaline)
2. Osmium tetroxide (OSO₄/CH₂Cl₂)
3. B₂H₆ and alk.H₂O₂
4. O₃/Zn

Answer: 3. B₂H₆ and alk.H₂O₂

Question 47. Phenol is heated with CHCl₃ and aqueous KOH when salicylaldehyde is produced. This reaction is known as

1. Rosenmund’s reaction
2. Reimer-Tiemann reaction
3. Friedel-Crafts reaction
4. Sommelet reaction.

Answer: 2. Reimer-Tiemann reaction

Treatment of phenol with CHCI₃ and aqueous hydroxide introduces the – CHO group, onto the aromatic ring generally ortho to the – OH group. This reaction is known as Reimer-Tiemann reaction.

Question 48. Lucas’s reagent is

1. cone. HCl and anhydrous ZnCl₂
2. cone. HNO₃ and hydrous ZnCl₂
3. cone. HCl and hydrous ZnCl₂
4. cone. HNO₃ and anhydrous ZnCl₂

Answer: 1. cone. HCl and anhydrous ZnCl₂

Question 49. Methanol is industrially prepared by

1. Oxidation of CH₄ by steam at 900°C
2. Reduction of HCHO using LiAIH₄
3. The reaction of HCHO with a solution of NaOH
4. Reduction of CO using H₂ and ZnO-Cr₂O₃.

Answer: 4. Reduction of CO using H₂ and ZnO-Cr₂O₃.

⇒ \(\mathrm{CO}+2 \mathrm{H}_2\) \(\underrightarrow{\mathrm{ZnO}-\mathrm{Cr}_2}\) \({\mathrm{O}_3} \mathrm{CH}_3 \mathrm{OH}\)

Question 50. Consider the following reaction

Answer: 1

Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to a more stable aryl-oxygen bond. The reaction yields phenol and alkyl halide.

Question 51. Anisole on cleavage with HI gives

Answer: 1

Question 52. The compound that is most difficult to protonate is

Answer: 1

The lone pair of oxygen is in conjugation with the phenyl group so, it is the least basic among the given compounds and is most difficult to protonate.

Question 53. The major products C and D formed in the following reactions respectively are \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\) \(\underrightarrow{\text { excess } \mathrm{HI}}\) C+D

1. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I}\) and \(\mathrm{I}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)
2. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}\) and \(\mathrm{I}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)
3. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I}\) and \(\mathrm{HO}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)
4. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}\) and \(\mathrm{HO}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)

Answer: 1. \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I}\) and \(\mathrm{I}-\mathrm{C}\left(\mathrm{CH}_3\right)_3\)

Ethers are readily attacked by HI to give an alkyl halide and alcohol. But when heated with excess of HI, the product alcohol first formed reacts further with HI to form the corresponding alkyl iodide.

⇒ \(\mathrm{R}-\mathrm{O}-\mathrm{R}^{\prime}+\underset{\text { (excess) }}{2 \mathrm{HI}}\) \(\underrightarrow{\text { (Heat) }}\) \(\mathrm{RI}+\mathrm{R}^{\prime} \mathrm{I}+\mathrm{H}_2 \mathrm{O}\)

Question 54. The heating of phenyl methyl ether with HI produces

1. Iodobenzene
2. Phenol
3. Benzene
4. Ethyl chloride.

Answer: 2. Phenol

In case of phenyl methyl ether, methyl phenyl oxonium ion is formed by protonation of ether. The O-CH₃ bond is weaker than O-C₆H₅ bond as O-C₆H₅ has a partial double bond character. Therefore, the attack by I^– ion breaks O-CH₃ bond to form CH₃l.

Question 55. The reaction can be classified as

1. Dehydration reaction
2. Williamson alcohol synthesis reaction
3. Williamson ether synthesis reaction
4. Alcohol formation reaction.

Answer: 3. Williamson ether synthesis reaction

Williamson ether synthesis reaction involves the treatment of sodium alkoxide with a suitable alkyl halide to form an ether.

Question 56. The reaction is called

1. Etard reaction
2. Gattermann-Koch reaction
3. Williamson synthesis
4. Williamson continuous etherification process.

Answer: 3. Williamson synthesis

Williamson synthesis is the best method for the preparation of ether.

Question 57. Among the following sets of reactants which one produces anisole?

1. \(\mathrm{CH}_3 \mathrm{CHO} ; \mathrm{RMgX}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} ; \mathrm{NaOH} ; \mathrm{CH}_3 \mathrm{I}\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\); neutral \(\mathrm{FeCl}_3\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3 ; \mathrm{CH}_3 \mathrm{COCl} ; \mathrm{AlCl}_3\)

Answer: 2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} ; \mathrm{NaOH} ; \mathrm{CH}_3 \mathrm{I}\)

Question 58. Identify Z in the sequence of reactions: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2\) \(\underrightarrow{\mathrm{HBr} / \mathrm{H}_2 \mathrm{O}_2}\) Y \(\underrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}} \) Z

1. \(\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_3-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)
2. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)
3. \(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4-\mathrm{O}-\mathrm{CH}_3\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)

Answer: 1. \(\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_3-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3\)

Question 59. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?

Answer: 1

Question 60. In the reaction, which of the following compounds will be formed?

Answer: 3

The alkyl iodide produced depends on the nature of the alkyl groups. If one group is Me and the other a primary or secondary alkyl group, it is methyl iodide that is produced. This can be explained by the assumption that the mechanism is S_N2, and because of the steric effect of the larger group, I^– attack the smaller methyl group.

When the substrate is a methyl t-alkyl ether, the products are t-RI and MeOH. This can be explained by S_N1 mechanism, the carbonium ion produced being the t-alkyl since a tertiary carbonium ion is more stable than a primary or secondary carbonium ion.

Question 61. The major organic product in the reaction is \(\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}\left(\mathrm{CH}_3\right)_2+\mathrm{HI} \longrightarrow\) products

Answer: 1

With cold HI, a mixture of alkyl iodide and alcohol is formed. In the case of mixed ethers, the halogen atom attaches to a smaller and less complex alkyl group.

CH₃OCH(CH₃)₂ + HI → CH₃I + (CH₃)₂CHOH

Question 62. Ethyl chloride is converted into diethyl ether by

1. Perkins reaction
2. Grignard reaction
3. Wurtz synthesis
4. Williamson synthesis.

Answer: 4. Williamson synthesis.

⇒ \(\mathrm{C}_2 \mathrm{H}_5-\mathrm{Cl}+\mathrm{Na}-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5 \longrightarrow \mathrm{C}_2 \mathrm{H}_5-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5+\mathrm{NaCl}\)

Question 63. Which one of the following compounds is resistant to nucleophilic attack by hydroxyl ions?

1. Diethyl ether
2. Acetonitrile
3. Acetamide
4. Methyl acetate

Answer: 1. Diethyl ether

Diethyl ether is a saturated compound, so it is resistant to nucleophilic attack by a hydroxyl ion (OH^–). Other compounds have unsaturation and the unsaturated ‘C’ atom bears partial +ve charge, therefore they undergo easy nucleophilic attack by OH^– ion.

Question 64. The compound which does not react with sodium is

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

Answer: 4. \(\mathrm{CH}_3 \mathrm{OCH}_3\)

Ethers are very inert. The chemical inertness of others is due to the absence of an active group in their molecules’ Since CH₃ – O – CH₃ is inert and does not contain an active group, therefore it does not react with sodium

Question 65. Which one is formed when sodium phenoxide is heated with ethyl iodide?

1. Phenetole
2. Ethyl phenyl alcohol
3. Phenol
4. None of these

Answer: 1. Phenetole

Phenetole is formed when sodium phenoxide is heated with ethyl iodide.

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{ONa}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}\) \(\underrightarrow{\Delta}\) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5\)

 

The Solid State

Question 1. The pure crystalline substance on being heated gradually first forms a turbid liquid at constant temperature and still at higher temperature turbidity completely disappears. The behaviour is a characteristic of substance-forming

1. Allotropic crystals
2. Liquid crystals
3. Isomeric crystals
4. Isomorphous crystals.

Answer: 2. Liquid crystals

Liquid crystals on heating first become turbid and then on further heating turbidity completely disappears.

Question 2. Glass is a

1. Liquid
2. Solid
3. Supercooled liquid
4. Transparent organic polymer.

Answer: 3. Supercooled liquid

Glass is a supercooled liquid which forms a noncrystalline solid without a regular lattice.

Question 3. Most crystals show good cleavage because their atoms, ions or molecules are

1. Weakly bonded together
2. Strongly bonded together
3. Spherically symmetrical
4. Arranged in planes.

Answer: 4. Arranged in planes.

Crystals show good cleavage because their constituent particles are arranged in planes.

Question 4. The ability of a substance to assume two or more crystalline structures is called

1. Isomerism
2. Polymorphism
3. Isomorphism
4. Amorphism.

Answer: 2. Polymorphism

The phenomenon of the existence of a substance in two or more crystalline structures is called polymorphism.

Question 5. Cation and anion combine in a crystal to form the following type of compound

1. Ionic
2. Metallic
3. Covalent
4. Dipole-dipole.

Answer: 1. The electrostatic force of attraction which exists between oppositely charged ions is called an ionic bond.

Question 6. For two ionic solids CaO and KI, identify the wrong statement among the following.

1. CaO has a high melting point.
2. The lattice energy of CaO is much larger than that of KI.
3. KI has a high melting point.
4. KI is soluble in benzene.

Answer: 4. KI is soluble in benzene.

KI is an ionic compound while benzene is not.

Read and Learn More NEET MCQs with Answers

Question 7. The correct option for the number of body-centred unit cells in all 14 types of Bravais lattice unit cells is

1. 3
2. 7
3. 5
4. 2

Answer: 1. 3

Out of 14 types of Bravaisiattice, three body-centred units crisis are there which are: orthorhombic, tetragonal and cubic

Question 8. For the orthorhombic system, axial ratios are α = β = γ = 90° and the axial angles are

1. α = β = γ ≠ 90°
2. α = β = γ = 90°
3. α = γ = 90°, β ≠ 90°
4. α ≠ β ≠ γ ≠ 90°

Answer: 2. α = β = γ = 90°

For orthorhombic system, α = β = γ = 90°

Question 9. The number of carbon atoms per unit cell of the diamond unit cell is

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

Answer: 4. 8

Diamond is like ZnS (Zincblende).

Carbon forms ccp (fcc) and also occupies half of the tetrahedral voids.

Total no. of carbon atoms per unit cell = 8 x 1/8 + 6 x 1/2 + 4 = 8

Question 10. In a face-centred cubic lattice, a unit cell is shared equally by how many unit cells?

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

Answer: 3. 6

Here given unit cell is shared equally by six faces in the fcc which is shared equally by six different unit cells.

Question 11. When Zn converts from its melted state to its solid state, it has a hcp structure and then finds the number of nearest atoms.

1. 6
2. 8
3. 12
4. 4

Answer: 3. 12

hcp is a closed-packed arrangement in which the unit cell is hexagonal and the coordination number is 12

Question 12. The fee crystal contains how many atoms in each unit cell?

1. 6
2. 8
3. 4
4. 5

Answer: 3. 4

The contribution of eight atoms of face-centred cubic unit cell = 8 x 1/8 = 1 atom. There is one atom at each of the six faces, which is shared by 2 unit cells each.

The contribution of 6 face-centred atoms = 6 x 1/2 = 3

Therefore n = L + 3 = 4

Question 13. The number of atoms contained in a fee unit cell of a monatomic substance is

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

Answer: 3. 4

fcc crystal contains = 8 x 1/8 + 6 x 1/2 = 4 atoms in a unit cell

Question 14. A compound is formed by two elements A and B. The element B forms a cubic close-packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A_x B_y, then the value of x + y is in the option

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

Answer: 3. 5

Let, the number of atoms in ccp unit cell = N

Thus, the number of B atoms = N

Number of tetrahedral voids = 2N

Number of A atoms = \(\frac{2 N}{3}\)

Molecular ratio A:B = \(\frac{2 N}{3}\)

The formula of the compound is A₂B₂

Hence, x+y=2+3=5

Question 15. What fraction of one edge-centred octahedral void lies in one unit cell of fcc?

1. 1/4
2. 1/12
3. 1/2
4. 1/3

Answer: 1. 1/4

There is an octahedral void at each edge of the unit cell such that 1/4th of the void is in one cell.

Question 16. The right option for the number of tetrahedral and octahedral voids in the hexagonal primitive unit cells are

1. 12,6
2. 8,4
3. 6,12
4. 2, 1

Answer: 1. 12,6

Number of atoms in hexagonal unit cell = N = 6

Number of octahedral void = N = 6

Number of tetrahedral void = 2N = 12

Question 17. A compound is formed by cation C and anion A. Hie anions form hexagonal close-packed (hep) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is

1. \(C_4 A_3\)
2. \(C_2 A_3\)
3. \(C_3 A_2\)
4. \(C_3 A_4\)

Answer: 4. \(C_3 A_4\)

Number of atoms per unit cell in hcp = 6

Number of octahedral voids in hcp = 6

Number of anions per unit cell = 6

Number of octahedral voids occupied by cations = 6 x \(\frac{75}{100}\) = \(\frac{9}{2}\)

∴ Formula of compound = C_9/2A₆ = C₃A₄

Question 18. In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca²⁺) and fluoride ion (F^–) are

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

Answer: 3. 8 and 4

In fluorite structure, Ca²⁺ ions are in the face-centred cubic arrangement. Each Ca²⁺. is connected to 4 F^– ions below it and to another set of 4 F^– ions above it i.e. Ca²⁺ has a coordination number of 8 and each F¹ ion has a coordination number 4.

Question 19. The ionic radii of A⁺ and B^– ions are 0.98 x 10^-10 m and 1.81 x 10^-10 m. The coordination number of each ion in AB is

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

Answer: 3. 6

Radius ration, \(\frac{r_{+}}{r_{-}}=\frac{0.98 \times 10^{-10}}{1.81 \times 10^{-10}}=0.54\)

It lies in the range of 0.414 to 0.732 hence, the coordination number of each ion will be 6 as the compound will have NaCl type structure i.e., octahedral arrangement.

Question 20. The number of octahedral voids (s) per atom present in a cubic close-packed structure is

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

Answer: 1. 1

The number of octahedral voids is the same as the number of atoms

Question 21. The structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is

1. \(\mathrm{ABO}_2\)
2. \(A_2 B_2\)
3. \(A_2 B_3 \mathrm{O}_4\)
4. \(A B_2 \mathrm{O}_2\)

Answer: 4. \(A B_2 \mathrm{O}_2\)

Number of atoms in ccp = 4= O^2-

Number of tetrahedral voids = 2 x N = 2 x 4 = B

Number of A²⁺ ions = 8 x 1/4 = 2

Number of octahedral voids = Number of B⁺ ions = N = 4

Ratio, O^2-: A²⁺ : Bn = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB₂A₂

Question 22. A solid compound XY has a NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y^–) will be

1. 275.1 pm
2. 322.5 pm
3. 241.5 pm
4. 165.7 pm

Answer: 3. 241.5 pm

For NaCl, \(\frac{r_{+}}{r_{-}}=0.414\)

Given: radius of cation =100pm

⇒ \(\frac{100}{r_{-}}=0.414 \Rightarrow \frac{100}{0.414}=r_{-} \Rightarrow r_{-}=241.5 \mathrm{pm}\)

Question 23. A compound formed by elements X and Y crystallises in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face- centres. The formula of the compound is

1. XY₃
2. X₃Y
3. XY
4. XY₂

Answer: 1. XY₃

In a unit cell, X atoms at the corners = \(\frac{1}{8}\) = 1

Y atoms at the face centres = \(\frac{1}{2}\) x 6 = 3

The ratio of X and Y = 1 : 3. Hence formula is XY

Question 24. In a cube of any crystal, an A-atom is placed at every corner and a B-atom is placed at every centre of the face. The formula of a compound is

1. AB
2. AB₃
3. A₂B₂
4. A₂B₃

Answer: 2. AB₃

‘A’ atoms are at ‘8’ corners of the cube.

Thus, no. of ‘A’atoms per unit cell = 8 x \(\frac{1}{8}\) = 1

‘B’ atoms are at the face centre of six faces. Thus, no. of ‘B’ atoms per unit cell = 6 x \(\frac{1}{2}\) = 3

The formula is AB₃

Question 25. In crystals of which one of the following ionic compounds would you expect the maximum distance between centres of cations and anions?

1. Csl
2. CsF
3. LiF
4. Lil

Answer: 1. Csl

As Cs⁺ ion has a larger size than Li⁺ and I^– has a larger size than F^–, so maximum distance between centres of cations and anions is in CsI.

Question 26. The second-order Bragg diffraction of X-rays with λ = 1.00 Å from a set of parallel planes in a metal occurs at an angle of 60°. Die distance between the scattering planes in the crystal is

1. 2.00 Å
2. 1.00 Å
3. 0.575 Å
4. 1.15 Å

Answer: 4. 1.15 Å

According to Bragg’s equation, nλ = 2d sin θ

As, n = 2λ = 1.00Å, θ = 60°, d = ?

2dsinθ = nλ

2dsin60°=2×1Å

2d x \(\frac{\sqrt{3}}{2}=2 \Rightarrow d=\frac{2}{\sqrt{3}}=1.15\) Å (because \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\))

Question 27. The intermetallic compound LiAg crystallizes in the cubic lattice in which both lithium and silver have a coordination number of eight. The crystal class is

1. Face-centred cube
2. Simple cube
3. Body-centred, cube
4. None of these.

Answer: 3. Body-centred, cube

A body-centred cubic unit cell consists of 8 atoms at the corners and one atom at the centre.

Question 28. In the fluorite structure, the coordination number of Ca²⁺ ions is

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

Answer: 3. 8

In fluorite (CaF₂) structure, C.N. of Ca²⁺ =8, C.N. of F^– = 4

Question 29. An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is

1. \(\frac{\sqrt{3}}{4} \times 288 \mathrm{pm}\)
2. \(\frac{\sqrt{2}}{4} \times 288 \mathrm{pm}\)
3. \(\frac{4}{\sqrt{3}} \times 288 \mathrm{pm}\)
4. \(\frac{4}{\sqrt{2}} \times 288 \mathrm{pm}\)

Answer: 1. \(\frac{\sqrt{3}}{4} \times 288 \mathrm{pm}\)

For bcc structure, \(r=\frac{\sqrt{3}}{4} a\), where a is the unit cell edge length and r is the radius of the sphere (atom).

r = \(\frac{\sqrt{3}}{4} \times 288 \mathrm{pm}\)

Question 30. The Vacant space in bcc lattice unit cell is

1. 48%
2. 23%
3. 32%
4. 26%

Answer: 3. 32%

Packing efficiency of bcc lattice = 68%

Hence, empathy space = 32%

Question 31. If a is the length of the side of a cube, the distance between the body-centred atom and one corner atom in the cube will be

1. \(\frac{2}{\sqrt{3}} a\)
2. \(\frac{4}{\sqrt{3}} a\)
3. \(\frac{\sqrt{3}}{4} a\)
4. \(\frac{\sqrt{3}}{2} a\)

Answer: 4. \(\frac{\sqrt{3}}{2} a\)

The distance between the body-centred atom and one corner atom is \(\frac{\sqrt{3} a}{2}\) i.e. half of the body diagonal.

Question 32. A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is

1. 288 pm
2. 408 pm
3. 144 pm
4. 204 pm

Answer: 1. 288 pm

For a face-centred cubic (fcc) structure.

r = \(\frac{a}{2 \sqrt{2}}, a=408 \mathrm{pm}, r=\frac{408}{2 \sqrt{2}}=144 \mathrm{pm}\)

Diameter = 2r = 2 x 144 = 288 Pm

Question 33. AB crystallizes in a body-centred cubic lattice with edge length ‘a’ equal to 387 pm. The distance between two oppositely charged ions in the lattice is

1. 335 pm
2. 250 pm
3. 200 pm
4. 300 pm

Answer: 1. 335 pm

For a bcc latlice, 2(r₊+r_–)= √3a

where r₊ = radius of cation, r_– = radius of anion

a = edge length

∴ \(\left(r_{+}+r_{-}\right)=\frac{\sqrt{3} \times 387}{2}=335.15 \mathrm{pm}\) = 335 Pm

Question 34. Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of lithium will be

1. 151.8 pm
2. 75.5 pm
3. 300.5 pm
4. 240.8 pm

Answer: 1. 151.8 pm

Since Li crystallises in body-centred cubic crystal atomic radius,

r = \(\frac{\sqrt{3} a}{4} \quad(a=\text { edge length })\)

∴ r = \(\frac{\sqrt{3}}{4} \times 351=151.8 \mathrm{pm}\)

Given: a=351 pm

Question 35. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of a copper atom in pm?

1. 157
2. 181
3. 108
4. 128

Answer: 4. 128

Since Cu crystallises in a face-centred cubic lattice.

Atomic radius r = \(\frac{a}{2 \sqrt{2}}\) (a = edge length = 361 pm)

∴ r = \(\frac{361}{2 \sqrt{2}}=127.6=128 \mathrm{pm}\)

Question 36. Which of the following statements is not correct?

1. The number of carbon atoms in a unit cell of diamond is 8.
2. The number of Bravais lattices in which a crystal can be categorized is 14.
3. The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.
4. Molecular solids are generally volatile.

Answer: 3. The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.

The packing fraction for a cubic unit cell is given by f = \(\frac{Z \times \frac{4}{3} \pi r^3}{a^3}\).

Question 37. If a stands for the edge length of the cubic systems: simple cubic, body-centred cubic and face-centred cubic, then the ratio of radii of the spheres in these systems will be respectively

1. \(\frac{1}{2} a: \frac{\sqrt{3}}{2} a: \frac{\sqrt{2}}{2} a\)
2. \(1 a: \sqrt{3} a: \sqrt{2} a\)
3. \(\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a\)
4. \(\frac{1}{2} a: \sqrt{3} a: \frac{1}{\sqrt{2}} a\)

Answer: 3. \(\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a\)

For simple cubic: r = a/2

For body centred: r = \(a \sqrt{3} / 4\)

For face-centred: r = \(\frac{a}{2 \sqrt{2}}\)

where a = edge length, r = radius

∴ The ratio of radii of the three will be \(\frac{a}{2}: \frac{a \sqrt{3}}{4}: \frac{a}{2 \sqrt{2}}\)

Question 38. The fraction of the total volume occupied by the atoms present in a simple cube is

1. \(\frac{\pi}{3 \sqrt{2}}\)
2. \(\frac{\pi}{4 \sqrt{2}}\)
3. \(\frac{\pi}{4}\)
4. \(\frac{\pi}{6}\)

Answer: 4. \(\frac{\pi}{6}\)

Question 39. The pyknometric density of sodium chloride crystal is 2.165 x 10³ kg m^-3 while its X-ray density is 2.178 x 10³ kg m^-3. The fraction of unoccupied sites in sodium chloride crystals is

1. 5.96
2. 5.96 x 10^-2
3. 5.96 x 10^-1
4. 5.96 x 10^-3

Answer: 4. 5.96 x 10^-3

Molar volume from psychometric density = \(\frac{M}{2.165 \times 10^3} \mathrm{~m}^3\)

Molar volume from X-ray density = \(\frac{M}{2.178 \times 10^3} \mathrm{~m}^3\)

Volume occupied = \(\frac{M}{10^3}\left(\frac{1}{2.165}-\frac{1}{2.178}\right) \mathrm{m}^3\)

Fraction unoccupied

= \(\left(\frac{0.013 M \times 10^{-3}}{2.165 \times 2.178}\right) /\left(\frac{M \times 10^{-3}}{2.165}\right)=5.96 \times 10^{-3}\)

Question 40. The edge length of face-centred unit cubic cells is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is

1. 144 pm
2. 398 pm
3. 288 pm
4. 618 pm

Answer: 1. 144 pm

In the face-centred cubic lattice, the unit cell, a = r + 2R + r

where r = Radius of cation, R = Radius of anion

⇒ 508 = 2 x 110 + 2R

∴ R = 144Pm

Question 41. Copper crystallises in a fcc unit cell with a cell edge length of 3.608 x 10^-8 cm. The density of copper is 8.92 g cm^-3. Calculate the atomic mass of copper.

1. 63.1 u
2. 31.55 u
3. 60 u
4. 65 u

Answer: 1. 63.1 u

Density of unit cell

d = \(\frac{Z \times M}{N_0 \times a^3}\)

Given, a = \(3.608 \times 10^{-8} \mathrm{~cm}\)

d = \(8.92 \mathrm{~g} / \mathrm{cm}^3\)

Z = 4(for fcc)

M = \(\frac{N_0 \times a^3 \times d}{Z}=\frac{6.023 \times 10^{23} \times\left(3.608 \times 10^{-8}\right)^3 \times 8.92}{4}\)

= \(63.08 \mathrm{u} \approx 63.1 \mathrm{u}^4\)

Question 42. Iron exhibits bee structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of the density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remain constant with temperature) is

1. \(\frac{\sqrt{3}}{\sqrt{2}}\)
2. \(\frac{4 \sqrt{3}}{3 \sqrt{2}}\)
3. \(\frac{3 \sqrt{3}}{4 \sqrt{2}}\)
4. \(\frac{1}{2}\)

Answer: 3. \(\frac{3 \sqrt{3}}{4 \sqrt{2}}\)

For bcc lattice: \(Z=2, a=\frac{4 r}{\sqrt{3}}\)

For fcc lattice : Z=4, a = \(2 \sqrt{2}r\)

∴ \(\frac{d_{R T}}{d_{900^{\circ} \mathrm{C}}}=\frac{\left(\frac{Z M}{N_A a^3}\right)_{b c c}}{\left(\frac{Z M}{N_A a^3}\right)_{f c c}}\)

Given, molar mass and atomic radii are constant.

= \(\frac{2}{4}\left(\frac{2 \sqrt{2} r}{\frac{4 r}{\sqrt{3}}}\right)^3=\frac{3 \sqrt{3}}{4 \sqrt{2}}\)

Question 43. Lithium has a bcc structure. Its density is 530 kg m^-3 and its atomic mass is 6.94 g mol^-1. Calculate the edge length of a unit cell of lithium metal. (N_A = 6.02 x 10²³ mol^-1)

1. 527 pm
2. 264 pm
3. 154 pm
4. 352 pm

Answer: 4. 352 pm

For bcc, Z = 2, ρ = 530 kg m^-3,

at, mass of Li = 6.94 g mol^-1, N_A = 6.02 x 10²³ mol^-1

⇒ \(\rho=530 \mathrm{~kg} \mathrm{~m}^{-3}=\frac{530 \times 1000 \mathrm{~g}}{1 \times(100)^3 \mathrm{~cm}^3}=0.53 \mathrm{~g} \mathrm{~cm}^{-3}\)

⇒ \(\rho=\frac{Z \times \text { At. mass }}{N_A \times a^3}\)

⇒ \(a^3=\frac{Z \times \text { At. mass }}{N_A \times \rho}=\frac{2 \times 6.94}{6.02 \times 10^{23} \times 0.53}\)

= \(43.5 \times 10^{-24} \mathrm{~cm}^3\)

a = \(352 \times 10^{-10} \mathrm{~cm}=352 \mathrm{pm}\)

Question 44. A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^-3. The molar mass of the metal is (N_A Avogadro’s constant = 6.02 x 10²³ mol^-1)

1. 27 g mol^-1
2. 20 g mol^-1
3. 40 g mol^-1
4. 30 g mol^-1

Answer: 1. 27 g mol^-1

d = \(\frac{Z M}{N_A a^3}(Z=4 \text { for } f c c)\)

M = \(\frac{d \times N_A \times a^3}{Z}=\frac{2.72 \times 6.023 \times 10^{23} \times\left(404 \times 10^{-10}\right)^3}{4}\)

M = \(27 \mathrm{~g} \mathrm{~mol}^{-1}\)

Question 45. CsBr crystallises in a body-centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 x 10²³ mol^-1, the density of CsBr is

1. 4.25 g/cm³
2. 42.5 g/cm³
3. 0.425 g/cm³
4. 8.25 g/cm³

Answer: 1. 4.25 g/cm³

Density of \(\mathrm{CsBr}=\frac{Z \times M}{a^3 \times N_A}\)

= \(\frac{1 \times 213}{\left(436.6 \times 10^{-10}\right)^3 \times 6.023 \times 10^{23}}=4.25 \mathrm{~g} / \mathrm{cm}^3\)

(It has one formula unit in the unit cell, so Z=1.)

Question 46. An element (atomic mass =100 g/mol) having bcc structure has a unit cell edge of 400 pm. The density of the element is

1. 7.289 g/cm³
2. 2.144 g/cm³
3. 10.376 g/cm³
4. 5.188 g/cm³

Answer: 4. 5.188 g/cm³

Cell edge = 400 pm; number of atoms in bcc (Z) = 2

and atomic mass = 100 g/mol.

Since atomic mass is 100 g/mol, therefore the mass of each atom (m) = \(\frac{100}{6.023 \times 10^{23}}=16.6 \times 10^{-23} \mathrm{~g}\)

We know that volume of unit cell = (400 pm)³

= (64 x 10⁶)pm³ = 64 x 10^-24 cm³ and

mass of unit cell= Z x m =2x (16.6 x 10^-23) = 33.2 x 10^-23 g

Therefore, density = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)

= \(\frac{33.2 \times 10^{-23}}{64 \times 10^{-24}}=5.188 \mathrm{~g} / \mathrm{cm}^3\)

Question 47. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): In a particular point defect, an ionic solid is electrically neutral, even if a few of its cations are missing from its unit cell.

Reason (R): In an ionic solid, Frenkel defect arises due to the dislocation of a cation from its lattice site to the interstitial site, maintaining overall electrical neutrality.

In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both (A) and (R) are correct and (R) is the correct explanation of (A).
2. Both (A) and (R) are correct but (R) is not the correct explanation of (A).
3. (A) is correct but (R) is not correct.
4. (A) is not correct but (R) is correct

Answer: 2. Both (A) and (R) are correct but (R) is not the correct explanation of (A).

Frenkel defect is shown by ionic solids. The smaller ion (usually a cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.

Question 48. The formula of nickel oxide with metal deficiency defect in its crystal is Ni_0-98O. The crystal contains Ni²⁺ and Ni³⁺ ions. The fraction of nickel existing as Ni²⁺ ions in the crystal is

1. 0.96
2. 0.04
3. 0.50
4. 0.3

Answer: 1. 0.96

Let the fraction of metal which exists as Ni²⁺ ion be x.

Then the fraction of metal as Ni³⁺ = 0.98 – x

∴ 2x + 3(0.98 – x) = 2

⇒ 2x+2.94-3x=2

⇒  x=0.94=0.96

Question 49. The correct statement regarding defects in crystalline solids is

1. Frenkel defects decrease the density of crystalline solids
2. Frenkel defect is a dislocation defect
3. Frenkel defect is found in halides of alkaline metals

Answer: 2. Frenkel defect is a dislocation defect

Frenkel defect is a dislocation defect as smaller ions (usually cations) are dislocated from normal sites to interstitial sites. Frenkel defect is shown by compounds having large differences in the size of cations and anions hence, alkali metal halides do not show Frenkel defect.

Also, Schottky’s defect decreases the density of the crystal while Frenkel’s defect has no effect on the density of crystals.

Question 50. The appearance of colour in solid alkali metal halides is generally due to

1. Interstitial positions
2. F-centres
3. Schottky defect
4. Frenkel defect.

Answer: 2. F-centres

F-centres are the sites where anions are missing and instead, electrons are present. They are responsible for colours.

Question 51. Schottky defect in crystals is observed when

1. The density of the crystal is increased
2. Unequal number of cations and anions are missing from the lattice
3. An ion leaves its normal site and occupies an interstitial site
4. An equal number of cations and anions are missing from the lattice.

Answer: 4. Equal number of cations and anions are missing from the lattice.

In the Schottky defect, an equal no. of cations and anions are missing from the lattice. So, the crystal remains neutral. Such defect is more common in highly ionic compounds of similar cationic and anionic size, i.e. NaCl

Question 52. Ionic solids, with Schottky defects, contain in their structure

1. Cation vacancies only
2. Cation vacancies and interstitial cations
3. An equal number of cation and anion vacancies
4. Anion vacancies and interstitial anions.

Answer: 3. Equal number of cation and anion vacancies

When an atom is missing from its normal lattice site, a lattice vacancy is created. Such a defect, which involves an equal number of cation and anion vacancies in the crystal lattice is called a Schottky defect

Question 53. Which is the incorrect statement?

1. Density decreases in the case of crystals with Schottky defect.
2. NaCl_(s) is an insulator, silicon is a semiconductor, silver is a conductor, and quartz is a piezoelectric crystal.
3. Frenkel defect is favoured in those ionic compounds in which sizes of cations and anions are almost equal.
4. FeO_0.98 has a non-stoichiometric metal deficiency defect.

Answer: 3. Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal. and

4. FeO_0.98 has a non-stoichiometric metal deficiency defect.

Frenkel defect is favoured in those ionic compounds in which there is a large difference in the size of cations and anions.

Non-stoichiometric defects due to metal deficiency are shown by  Fe_xO where x = 0.93 to 0.96.

Question 54. With which one of the following elements silicon should be doped so as to give a p-type of semiconductor?

1. Selenium
2. Boron
3. Germanium
4. Arsenic

Answer: 2. Boron

If silicon is doped with any of the elements of group 13 (B, Al, Ga, In, Tl) of the periodic table, the p-type of the semiconductor will be obtained.

Question 55. If NaCl is doped with 10^-4 mol % of SrCl₂, the concentration of cation vacancies will be (N_A = 6.02 x 10²³ mol^-1)

1. 6.02 x 10¹⁶ mol^-1
2. 6.02 x 10¹⁷ mol^-1
3. 6.02 x 10¹⁴ mol^-1
4. 6.02 x 10¹⁵ mol^-1

Answer: 2. 6.02 x 10¹⁷ mol^-1

As each Sr²⁺ ion introduces one cation vacancy, therefore, the concentration of cation vacancies = mole % of SrCI₂ added.

∴ The concentration of cation vacancies = 10^-4 mole%

= \(\frac{10^{-4}}{100} \times 6.023 \times 10^{23}=6.023 \times 10^{17}p\)

Question 56. If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor formation will occur?

1. n-type semiconductor
2. p-type semiconductor
3. Both (1) and (2)
4. None of these

Answer: 1. n-type semiconductor

When an impurity atom with 5 valence electrons (as arsenic) is introduced in a germanium crystal, it replaces one of the germanium atoms. Four of the five valence electrons of the impurity atom form covalent bonds with each valence electron of four geranium atoms and the fifth valence electron becomes free to move in the crystal structure.

This free electron acts as a charge carrier. Such as an impure germanium crystal is called an n-type semiconductor because in it charge carriers are negative (free electrons)

Question 57. On doping Ge metal with a little of In or Ga, one gets

1. p-type semiconductor
2. n-type semiconductor
3. Insulator
4. Rectifier.

Answer: 1. p-type semiconductor

p-type of semiconductors are produced

Due to metal deficiency defects

By adding impurities containing fewer electrons (i.e. atoms of group 13). Ge belongs to Group 14 and In or Ga to Group 13. Hence on doping-type semiconductor is obtained. This doping of Ge with In increases the electrical conductivity of the Ge crystal.

 

Solutions

Question 1. In one molal solution that contains 0.5 moles of a solute, there is

1. 500 mL of solvent
2. 500 g of solvent
3. 100 mL of solvent
4. 1000 g of solvent.

Answer: 2. 500 g of solvent

Molality (m) = \(\frac{\text { Moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)

Let the amount of solvent be \(x \mathrm{~g}\).

1 = \(\frac{0.5}{\frac{x}{1000}}\)

x = \(500 \mathrm{~g}\)

Question 2. Which of the following is dependent on temperature?

1. Molarity
2. Mole fraction
3. Weight percentage
4. Molality

Answer: 1. Molarity

Molarity is a function of temperature as volume depends on temperature.

Question 3. What is the mole fraction of the solute in a 1.00 m aqueous solution?

1. 1.770
2. 0.0354
3. 0.0177
4. 0.177

Answer: 3. 0.0177

1 molal aqueous solution means 1 mole of solute is present in 1000 g of water.

∴ \(x_{\text {solute }}=\frac{1}{1+\frac{1000}{18}}=\frac{1}{56.5}=0.0177\)

Question 4. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO₃? The concentrated acid is 70% HNO₃.

1. 70.0 g cone. HNO₃
2. 54.0 g cone. HNO₃
3. 45.0 g cone. HNO₃
4. 90.0 g cone. HNO₃

Answer: 3. 45.0 g cone. HNO₃

Molarity = \(\frac{w_{\mathrm{HNO}_3} \times 1000}{M_{\mathrm{HNO}_3} \times V_{\mathrm{sol}(\mathrm{mL})}}\)

or \(2=\frac{w_{\mathrm{HNO}_3}}{63} \times \frac{1000}{250} \Rightarrow w_{\mathrm{HNO}_3}=\frac{63}{2} \mathrm{~g}\)

Mass of acid \(\times \frac{70}{100}=\frac{63}{2}\)

Mass of acid \(=45 \mathrm{~g}\)

Read and Learn More NEET MCQs with Answers

Question 5. Which of the following compounds can be used as antifreeze in automobile radiators?

1. Methyl alcohol
2. Glycol
3. Nitrophenol
4. Ethyl alcohol

Answer: 2. Glycol

A 35% (V/V) solution of ethylene glycol is used as an antifreeze in cars for cooling the engine’ At this concentration’ the antifreeze lowers the freezing point of water to 255°4 K (-17.6°C).

Question 6. Concentrated aqueous sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.80 g mol^-1. The volume of acid required to make one litre of 0.1 M H₂SO₄ solution is

1. 16.65 mL
2. 22.20 mL
3. 5.55 mL
4. 11.10 mL

Answer: 3. .55 mL

⇒ \(\mathrm{H}_2 \mathrm{SO}_4\) is \(98 \%\) by weight.

Weight of \(\mathrm{H}_2 \mathrm{SO}_4=98 \mathrm{~g}\), Weight of solution \(=100 \mathrm{~g}\)

∴ Volume of solution = \(\frac{\text { mass }}{\text { density }}=\frac{100}{1.80} \mathrm{~mL}\)

= \(55.55 \mathrm{~mL}=0.0555 \mathrm{~L}\)

Molarity of solution = \(\frac{98}{98 \times 0.0555}=18.02 \mathrm{M}\)

Let \(V \mathrm{~mL}\) of this \(\mathrm{H}_2 \mathrm{SO}_4\) is used to prepare 1 litre of 0.1 M \(\mathrm{H}_2 \mathrm{SO}_4\).

∴ mM of concentrated \(\mathrm{H}_2 \mathrm{SO}_4=\mathrm{mM}\) of dilute \(\mathrm{H}_2 \mathrm{SO}_4\)

or, \(V \times 18.02=1000 \times 0.1\)

⇒ V = \(\frac{1000 \times 0.1}{18.02}=5.55 \mathrm{~mL}\)

Question 7. The mole fraction of the solute in one molal aqueous solution is

1. 0.009
2. 0.018
3. 0.027
4. 0.036

Answer: 2. 0.018

1 molal aqueous solution means I mole of solute present in 1 kg of H₂O.

1 mole of solute present \(\frac{1000}{18}\) mole of H₂O

⇒ \(x_{\text {solute }}=\frac{1}{\frac{1000}{18}+1}=\frac{18}{1018}=0.01768=0.018\)

Question 8. 2.5 litres of 1 M NaOH solution is mixed with another 3 litres of 0.5 M NaOH solution. Then find out the arity of the resultant solution.

1. 0.80 M
2. 1.0 M
3. 0.73 M
4. 0.50 M

Answer: 3. 0.73 M

Molecular weight of NaOH =40 g mol^-1

2.5 litre of 1 M NaOH solution contain 40 g mol^-1 x 1 mol L^-1 x 2.5 L = 40 x 2.5 g of NaOH

3 litre of 0.5 M NaOH solution contain

40 g mol^-1 x 0.5 mol L^-1 x 3 L= 40 x 0.5 x 3 g of NaOH

If these two solutions are mixed, the volume of the resultant solution = (2.5 + 3) = 5.5 litre.

5.51itre of the resultant solution contain = 40(2.5 + 1.5) g of NaOH

1 litre of the resultant solution contain \(\frac{40 \times 4}{5.5} \text { g of } \mathrm{NaOH}=\frac{40 \times 4}{5.5 \times 40} \text { mole of } \mathrm{NaOH}\)

The molarity of the resultant solution = 0.727 = 0.73 M

Question 9. How many grams of dibasic acid (mol. weight 200) should be present in 100 mL of the aqueous solution to give a strength of 0.1 N?

1. 10 g
2. 2 g
3. 1 g
4. 20 g

Answer: 3. 1 g

The strength of the solution is 0.1 N.

⇒ \(\frac{w}{E}=\frac{V \times N}{1000}\) (Equivalent weight = \(\frac{200}{2}=100\))

⇒ w = \(\frac{100 \times 0.1 \times 100}{1000}=1 \mathrm{~g}\)

Question 10. What is the molarity of H₂SO₄ solution, which has a density 1.84 g/cc at 35°C and contains 98% by weight?

1. 18.4 M
2. 18 M
3. 4.18 M
4. 8.14 M

Answer: 1. 18.4 M

We know that 98% H₂SO₄ by weight means 98 g of H₂SO₄ is present in 100 g of solution

Therefore, its weight is 98 and moles of \(\mathrm{H}_2 \mathrm{SO}_4\)

= \(\frac{\text { Weight of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Molecular weight }}=\frac{98}{98}=1\)

and volume of solution = \(\frac{\text { Mass }}{\text { Density }}\)

= \(\frac{100}{1.84}=54.35 \mathrm{~mL}=\frac{54.35}{1000} \mathrm{~L}\)

Therefore, molarity of \(\mathrm{H}_2 \mathrm{SO}_4\)

= \(\frac{\text { Moles of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Volume (in litres) }}=\frac{1 \times 1000}{54.35}=18.4 \mathrm{M}\)

Question 11. The concentration unit, independent of temperature, would be

1. Normality
2. Weight volume per cent
3. Molality
4. Molarity.

Answer: 3. Molality

The molality involves the weights of the solute and the solvent. Since the weight does not change with the temperature. therefore molality does not depend upon the temperature.

Question 12. How many grams of CH₃OH should be added to water to prepare a 150 mL solution of 2 M CH₃OH?

1. 9.6 x 10³
2. 2.4 x 10³
3. 9.6
4. 2.4

Answer: 3. 9.6

Since the molecular mass of CH₃OH is 32 therefore the quantity of CH₃OH to prepare 150 mL solution of 2 M CH₃OH

= \(\left(\frac{2}{1000}\right) \times 150 \times 32=9.6 \mathrm{~g}\)

Question 13. The correct option for the value of vapour pressure of a solution at 45° C with benzene to octane in a molar ratio 3: 2 is

[At 45° C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume ideal gas]

1. 350 mm of Hg
2. 160 mm of Hg
3. 168 mm of Hg
4. 336 mm of Hg

Answer: 4. 336 mm of Hg

⇒ \(P_{\text {total }}=P_{\text {bemone }} x_{\text {benzene }}+P_{\text {ocane }} x_{\text {octane }}\)

= \(280 \times \frac{3}{5}+420 \times \frac{2}{5}=168+168\)

= \(336 \mathrm{~mm} \text { of } \mathrm{Hg}\)

Question 14. In water-saturated air, the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is

1. 1.18 atm
2. 1.76 atm
3. 1.176 atm
4. 0.98 atm.

Answer: 3. 1.176 atm

p_{water vapour} = x_{water vapour} x P_{water vapour}

= 0.02 x 1.2 = 0.024atm

P_total = P_{water vapour}+ P_{dry air}

1.2 = 0.024 + p_{dry air}

Pdry air= 1.2 – 0.024 = 1.176 atm

Partial vapour pressure is directly proportional to mole fraction, p ∝ x.

Question 15. p_A and p_B are the vapour pressures of pure liquid components, A and B, respectively of an ideal binary solution. If x_A represents the mole fraction of component A, the total pressure of the solution will be

1. p_A + x_A(p_B – p_A)
2. p_A + x_A(p_A – p_B)
3. p_B + x_A(p_B – p_A)
4. p_B + x_A(p_A– p_B)

Answer: 4. p_B + x_A(p_A– p_B)

According to Raoult’s law, \(p=x_A p_A+x_B p_B\)….(1)

For binary solutions, \(x_A+x_B=1, x_B=1-x_A\)….(2)

Putting the value of \(x_B\) from eqn. (2) to eqn. (1)

P = \(x_A p_A+\left(1-x_A\right) p_B=x_A p_A+p_B-x_A p_B\)

P = \(p_B+x_A\left(p_A-p_B\right)\)

Question 16. The vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 25°C is 200 mm Hg and 41.5 mm Hg respectively. The vapour pressure of the solution obtained by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at the same temperature will be

(Molecular mass of CHCl₂ = 119.5 u and molecular mass of CH₂Cl₂ = 85 u)

1. 173.9 mm Hg
2. 615.0 mm Hg
3. 347.9 mm Hg
4. 285.5 mm Hg

Answer: None

⇒ \(p_{\mathrm{CHCl}_5}^{\circ}=200 \mathrm{~mm} \mathrm{Hg}, p^{\mathrm{a}} \mathrm{CH}_2 \mathrm{Cl}_2=41.5 \mathrm{~mm} \mathrm{Hg}\)

Moles of \(\mathrm{CHCl}_3=\frac{\text { Weight }}{\text { Molecular weight }}=\frac{25.5}{119.5}=0.2132111\)

Moles of \(\mathrm{CH}_2 \mathrm{Cl}_2=\frac{40}{85}=0.470\)

⇒ \(x_{\mathrm{CHCl}_3}=\frac{0.213}{0.213+0.470}=0.31\)

⇒ \(x_{\mathrm{CH}_2 \mathrm{Cl}_2}=\frac{0.470}{0.213+0.470}=0.69\)

⇒ \(P_{\text {tocal }}=p^{\circ}{ }_{\mathrm{CHC}_3} x_{\mathrm{CHCl}_3}+p^{\circ}{ }_{\mathrm{CH}_2 \mathrm{Cl}_2} x_{\mathrm{CH}_2 \mathrm{C}_2}\)

= \(200 \times 0.31+41.5 \times 0.69=62+28.63=90.63 \mathrm{~mm} \mathrm{Hg}\)

Question 17. A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20 °C are 440 mm Hg for pentane and 120 mm Hg for hexane. Tire mole fraction of pentane in the vapour phase would be

1. 0.200
2. 0.549
3. 0.786
4. 0.478

Answer: 4. 0.478

⇒ \(\frac{n_{C_6} H_{12}}{n_{C_6} H_{14}}=\frac{1}{4}\)

⇒ \(x_{\mathrm{C}_5 \mathrm{H}_{12}}=\frac{1}{5} \text { and } x_{\mathrm{C}_6 \mathrm{H}_{14}}=\frac{4}{5}\)

⇒ \(p_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ}=440 \mathrm{~mm} \mathrm{Hg} ; P_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ}=120 \mathrm{~mm} \mathrm{Hg}\)

⇒ \(P_{\text {total }}=p^{\circ}{ }_{\mathrm{C}_5 \mathrm{H}_{12}} x_{\mathrm{C}_5 \mathrm{H}_{12}}+p^{\circ}{ }_{\mathrm{C}_6 \mathrm{H}_{14}} x_{\mathrm{C}_6 \mathrm{H}_{14}}^5\)

= \(440 \times \frac{1}{5}+120 \times \frac{4}{5}=88+96=184 \mathrm{~mm} \text { of } \mathrm{Hg}\)

By Raoult’s law, \(p_{\mathrm{C}_5 \mathrm{H}_{12}}=p^0 \mathrm{C}_5 \mathrm{H}_{12} x_{\mathrm{C}_5 \mathrm{H}_{12}}\)….(1)

⇒ \(x_{\mathrm{C}_5 \mathrm{H}_{12}} \rightarrow\) mole fraction of pentane in solution.

By Dalton’s law, \(P_{\mathrm{C}_3 \mathrm{H}_{12}}=x_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} P_{\text {total }}\)……(2)

⇒ \(x^{\prime}{ }_{\mathrm{C}_5 \mathrm{H}_{12}} \rightarrow\) mole fraction of pentane above the solution.

From (1) and (2), \(P_{\mathrm{C}_5 \mathrm{H}_{12}}=440 \times \frac{1}{5}=88 \mathrm{~mm} \text { of } \mathrm{Hg}\)

⇒ 88 = \(x^{\prime}{ }_{\mathrm{C}_5 \mathrm{H}_{12}} \times 184 \Rightarrow x^{\prime}{ }_{\mathrm{C}_5 \mathrm{H}_{12}}=\frac{88}{184}=0.478\)

Question 18. The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of the solution obtained by mixing 3 moles of P and 2 mol of Q would be

1. 72 torr
2. 140 torr
3. 68 torr
4. 20 torr

Answer: 1. 72 torr

By Raoult’s Law, \(P_T=p_p^0 x_p^0+p_Q^{\circ} x_Q\) where \(p_P^{\circ}=80\) torr, \(p_Q^{\circ}=60\) torr, \(x_P=\frac{3}{5} ; x_Q=\frac{2}{5}\)

⇒ \(P_T=80 \times \frac{3}{5}+60 \times \frac{2}{5}=48+24=72 \text { torr }\)

⇒ \(P_T=80 \times \frac{3}{5}+60 \times \frac{2}{5}=48+24=72 \text { torr }\)

Question 19. The mixture which shows positive deviation from Raoult’s law is

1. Ethanol + acetone
2. Benzene + toluene
3. Acetone + chloroform
4. Chloroethane + bromoethane.

Answer: 1. Ethanol + acetone

A mixture of ethanol and acetone shows a positive deviation from Raoult’s law.

In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of, the hydrogen bonds between them. Due to the weakening of interactions, the solution shows a positive deviation from Raoult’s law.

Question 20. For an ideal solution, the correct option is

1. Δ_mix G = 0 at constant T and P
2. Δ_mix S = 0 at constant T and P
3. Δ_mix V ≠ 0 at constant Tand P
4. Δ_mix H = 0 at constant T and P.

Answer: 4. Δ_mixH = 0 at constant T and P.

For an ideal solution, Δ_mix H = 0 and Δ_mix V = 0 at constant T and P.

Question 21. The mixture that forms maximum boiling azeotrope is

1. Heptane + octane
2. Water + nitric acid
3. Ethanol + water
4. Acetone + carbon disulphide.

Answer: 2. Water + nitric acid

Maximum boiling azeotropes are formed by those solutions which show negative deviations from Raoult’s law. H₂O and HNO₃ mixture show negative deviations.

Question 22. Which of the following statements is correct regarding a solution of two components A and B exhibiting positive deviation from ideal behaviour?

1. Intermolecular attractive forces between A-A and B-B are stronger than those between A-B.
2. Δ_mix H = 0 at constant T and p.
3. Δ_mix V= 0 at constant T and P.
4. Intermolecular attractive forces between A-A and B-B are equal to those between A-B

Answer: 1. Intermolecular attractive forces between A-A and B-B are stronger than those between A-B.

In case of positive deviation from Raoult’s Law, A-B interactions are weaker than those between A-A or B-B, i.e., in this case, the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules.

This means that in such a solution, molecules of A (or B) will find it easier to escape than in a pure state. This will increase the vapour pressure and result in positive deviation.

Question 23. Which one of the following is incorrect for an ideal solution?

1. ΔH_mix= 0
2. ΔU_mix = 0
3. ΔP = P_obs — P_{calculated by Raoults’s law-} = 0
4. ΔG_mix= 0

Answer: 4. ΔH_mix= 0

For an ideal solution, \(\Delta H_{\text {mix }}=0, \Delta V_{\mathrm{mix}}=0 \text {, }\)

Now, \(\Delta U_{\text {mix }}=\Delta H_{\text {mix }}-P \Delta V_{\text {mix }}\)

∴ \(\Delta U_{\mathrm{mix}}=0\)

Also, for an ideal solution, \(P_A=x_A P_A^{\circ}, P_B=x_B P_B^{\circ}\)

∴ \(\Delta P=P_{\text {cbs }}-P_{\text {calculaned by Rooult’s law }}=0\)

∴ \(\Delta G_{\text {mix }}= \Delta H_{\text {mix }}-T \Delta S_{\text {mix }}\)

For an ideal solution, \(\Delta S_{\text {mis }} \neq 0\)

∴ \(\Delta G_{\text {mix }} \neq 0\)

Question 24. Which of the following statements about the composition of the vapour over an ideal 1: 1 molar mixture of benzene and toluene is correct?

Assume that the temperature is constant at 25°C. (Given, vapour pressure data at 25°C, benzene = 12.8 kPa, toluene = 3.85 kPa)

1. The vapour will contain equal amounts of benzene and toluene.
2. Not enough information is given to make a prediction.
3. The vapour will contain a higher percentage of benzene
4. The vapour will contain a higher percentage of toluene.

Answer: 3. The vapour will contain a higher percentage of benzene

⇒ \(p_{\text {Bewaene }}\) = \(x_{\text {Benzene }}p_{\text {Benaene }}^{\circ}\)

⇒ \(P_{\text {Taluene }}\) = \(x_{\text {Talsene }}p_{\text {Toluene }}^{\circ}\)

For an ideal 1: 1 molar mixture of benzene and toluene, \(x_{\text {Benzene }}=\frac{1}{2} \text { and } x_{\text {Toluene }}=\frac{1}{2}\)

⇒ \(P_{\text {Benzene }}=\frac{1}{2} p_{\text {Benzene }}^{\text {o }}=\frac{1}{2} \times 12.8 \mathrm{kPa}=6.4 \mathrm{kPa}\)

⇒ \(P_{\text {Toluene }}=\frac{1}{2} \rho_{\text {Toluene }}^{\mathrm{a}}=\frac{1}{2} \times 3.85 \mathrm{kPa}=1.925 \mathrm{kPa}\)

Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene.

Question 25. Which condition is not satisfied by an ideal solution?

1. Δ_mixV = 0
2. Δ_mixS = 0
3. Obeyance to Raoulfs Taw
4. Δ_mixH = 0

Answer: 2. Δ_mixS = 0

For an ideal solution:

• Volume change (ΔV) on mixing should be zero.
• Heat change (ΔH) on mixing should be zero.
• Obeys Raouitt law at every range of concentration.
• Entropy change ((ΔS) on mixing ≠ 0.

Question 26. A solution of acetone in ethanol

1. Obeys Raoult’s law
2. Shows a negative deviation from Raoult’s law
3. Shows a positive deviation from Raoult’s law
4. Behaves like a near-ideal solution.

Answer: 3. Shows a positive deviation from Raoult’s law

Both the components escape easily showing higher vapour pressure than the expected value. This is due to the breaking of some hydrogen bonds between ethanol molecules.

Question 27. A solution containing components A and B follows Raoulf’s law

1. A – B attraction force is greater than A – A and B-B
2. The a-B attraction force is less than A-A and B-B
3. A-B attraction force remains the same as A – A and B-B
4. The volume of the solution is different from the sum of the volume of the solute and solvent.

Answer: 3. A-B attraction force remains the same as A – A and B-B

Raoultt’s law is valid for ideal solutions only. The element of non-ideality enters into the picture when the molecules of the solute and solvent affect each other’s intermolecular forces. A solution containing components of A and B behaves as an ideal solution when the A – B attraction force remains the same as the A – A and B – B attraction forces.

Question 28. All form ideal solutions except

1. C₆H₆ and C₆H₅CH₃
2. C₂H₅Br and C₂H₅I
3. C₆H₅Cl and C₆H₅Br
4. C₂H₅I and C₂H₅OH

Answer: 4. C₂H₅I and C₂H₅OH

C₂H₅OH is associated through H-bonding while C₂H₅I does not show H-bonding.

Question 29. An ideal solution is formed when its components

1. Have no volume change on mixing
2. Have no enthalpy change on mixing
3. Have both the above characteristics
4. Have high solubility.

Answer: 3. Have both of the above characteristics

For the ideal solution, ΔV_mixing = 0 and ΔH_mixing = 0

Question 30. The following solutions were prepared by dissolving 10 g of glucose (C₆H₁₂O₆) in 250 ml of water (P₁). 10 g of urea (CH₄N₂O) in 250 ml of water (P₂) and 10 g of sucrose (C₁₂H₂₂O₁₁) in 250 ml of water (P₃). The right option for the decreasing order of osmotic pressure of these solutions is

1. P₃ > P₁ > P₂
2. P₂ > P₁ > P₃
3. P₁ >P₂> P₃
4. P₂ > P3 > P₁

Answer: 2. P₂ > P₁ > P₃

Osmotic pressure \((\pi)=C R T\)

∴ \(\pi \propto \mathrm{C}\)

For glucose solution, \(C_1=\frac{10}{180} \times \frac{1000}{250}=0.22 M\)

For urea solution, \(C_2=\frac{10}{60} \times \frac{1000}{250}=0.66 \mathrm{M}\)

For sucrose solution, \(C_3=\frac{10}{342} \times \frac{1000}{250}=0.117 \mathrm{M}\)

Hence, the order of osmotic pressure is \(P_2>P_1>P_3\).

Question 31. The freezing point depression constant (K_f) of benzene is 5.12 K kg mol^-1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places)

1. 0.20 K
2. 0.80 K
3. 0.40 K
4. 0.60 K

Answer: 3. 0.40 K

Given K_f = 5.12 K kg mol^-1, m =0.078m

ΔT_f = K_f x m = 5.12 x 0.078 = 0.39936 ≈ 0.40 K

Question 32. If the molality of the dilute solution is doubled, the value of the molal depression constant K_f will be

1. Halved
2. Tripled
3. Unchanged
4. Doubled.

Answer: 3. Unchanged

The value of molal depression-constant, K_f is constant for a particular solvent, thus, it will be unchanged when the molality of the dilut. the solution is doubled.

Question 33. At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be

1. 102 °C
2. 103 °C
3. 101 °C
4. 100 °C

Answer: 3. 101 °C

Given: \(W_B=6.5 \mathrm{~g}, W_A=100 \mathrm{~g}\), \(p_{\mathrm{s}}=732 \mathrm{~mm}, K_b=0.52, T_b^{\circ}=100^{\circ} \mathrm{C}, p^{\circ}=760 \mathrm{~mm}\)

∴ \(\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n_2}{n_1} \Rightarrow \frac{760-732}{760}=\frac{n_2}{100 / 18}\)

⇒ \(n_2=\frac{28 \times 100}{760 \times 18}=0.2046 \mathrm{~mol}\)

⇒ \(\Delta T_b=K_b \times m\)

⇒ \(T_b-T_b^{\circ}=K_b \times \frac{n_2 \times 1000}{W_A(\mathrm{~g})}\)

⇒ \(T_b-100^{\circ} \mathrm{C}=\frac{0.52 \times 0.2046 \times 1000}{100}=1.06\)

⇒ \(T_b=100+1.06=101.06^{\circ} \mathrm{C}\)

Question 34. 200 mL of an aqueous solution of a protein contains 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 x 10^-3 bar. The molar mass of the protein will be (R = 0.083 L bar mol^-1 K^-1)

1. 51022 g mol^-1
2. 122044 g mol^-1
3. 31011 g mol^-1
4. 61038 g mol^-1

Answer: 4. 61038 g mol^-1

We know that \(p V=n R T\), where\(n=\frac{w}{M}\)

πV = \(\frac{w}{M} R T\)

M = \(\frac{w R T}{\pi V}=\frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times \frac{200}{1000}}\)

= \(\frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.2}=61038 \mathrm{~g} \mathrm{~mol}^{-1}\)

Question 35. A solution of sucrose (molar mass = 342 g mol^-1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (K_f for water = 1.86 K kg mol^-1)

1. -0.372 °C
2. -0.520 °C
3. +0.372 °C
4. -0.570 °C

Answer: 1. -0.372 °C

We know, \(\Delta T_f=K_f m\)

m = \(\frac{w_B}{M_B} \times \frac{1000}{W_A}=\frac{68.5 \times 1000}{342 \times 1000}=\frac{68.5}{342}\)

⇒ \(\Delta T_f=1.86 \times \frac{68.5}{342}=0.372^{\circ} \mathrm{C}\)

⇒ \(T_f=0-0.372^{\circ} \mathrm{C}=-0.372^{\circ} \mathrm{C}\)

Question 36. During osmosis, the flow of water through a semipermeable membrane is

1. From solution having lower concentration only
2. From solution having higher concentration only
3. From both sides of the semipermeable membrane with equal flow rates
4. From both sides of the semipermeable membrane with unequal flow rates.

Answer: 1. From solution having lower concentration only

Question 37. 1.00 g of a non-electrolyte solute (molar mass 250 g mol^-1) was dissolved in 51.2 g of benzene. If the freezing point constant, Kf of benzene is 5.12 K_f kg mol^-1, the freezing point of benzene will be lowered by

1. 0.2 K
2. 0.4 K
3. 0.3 K
4. 0.5 K

Answer: 2. 0.4 K

= \(M_B=\frac{1000 \times K_f \times w_B}{W_A \times \Delta T_f}\)

or, \(250=\frac{1000 \times 5.12 \times 1}{51.2 \times \Delta T_f}\)

∴ \(\Delta T_f=\frac{1000 \times 5.12 \times 1}{51.2 \times 250}=0.4 \mathrm{~K}\)

Question 38. A solution containing 10 g per dm³ of urea (molecular mass = 60 g mol^-1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is

1. 200 g mol^-1
2. 250 g mol^-1
3. 300 g mol^-1
4. 350 g mol^-1

Answer: 3. 300 g mol^-1

Molar concentration of urea = \(\frac{10}{60}\) dm3

Molar concentration of non-volatile solution = \(\frac{50}{M_B} \mathrm{~L}^{-1}=\frac{50}{M_B} \mathrm{dm}^{-3}\)

For isotonic solutions, \(\frac{10}{60}=\frac{50}{M_B}\)

M_B =300 g mol^-1

Question 39. A solution of urea (mol. mass 56 g mol^-1) boils at 100.18°C at the atmospheric pressure. If K_f and K_b for water are 1.86 and 0.512 K kg mol^-1 respectively, the above solution will freeze at

1. 0.654°C
2. – 0.654°C
3. 6.54°C
4. -6.54°C

Answer: 2. – 0.654°C

⇒ \(\Delta T_f=K_f m\)….(1)

⇒ \(\Delta T_b=K_b m\)….(2)

⇒ \(\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}\)….(3)

⇒ \(\Delta T_f\) depression in freezing point

⇒ \(\Delta T_b \rightarrow\) elevation in boiling point

⇒ \(K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)

⇒ \(K_b=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \Delta T_b=100.18-100=0.18\)

From eq. (3), \(\frac{\Delta T_f}{0.18}=\frac{1.86}{0.512}\)

⇒ \(\Delta T_f=0.654=T_f^{\circ}-T_f=0-T_f \Rightarrow T_f=-0.654^{\circ} \mathrm{C}\)

Freezing point of urea in water \(=-0.654^{\circ} \mathrm{C}\)

Question 40. A solution contains a non-volatile solute of molecular mass M₂. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure? Pure water can be obtained from seawater by

1. \(M_2=\left(\frac{m_2}{\pi}\right) V R T\)
2. \(M_2=\left(\frac{m_2}{V}\right) \frac{R T}{\pi}\)
3. \(M_2=\left(\frac{m_2}{V}\right) \pi R T\)
4. \(M_2=\left(\frac{m_2}{V}\right) \frac{\pi}{R T}\)

Answer: 2. \(M_2=\left(\frac{m_2}{V}\right) \frac{R T}{\pi}\)

For dilute solution, \(\pi=\frac{n}{V} R T\)

⇒ \(\pi V=\frac{m_2}{M_2} R T \Rightarrow M_2=\frac{m_2 R T}{\pi V}\)

Question 41. Pure water can be obtained from seawater by

1. Centrifugation
2. Plasmolysis
3. Reverse osmosis
4. Sedimentation

Answer: 3. Reverse osmosis

Question 42. From the colligative properties of the solution, which one is the best method for the determination of the molecular weight of proteins and polymers?

1. Osmotic pressure
2. Lowering in vapour pressure
3. Lowering in freezing point
4. Elevation in boiling point

Answer: 1. Osmotic pressure

Question 43. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte solid, weighing 2.175 g is added to 39.08 of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of a solid substance?

1. 69.5
2. 59.6
3. 49.50
4. 79.8

Answer: 1. 69.5

p°_benzene = 640 mm Hg, p_s = 600 mm Hg

w_B = 2.175 g, W_benzene = 39.08 g

From Raoultt law \(\frac{p^{\circ}-p_8}{p^{\circ}}=\frac{w_B \times M_{\text {benzene }}}{W_{\text {benzene }} \times M_B} \Rightarrow \frac{640-600}{640}=\frac{2.175 \times 78}{39.08 \times M_B}\)

⇒ M_B = 69.5

Question 44. If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by 0.216°C, than that of the pure solvent. The molecular weight of the substance (Molal elevation constant for the solvent is 2.16°C) is

1. 10.1
2. 100
3. 1.01
4. 1000

Answer: 1. 10.1

⇒ \(w_B=0.15 \mathrm{~g}, W_A=15 \mathrm{~g}, \Delta T_b=0.216^{\circ} \mathrm{C}\)

⇒ \(K_b=2.16, m=? \)

As \(\Delta T_b=\frac{1000 \times K_b \times w_B}{M_B \times W_A}\)

⇒ \(M_B=\frac{1000 \times 2.16 \times 0.15}{0.216 \times 15}=100\)

Question 45. A 5% solution of cane sugar (mol. wt. = 342) is isotonic with 1% solution of a substance X. The molecular weight of X is

1. 68.4
2. 171.2
3. 34.2
4. 136.8

Answer: 1. 68.4

For isotonic solutions, \(C_1=C_2\) \(\frac{W_1}{M_1 V_1}=\frac{W_2}{M_2 V_2} \Rightarrow \frac{5}{342 \times 0.1}=\frac{1}{M_2 \times 0.1}\)

⇒ \(M_2=\frac{342}{5}=68.4\)

Question 46. The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?

1. 0.4
2. 0.6
3. 0.8
4. 0.2

Answer: 2. 0.6

x₂ (mole fraction of solute) = 0.2

From Raoults law \(\frac{p^a-p_8}{p^{\circ}}=x_2 \Rightarrow \frac{10}{p^{\circ}}=0.2 \Rightarrow p^{\circ}=50 \mathrm{~mm} \mathrm{Hg}\)

Again, when \(p^{\circ}-p_s=20 \mathrm{~mm} \mathrm{Hg}\), then \(\frac{p^{\circ}-p_S}{p^{\circ}}=\) mole fraction of solute \(=\frac{20}{50}=0.4\)

⇒ mole fraction of solvent \(=1-0.4=0.6\)

Question 47. The vapour pressure of CCl₄ at 25°C is 143 mm Hg If 0.5 g of a non-volatile solute (mol. weight = 65) is dissolved in 100 g CCl₄, the vapour pressure of the solution will be

1. 199.34 mm Hg
2. 143.99 mm Hg
3. 141.43 mm Hg
4. 94.39 mm Hg.

Answer: 3. 141.43 mm Hg

Vapour pressure of pure solvent (p°_A) = 143 mm Hg, weight of solute (w_B) = 0.5 g, weight of solvent (W_A) = 100 g, the molecular weight of solute M_B = 65 and molecular weight of solvent (M_A) = 154.

⇒ \(\frac{p_A^{\circ}-p_s}{p_A^a}=\frac{w_B M_A}{M_B W_A} \text { or } \frac{143-p_s}{143}=\frac{0.5 \times 154}{65 \times 100}\)

or \(p_s=141.31 \mathrm{~mm} \mathrm{Hg}\)

Question 48. The relationship between osmotic pressure at 273 K when 10 g glucose (p₁), 10 g urea (p₂), and 10 g sucrose (p₃) are dissolved in 250 mL of water is

1. p₂>p₁ > p₃
2. p₂>p₃>p₁
3. p₁>p₂>p₃
4. p₃>p₁>p₂

Answer: 1. p₂>p₁> p₃

Weight of glucose = 10 g,

Weight of urea = 10 g and weight of sucrose = 10 g

The number of moles of glucose, \(\left(n_1\right)=\frac{\text { Weight }}{\text { Molecular weight }}=\frac{10}{180}=0.05\)

Similarly, number of moles of urea \(\left(n_2\right)=\frac{10}{60}=0.16\) and the number of moles of sucrose \(\left(n_3\right)=\frac{10}{342}=0.03\)

Question 49. According to Raoult’s law, the relative lowering of vapour pressure for a solution is equal to

1. Mole fraction of solute
2. Mole fraction of solvent
3. Moles of solute
4. Moles of solvent.

Answer: 1. Mole fraction of solute

Question 50. If 0.1 M solution of glucose and 0.1 M solution of urea are placed on two sides of the semipermeable membrane to equal heights, then it will be correct to say that

1. There will be no net movement across the membrane
2. Glucose will flow towards the glucose solution
3. Urea will flow towards the glucose solution
4. Water will flow from the urea solution to glucose.

Answer: 1. There will be no net movement across the membrane

There is no net movement of the solvent through the semipermeable membrane between two solutions of equal concentration

Question 51. Which one is a colligative property?

1. Boiling point
2. Vapour pressure
3. Osmotic pressure
4. Freezing point

Answer: 3. Osmotic pressure

The properties which depend only upon the number of solute particles present in the solution irrespective of their nature are called colligative properties. Lowering in vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure are colligative properties.

Question 52. Blood cells retain their normal shape in solution which are

1. Hypotonic to blood
2. Isotonic to blood
3. Hypertonic to blood
4. Equinormal to blood.

Answer: 2. Isotonic to blood

Blood cells neither swell nor shrink in isotonic solution. The solutions having the same osmotic pressure are called isotonic solutions.

Question 53. The relative lowering of the vapour pressure is equal to the ratio between the number of

1. Solute molecules to the solvent molecules
2. Solute molecules to the total molecules in the solution
3. Solvent molecules to the total molecules in the solution
4. Solvent molecules to the total number of ions of the solute.

Answer: 2. Solute molecules to the total molecules in the solution

Relative lowering of vapour pressure is equal to the mole fraction of solute which is the ratio of solute molecules to the total molecules in the solution

Question 54. The van’t Hoff factor (1) for a dilute aqueous solution of the strong electrolyte barium hydroxide is

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

Answer: 4. 3

Being a strong electrolyte, Ba(OH)₂ undergoes 100% dissociation in a dilute aqueous solution, \(\mathrm{Ba}(\mathrm{OH})_{2(a q)} \rightarrow \mathrm{Ba}^{2+}{ }_{\langle(q)}+2 \mathrm{OH}^{-}{ }_{(a q)}\)

Thus, vant Hoff factor i = 3

Question 55. The boiling point of 0.2 mol kg^-1 solution of X in water is greater than the equimolal solution of Y in water. Which one of the following statements is true in this case?

1. The molecular mass of X is less than the molecular mass of Y.
2. Y is undergoing dissociation in water while X undergoes no change.
3. X is undergoing dissociation in water.
4. The molecular mass of X is greater than the molecular mass of Y.

Answer: 3. X is undergoing dissociation in water.

ΔT_b = i x K_b m
For equimolal solutions, elevation in boiling point will be higher if the solution undergoes dissociation i.e., i > l.

Question 56. Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?

1. KCl
2. C₆H₁₂O₆
3. AI₂(SO₄)₃
4. K₂SO₄

Answer: 3. AI₂(SO₄)₃

⇒ \(\Delta T_f=i \times K_f \times m\)

So, \(\Delta T_f \propto i\) (van’t Hoff factor)

Hence, I am the maximum i.e., 5 for AI₂(SO₄)₃

Question 57. The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in another solvent is respectively

1. Less than one and greater than one
2. Less than one and less than one
3. Greater than one and less than one
4. Greater than one and greater than one.

Answer: 3. Greater than one and less than one

From the value of van t Hoff factor I it is possible to determine the degree of dissociation or association. In the case of dissociation, i is greater than I and in the case of an association, i is less than 1.

Question 58. The freezing point depression constant for water is -1.86 °Cm^-1, If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by -3.82 °C. Calculate the van’t Hoff factor for Na₂SO₄.

1. 2.05
2. 2.63
3. 3.11
4. 0.381

Answer: 2. 2.63

We know that, \(\Delta T_f=i \times K_f \times \frac{w_B \times 1000}{M_B \times W_A}\)

Given: \(\Delta T_f=3.82, K_f=1.86,\)

⇒ \(w_B=5, M_B=142, W_A=45\)

i = \(\frac{\Delta T_f \times M_B \times W_A}{K_f \times w_B \times 1000}=\frac{3.82 \times 142 \times 45}{1.86 \times 5 \times 1000}=2.63\)

Question 59. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K_f for water is 1.86°C/m, the freezing point of the solution will be

1. – 0.18 °C
2. – 0.54 °C
3. – 0.36 °C
4. – 0.24 °C

Answer: 4. – 0.24 °C

We know that ΔT_f =i x K_f x m

Here i is vant Hoff factor.

i for weak acid is 1 + α.

Here α is the degree of dissociation i,e., 30/100 = 0.3

∴ i = 1 + α =1+0.3=1.3

ΔT_f = i x K_f x m= 1.3 x 1.86 x 0.1 = 0.24

∴ Freezing point of the solution, T_f= T°_f – ΔT_f =0-0.24=-0.24°C

Question 60. An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?

1. Addition of NaCl
2. Addition of Na₂SO₄ m
3. Addition of 1.00 molal KI
4. Addition of water

Answer: 4. Addition of water

The addition of water to an aqueous solution of KI causes the concentration of the solution to decrease thereby increasing the vapour pressure. In the other three options, the electrolytes undergo ionization, which leads to a lowering of vapour pressure.

Question 61. A 0.0020 m aqueous solution of an ionic compound [CO(NH₃)₅(NO₂)]Cl freezes at -0.00732 °C. The number of moles of ions that 1 mol of ionic compound produces on being dissolved in water will be (K_f = -1.86 °C/m)

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

Answer: 4. 2

The number of moles of ions produced by 1 mol of ionic compound = i

Applying, ΔT_f =i x K_f x m

0.00732 = i x 1.86 x 0.002

⇒ i = \(\frac{0.00732}{1.86 \times 0.002}=1.96 \div 2\)

Question 62. 0. 5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K_f for water is 1.86 K kg mol^-1, the lowering in freezing point of the solution is

1. 0.56 K
2. 1.12 K
3. -0.56 K
4. -1.12 K

Answer: 2. 1.12 K

HX ⇔ H⁺ + x^–

Total=1+α

∴ i=1 + α = 1+0.2 = 1.2

ΔT_f = i x K_f x m = 1.2 x 1.86 x 0.5 = 1.116 K≈ 1.12 K

Question 63. Which of the following 0.10m aqueous solution will have the lowest freezing point?

1. KI
2. C₁₂H₂₂O₁₁
3. Al₂(SO₄)₃
4. C₅H₁₀O₅

Answer: 3. Al₂(SO₄)₃

Since Al₂ (SO₄)₃ gives a maximum number of ions on dissociation, therefore it will have the lowest freezing point.

ΔT_f = i K_f m

Question 64. Which of the following salts has the same value of van’t Hoff factor (1) as that of K₃[Fe(CN)₆]?

1. Na₂SO₄
2. Al(NO₃)₃
3. Al₂(SO₄)₄
4. NaCl

Answer: 2. Al(NO₃)₃

⇒ \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 3 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\) and \(\mathrm{Al}\left(\mathrm{NO}_3\right)_3 \rightleftharpoons \mathrm{Al}^{3+}+3 \mathrm{NO}_3^{-}\)

Since both AI(NO₃)₃ and K₃[Fe(CN)₆] give the same number of ions, therefore they have the same van t Hoff factor.

Question 65. At 25°C, the highest osmotic pressure is exhibited by 0.1 M solution of

1. Glucose
2. Urea
3. CaCl₂
4. KCl.

Answer: 3. CaCl₂

In solution, CaCl_f gives three ions, KCl gives two ions while glucose and urea are covalent molecules so they do not undergo ionisation. Since osmotic pressure is a colligative property and depends upon the number of solute particles (ions), therefore, 0.1 M solution of CaCI₂ exhibits the highest osmotic pressure.

Question 66. Which of the following aqueous solutions has a minimum freezing point?

1. 0.01 m NaCl
2. 0.005 m C₂H₅OH
3. 0.005 mMgI₂
4. 0.005 mMgSO₄

Answer: 1. 0.01 m NaCl

Here, ΔT_f = i x K_f x m

vant Hoff factor, i = 2 for NaCl, so conc. = 0.02, which is the maximum in the present case.

Hence, ΔT_f is the maximum or freezing point is minimum in 0.01 m NaCI.

Chemical Kinetics

Question 1. For the chemical reaction, \(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}\) the correct option is

1. \(3 \frac{d\left[\mathrm{H}_2\right]}{d t}=2 \frac{d\left[\mathrm{NH}_3\right]}{d t}\)
2. \(-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}=-\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)
3. \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=2 \frac{d\left[\mathrm{NH}_3\right]}{d t}\)
4. \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)

Answer: 4. \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)

For the given chemical reaction,

Rate of reaction = \(-\frac{d\left[\mathrm{~N}_2\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}\)

Question 2. The rate of the reaction \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\) can be written in three ways.

• \(\frac{-d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=k\left[\mathrm{~N}_2 \mathrm{O}_5\right]\)
• \(\frac{d\left[\mathrm{NO}_2\right]}{d t}=k^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right]\); \(\frac{d\left[\mathrm{O}_2\right]}{d t}=k^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right]\)

The relationship between k and k’ and between k and k” are

1. k’ = 2k, k” = k
2. k’ = 2k, k” = k/2
3. k’ = 2k, k” = 2k
4. k’ = k, k” = k

Answer: 2. k’ = 2k, k” = k/2

For the reaction, \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\)

⇒ \(-\frac{1}{2} (k) \frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t} (k’) =+\frac{1}{4} \frac{d\left[\mathrm{NO}_2\right]}{d t}=+\frac{d\left[\mathrm{O}_2\right]}{d t}(k”)\)

⇒ \(\frac{1}{2} k=\frac{1}{4} k^{\prime}=k^{\prime \prime}, k^{\prime}=2 k ; k^{\prime \prime}=\frac{1}{2} k\)

Question 3. For the reaction \(\mathrm{N}_2 \mathrm{O}_{5(g)} \rightarrow 2 \mathrm{NO}_{2(g)}+1 / 2 \mathrm{O}_{2(g)}\) the value of rate of disappearance of N₂O₅ is given as 6.25 x 10^-3 mol L^-1s^-1 The rate of formation of N02 and 02 is given respectively as

1. \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
2. \(1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
3. \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
4. \(1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(6.25 \times 10^{-3} \mathrm{moll}^{-1} \mathrm{~s}^{-4}\)

Answer: 2. \(1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

⇒ \(\mathrm{N}_2 \mathrm{O}_{5(g)} \rightarrow 2 \mathrm{NO}_{2(g)}+1 / 2 \mathrm{O}_{2(g)}\)

For the given reaction the rate is written as \(\frac{-d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{2 d\left[\mathrm{O}_2\right]}{d t}\)

Given that \(\frac{-d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

∴ \(\frac{d\left[\mathrm{NO}_2\right]}{d t}=2 \times 6.25 \times 10^{-3}=1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

and \(\frac{d\left[\mathrm{O}_2\right]}{d t}=\frac{6.25 \times 10^{-3}}{2}=3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Question 4. For the reaction, \(\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3\) if \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)  the value of \(\frac{-d\left[\mathrm{H}_2\right]}{d t}\) would be

1. \(4 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
2. \(6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
3. \(1 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
4. \(3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Answer: 4. \(3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

For reaction, \(\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3\)

Rate = \(\frac{1}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}=-\frac{d\left[\mathrm{~N}_2\right]}{d t}\)

Given, \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

∴ \(-\frac{d\left[\mathrm{H}_2\right]}{d t}=\frac{3}{2} \frac{d\left[\mathrm{NH}_3\right]}{d t}=\frac{3}{2} \times 2 \times 10^{-4}\)

⇒ \(-\frac{d\left[\mathrm{H}_2\right]}{d t}=3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Question 5. In the reaction: \(\mathrm{BrO}_{3(\text { aq) }}^{-}+5 \mathrm{Br}_{(\text {aq) }}+6 \mathrm{H}_{(\text {aq) }}^{+} \rightarrow 3 \mathrm{Br}_{2(t)}+3 \mathrm{H}_2 \mathrm{O}_{(l)}\). The rate of disappearance of bromide ions as

1. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{5}{3} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)
2. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=\frac{5}{3} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)
3. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)
4. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)

Answer: 4. \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)

For the given reaction, \(\mathrm{BrO}_3^{-}{ }_{(a q)}^{-}+5 \mathrm{Br}_{(a q)}^{-}+6 \mathrm{H}^{+}{ }_{(a q)} \rightarrow 3 \mathrm{Br}_{2(l)}+3 \mathrm{H}_2 \mathrm{O}_{(l)}\)

Rate of reaction in terms of \(\mathrm{Br}_2\) and \(\mathrm{Br}^{-}\) is,

Rate = \(\frac{1}{3} \frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{1}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t} \)

∴ \(\frac{d\left[\mathrm{Br}_2\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}\)

Question 6. Consider the reaction: \(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)}\) The equality relationship between, \(\frac{d\left[\mathrm{NH}_3\right]}{d t}\) and \(-\frac{d\left[\mathrm{H}_2\right]}{d t}\) is

1. \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{d\left[\mathrm{H}_2\right]}{d t}\)
2. \(\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)
3. \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)
4. \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{3}{2} \frac{d\left[\mathrm{H}_2\right]}{d t}\)

Answer: 3. \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)

⇒ \(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)}\)

Rate = \(\frac{-d\left[\mathrm{~N}_2\right]}{d t}=-\frac{d\left[\mathrm{H}_2\right]}{3 d t}=+\frac{d\left[\mathrm{NH}_3\right]}{2 d t}\)

Hence, \(+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}\)

Question 7. For the reaction, 2A + B → 3C + D,  which of the following does not express the reaction rate?

1. \(-\frac{d[A]}{2 d t}\)
2. \(-\frac{d[C]}{3 d t}\)
3. \(-\frac{d[B]}{d t}\)
4. \(\frac{d[D]}{d t}\)

Answer: 2. \(-\frac{d[C]}{3 d t}\)

2 A+B → 3 C+D

rate = \(\frac{-d[A]}{2 d t}=-\frac{d[B]}{d t}=\frac{d[C]}{3 d t}=\frac{d[D]}{d t}\)

A negative sign shows a decrease in concentration.

Question 8. 3A → 2B, rate of reaction \(\frac{+d[B]}{d t}\) is equal to

1. \(-\frac{3}{2} \frac{d[A]}{d t}\)
2. \(-\frac{2}{3} \frac{d[A]}{d t}\)
3. \(-\frac{1}{3} \frac{d[A]}{d t}\)
4. \(+2 \frac{d[A]}{d t}\)

Answer: 2. \(-\frac{2}{3} \frac{d[A]}{d t}\)

3 A → 2 B

Rate of the reaction = \(\frac{1}{2} \frac{d[B]}{d t}=-\frac{1}{3} \frac{d[A]}{d t}\)

⇒ \(\frac{d[B]}{d t}=-\frac{2}{3} \frac{d[A]}{d t}\)

Question 9. For the reaction, \(\mathrm{H}^{+}+\mathrm{BrO}_3^{-}+3 \mathrm{Br}^{-} \longrightarrow 5 \mathrm{Br}_2+\mathrm{H}_2 \mathrm{O}\) which of the following relations correctly represents the consumption and formation of products?

1. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)
2. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)
3. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{5}{3} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)
4. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=\frac{5}{3} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

Answer: 1. \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

Rate of reaction = \(-\frac{1}{3} \frac{d\left[\mathrm{Br}^{-}\right]}{d t}=+\frac{1}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

∴ \(\frac{d\left[\mathrm{Br}^{-}\right]}{d t}=-\frac{3}{5} \frac{d\left[\mathrm{Br}_2\right]}{d t}\)

Question 10. For the reaction, \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}\) the rate of reaction is expressed as

1. \(\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=-\frac{\Delta[\mathrm{HI}]}{\Delta t}\)
2. \(-\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{HI}]}{\Delta t}\)
3. \(\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{\Delta[\mathrm{HI}]}{2 \Delta t}\)
4. None of these.

Answer: 2. \(-\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{HI}]}{\Delta t}\)

For \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(g)} \rightarrow 2 \mathrm{HI}_{(g)}\), the rate of reaction is

⇒ \(-\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{I}_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{HI}]}{\Delta t}\)

The negative sign shows the disappearance of the reactant and the positive sign shows the appearance product

Question 11. For a certain reaction, the rate = k[A]²[B], when the initial concentration of A is tripled keeping the concentration of B constant, the initial rate would

1. Increase by a factor of nine
2. Increase by a factor of three
3. Decrease by a factor of nine
4. Increase by a factor of six.

Answer: 1. Increase by a factor of nine

⇒ \(r_1=k[A]^2[B]\)

Keeping concentration of B constant and tripling conc. of A, new rate would be \(r_2=k[3 A]^2[B]=9 k[A]^2[B]\)

∴ \(r_2=9 \times r_1\)

Question 12. Mechanism of a hypothetical reaction, \(X_2+Y_2 \longrightarrow 2 X Y\), is given below:

1. \(X_2 \rightarrow X+X\) (fast)
2. \(X+Y_2 \rightleftharpoons X Y+Y\) (slow)
3. \(X+Y \longrightarrow X Y\) (fast)

The overall order of the reaction will be

1. 2
2. 0
3. 1.5
4. 1

Answer: 3. 1.5

Correct the reactions given in question as \(X_2 \rightleftharpoons{ } X+X\) fast; \(X+Y_2 \longrightarrow X Y+Y\) (slow)

The slow step is the rate-determining step.

Rate \(=k[X]\left[Y_2\right]\)…(1)

Equilibrium constant for fast step, \(K=\frac{[X]^2}{\left[X_2\right]}\)

[X] = \(\sqrt{K\left[X_2\right]}\)

By substituting [X] in equation (1), we get

Rate = \(k \sqrt{K\left[X_2\right]}\left[Y_2\right]=k\left[X_2\right]^{1 / 2}\left[Y_2\right]\)

∴ Order of reaction = \(\frac{1}{2}+1=\frac{3}{2}=1.5\)

Question 13. The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. It is because the

1. The rate is proportional to the surface coverage
2. Rate is inversely proportional to the surface coverage
3. Rate is independent of the surface coverage
4. The rate of decomposition is very slow.

Answer: 1. Rate is proportional to the surface coverage

At low pressure, the rate is proportional to the surface coverage and is of first order while at high pressure, it follows zero-order kinetics due to the complete coverage of the surface area.

Question 14. The rate constant of the reaction A → B is 0. 6 x 10^-3 mol L^-1 s^-1. If the concentration of A is 5 M, then the concentration of B after 20 minutes is

1. 3.60 M
2. 0.36 M
3. 0.72 M
4. 1.08 M

Answer: 3. 0.72 M

The reaction is of zero order as the unit of rate constant is mol L^-1 s^-1.

Concentration of B =k x t = 0.6 x 10^-3 x 20 x 60 =0.72M

Question 15. For a reaction between A and B, the order with respect to A is 2 and the order with respect to B is 3. The concentrations of both A and B are doubled, the rate will increase by a factor of

1. 12
2. 16
3. 32
4. 10

Answer: 3. 32

Rate \(_1=k[A]^2[B]^3\)

Rate \(_2=k[2 A]^2[2 B]^3\)

Rate \(_2=32 k[A]^2[B]^3\)

∴ Rate \(_2=32\left(\text { Rate }_1\right)\)

Question 16. In a reaction, A + B → product, the rate is doubled when the concentration of B is doubled, and the rate increases by a factor of 8 when the concentration of both the reactants (A and B) are doubled, the rate law for the reaction can be written as

1. Rate = k[A][B]²
2. Rate = k[A]²[B]²
3. Rate = k[A][B]
4. Rate = k[A]²[B]

Answer: 4. Rate = k[A]²[B]

⇒ \(\begin{array}{ccc}
{[A]} & {[B]} & \text { Rate } \\
x & y & R …….(1)\\
x & 2 y & 2 R ……(2)\\
2 x & 2 y & 8 R….(3)
\end{array}\)

Let the rate law; rate \(=k[A]^a[B]^b\)

From data given, \((x)^a(y)^b=R\)….(4)

⇒ \((x)^a(2 y)^b=2 R\)…….(5)

Dividing eqn. (5) by (4), \(\frac{(2 y)^b}{(y)^b}=\frac{2 R}{R} \Rightarrow(2)^b=2=2^1\)

Thus, b=1

From data of (3) experiment, \((2 x)^a(2 y)^b=8 R\)….(6)

From equation (5) and (6), \(\frac{(2 x)^a}{(x)^a}=\frac{8 R}{2 R} \Rightarrow(2)^a=4=2^2\)

Thus, a=2. By replacing the values of a and b in rate law; rate \(=k[A]^2[B]\)

Question 17. Which one of the following statements for the order of a reaction is incorrect?

1. Order can be determined only experimentally.
2. Order is not influenced by the stoichiometric coefficient of the reactants.
3. The order of a reaction is the sum of power to the concentration terms of reactants to express the rate of reaction.
4. The order of reaction is always a whole number.

Answer: 4. Order of reaction is always a whole number.

The order of a reaction is not always a whole number. It can be zero, or fractional also.

Question 18. The unit of rate constant for a zero-order reaction is

1. mol L^-1 s^-1
2. L mol^-1 s^-1
3. L² mol^-2 s^-1
4. s^-1

Answer: 1. mol L^-1 s^-1

Rate = k[A]⁰

mol L^-1 s^-1 = k

Thus, the unit of rate constant is mol L^-1 s^-1

Question 19. During the kinetic study of the reaction, 2A + B → C + D, following results were obtained:

Based on the above data which one of the following is correct?

1. Rate = k[A]²[B]
2. Rate = k[A][B]
3. Rate = k[A]²[B]²
4. Rate = k[A][B]²

Answer: 4. Rate = k[A][B]²

Let the rate of reaction be given by rate = \(k[A]^a[B]^b\)

Now consider 2 and 3 where [A] is constant, \(\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}=\frac{[0.3]^a[0.2]^b}{[0.3]^a[0.4]^b}\)

⇒ \(\frac{1}{4}=\left(\frac{1}{2}\right)^b \Rightarrow b=2\)

Now consider 1 and 4, \(\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}=\frac{[0.1]^a[0.1]^b}{[0.4]^a[0.1]^b}\)

⇒ \(\frac{1}{4}=\left(\frac{1}{4}\right)^a \Rightarrow a=1\)

Rate = \(k[A][B]^2\)

Question 20. For the reaction, A + B → products, it is observed that

1. On doubling the initial concentration of an only, the rate of reaction is also doubled and
2. On doubling the initial concentration of both a and b, there is a change by a factor of 8 in the rate of the reaction.

The rate of this reaction is given by

1. Rate = k[A] [B]²
2. Rate = k[A]²[B]²
3. Rate = k[A][B]
4. Rate = k[A]²[B]

Answer: 1. Rate = k[A][B]²

R = \(k[A]^m[B]^n\)….(1)

2R = \(k[2 A]^m[B]^n\)…..(2)

from (1), (2) and (3), m=1, n=2….(3)

So, rate \(=k[A][B]^2\)

Question 21. The bromination of acetone that occurs in an acid solution is represented by this equation. \(\mathrm{CH}_3 \mathrm{COCH}_{3(a q)}+\mathrm{Br}_{2(a q)} \longrightarrow \mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{Br}_{(a q)}\) + \(\mathrm{H}_{(a q)}^{+}+\mathrm{Br}_{(a q)}^{-}\)

These kinetic data were obtained for given reaction concentrations. Initial concentrations, M

Based on these data, the rate equation is

1. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\left[\mathrm{H}^{+}\right]^2\)
2. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\left[\mathrm{H}^{+}\right]\)
3. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{H}^{+}\right]\)
4. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\)

Answer: 3. Rate \(=k\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{H}^{+}\right]\)

From the first two experiments, it is clear that when the concentration of Br₂ is doubled, the initial rate of disappearance of Br₂ remains unaltered. So, the order of reaction with respect to Br₂ is zero. Thus, the probable rate law for the reaction will be k[CH₃COCH₃][H⁺]

Question 22. The reaction of hydrogen and iodine monochloride is given as: \(\mathrm{H}_{2(g)}+2 \mathrm{ICl}_{(g)} \rightarrow 2 \mathrm{HCl}_{(g)}+\mathrm{I}_{2(g)}\) This reaction is of first order with respect to H_2(g) and ICl_(g), following mechanisms were proposed.

Mechanism A: \(\mathrm{H}_{2(g)}+2 \mathrm{ICl}_{(g)} \rightarrow 2 \mathrm{HCl}_{(g)}+\mathrm{I}_{2(g)}\)

Mechanism B: \(\mathrm{H}_{2(g)}+\mathrm{ICl}_{(g)} \rightarrow \mathrm{HCl}_{(g)}+\mathrm{HI}_{(g)} \text {; slow } \); \(\mathrm{HI}_{(g)}+\mathrm{ICl}_{(g)} \rightarrow \mathrm{HCl}_{(g)}+\mathrm{I}_{2(g)} \text {; fast }\)

Which of the above mechanism(s) can be consistent with the given information about the reaction?

1. A and B both
2. Neither A nor B
3. A only
4. B only

Answer: 4. B only

The slow step is the rate-determining step and it involves 1 molecule of H_2(g) and 1 molecule of ICI_(g). Hence, the rate will be, r = k[H_2(g)] [ICl_(g)]

i.e., the reaction is 1st order with respect to H_2(g) and ICl_(g).

Question 23. The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is

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

Answer: 2. -2

Rate of reaction \(=k[A] \alpha[B] \beta\)

α → order of reaction with respect to A

β → order of reaction with respect to B

⇒ \(r_1=k[A]^\alpha[B]^\beta\)

⇒ \(r_2=r_1 / 4=k[A]^\alpha[2 B]^\beta\)

⇒ \(\frac{r_1}{r_2}=\frac{k[A]^\alpha[B]^\beta}{k[A]^\alpha[2 B]^\beta} \Rightarrow 4=\left(\frac{1}{2}\right)^\beta \Rightarrow \beta=-2\)

Question 24. If the rate of the reaction is equal to the rate constant, the order of the reaction is

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

Answer: 1. 0

A → products

If \(-\frac{d x}{d t}=k\), it means \(-\frac{d x}{d t}=k[A]^0=k\)

Hence, the order of reaction must be zero.

Question 25. 2A → B + C, It would be a zero-order reaction when

1. The rate of reaction is proportional to the square of the concentration of A
2. The rate of reaction remains the same at any concentration of A
3. The rate remains unchanged at any concentration of B and C
4. The rate of reaction doubles if the concentration of B is increased to double.

Answer: 2. The rate of reaction remains the same at any concentration of A

2A → B+C

The rate equation of this reaction may be expressed as r = k[A]⁰, Order= 0, r=k

∴ The rate is independent of the concentration of the reactant A

Question 26. For the reaction; \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\) rate and rate constant are 1.02 x 10⁴ and 3.4 x 10^-5 sec^-1 respectively, then concentration of N₂O₅ at that time will be

1. 1.732
2. 3
3. 1.02 x 10^-4
4. 3.4 x 10⁵

Answer: 2. 3

⇒ \(2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\)

This is a first-order reaction.

∴ rate = \(k\left[\mathrm{~N}_2 \mathrm{O}_5\right]\)

⇒ \([\mathrm{N}_2 \mathrm{O}_5]=\text { rate } / k=\frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}=3\)

Question 27. The experimental data for the reaction, 2A + B₂ →2AB is

⇒ \(\begin{array}{cccl}
\text { Experiment } & {[\boldsymbol{A}]} & {\left[\boldsymbol{B}_2\right]} & \text { Rate }\left(\text { mole s }^{-1}\right) \\
1 & 0.50 & 0.50 & 1.6 \times 10^{-4} \\
2 & 0.50 & 1.00 & 3.2 \times 10^{-4} \\
3 & 1.00 & 1.00 & 3.2 \times 10^{-4}
\end{array}\)

The rate equation for the above data is

1. Rate = k [A]²[B]²
2. Rate = k [A]²[B]
3. Rate = k [B₂]
4. Rate = k [B₂]²

Answer: 3. Rate = k [B₂]

For the reaction, \(2 A+B_2 \rightleftharpoons 2 A B\),

Rate \(\propto[A]^x\left[B_2\right]^y\).

From experiment 1, \(1.6 \times 10^{-4} \propto[0.50]^x[0.50]^y\)….(1)

From experiment 2, \(3.2 \times 10^{-4} \propto[0.50]^x[1.00]^y\)…(2)

From experiment 3, \(3.2 \times 10^{-4} \propto[1.00]^x[1.00]^y\)….(3)

On dividing equation (3) by (2), we get, 1 = \(\left[\frac{1.00}{0.50}\right]^x \Rightarrow 1=2^x \Rightarrow 2^0=2^x \Rightarrow x=0\)

Now, divide equation (2) by equation (1) we get, 2 = \(\left[\frac{1.00}{0.50}\right]^y \Rightarrow 2=2^y \Rightarrow y=1\)

Thus, rate equation is : Rate \(=k[A]^0\left[B_2\right]^1=k\left[B_2\right]\)

Question 28. The given reaction, \(2 \mathrm{FeCl}_3+\mathrm{SnCl}_2 \rightarrow 2 \mathrm{FeCl}_2+\mathrm{SnCl}_4 \) is an example of

1. Third order reaction
2. First order reaction
3. Second order reaction
4. None of these.

Answer: 1. Third-order reaction

For a general reaction, xA + yB + zC → product, the order of the reaction is x + y + z.

Since three molecules undergo a change in concentration, therefore it is a third-order reaction

Question 29. The data for the reaction A + B → C, is

⇒ \(\begin{array}{lccc}
\text { Exp. } & {[A]_0} & {[B]_0} & \text { Initial rate } \\
1 & 0.012 & 0.035 & 0.10 \\
2 & 0.024 & 0.070 & 0.80 \\
3 & 0.024 & 0.035 & 0.10 \\
4 & 0.012 & 0.070 & 0.80
\end{array}\)

The rate law corresponds to the above data is

1. Rate = k[A][B]³
2. Rate = k[A]²[B]²
3. Rate = k[B]³
4. Rate = k[B]⁴.

Answer: 3. Rate = k[B]³

A+B → C

Let rate = k[A]^x [B]^y

where order ofreaction is (x + y).

Putting the values of exp. 1, 2, and 3, we get the following equations.

0.10 = k [0.012]^x [0.035]^y ….(1)

0.80=k [0.024]^x [0.070]^y ……(2)

0.10 = k [0.024]^x [0.035]^y …..(3)

Dividing (2) by (3), we get \(\frac{0.80}{0.10}=\left(\frac{0.070}{0.035}\right)^y \Rightarrow 2^y=8 \Rightarrow y=3\)

Keeping [A] constant, [B] is doubled, and the rate becomes B times. Dividing eq. (3) by eq. (1), we get \(\frac{0.10}{0.10}=\left(\frac{0.024}{0.012}\right)^x \Rightarrow 2^x=1 \Rightarrow x=0\)

Keeping [B] constant, [A] is doubled, and the rate remains unaffected. Hence, the rate is independent of [A].

rate ∝ [B]³.

Question 30. For a first-order reaction A → Products, the initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. The rate constant for the reaction in min^-1 is

1. 1.3818
2. 0.9212
3. 0.4606
4. 0.2303

Answer: 2. 0.9212

For a first-order reaction

k = \(\frac{2.303}{t} \log \frac{a}{a-x}\)

k = \(\frac{2.303}{5} \log \frac{0.1}{0.001}\)

k = \(\frac{2.303}{5} \log 10^2=\frac{2.303 \times 2}{5}=0.9212 \mathrm{~min}^{-1}\)

Question 31. The given graph is a representation of the kinetics of a reaction.

The y and x axes for zero and first-order reactions, respectively are

1. Zero order (y = concentration and x = time), first order (y = t_1/2 and x = concentration)
2. Zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
3. Zero order (y = rate and x = concentration), first order (y = t_1/2 and x = concentration)
4. Zero order (y = rate and x = concentration), first order (y = rate and x = t_1/2)

Answer: 3. Zero order (y = rate and x = concentration), first order (y = t_1/2 and x = concentration)

Question 32. The rate constant for a first-order reaction is 4.606 x 10^-3 s^-1 The time required to reduce 2.0 g of the reactant to 0.2 g is

1. 100 s
2. 200s
3. 500 s
4. 1000 s

Answer: 3. 500 s

For a first-order reaction,

k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)

t = \(\frac{2.303}{4.606 \times 10^{-3} \mathrm{~s}^{-1}} \log \left(\frac{2}{0.2}\right)=\frac{2.303 \times 10^3}{4.606}=500 \mathrm{~s}\)

Question 33. If the rate constant for a first-order reaction is k, the time (t) required for the completion of 99% of the reaction is given by

1. t = 2303/k
2. t = 0.693/k
3. t = 6.909/k
4. t = 4.606/k

Answer: 4. t = 4.606/k

For 1 st order reaction,

t = \(\frac{2.303}{k} \log \frac{a}{a-x}=\frac{2.303}{k} \log \frac{100}{100-99}\)

= \(\frac{2.303}{k} \log 10^2=\frac{2.303}{k} \times 2 \times \log 10=\frac{4.606}{k}\)

Question 34. A first-order reaction has a rate constant of 2.303 x 10^-3 s^-1. The time required for 40 g of this reactant to reduce to 10 g will be [Given that log₁₀ 2 = 0.3010]

1. 230.3 s
2. 301 s
3. 2000 s
4. 602 s

Answer: 4. 602 s

For a first-order reaction, k = \(\frac{2.303}{t} \log \frac{[A]_0}{[A]_t}\)

⇒ \(2.303 \times 10^{-3}=\frac{2.303}{t} \log \frac{40}{10}\)

t = \(\frac{1}{10^{-3}} \log 2^2=\frac{2}{10^{-3}} \log 2=\frac{2}{10^{-3}} \times 0.3010=602 \mathrm{~s}\)

Question 35. The correct difference between first and second-order reactions is that

1. The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
2. The half-life of a first-order reaction does not depend on [A]0₂; the half-life of a second-order reaction does depend on [A]₀
3. A first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
4. The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations.

Answer: 2. The half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀

For the first order reaction, \(t_{1 / 2}=\frac{0.693}{k}\) independent of initial concentration [A]₀

For second order reaction, \(t_{1 / 2}=\frac{1}{k[A]_0}\)

Half-life depends on the initial concentration of the reactant

Question 36. When the initial concentration of the reactant is doubled, the half-life period of a zero-order reaction

1. Is halved
2. Is doubled
3. Is tripled
4. Remains unchanged.

Answer: 2. Is doubled

⇒ \(\left(t_{1 / 2}\right)_{\text {zero }}=\frac{[A]_0}{2 k}\)

The half-life of a zero-order reaction is directly proportional to the initial concentration.

∴ If [A]₂ = doubled then, t_1/2 = doubled

Question 37. A first-order reaction has a specific reaction rate of 10^-2 sec^-1. How much time will it take for 20 g of the reactant to reduce to 5 g?

1. 138.6 sec
2. 346.5 sec
3. 693.0 sec
4. 238.6 sec

Answer: 1. 138.6 sec

For a first-order reaction, k = \(\frac{2.303}{t} \log \frac{[A]_0}{[A]_t}\) or \(10^{-2}=\frac{2.303}{t} \log \frac{20}{5}\)

⇒ \(10^{-2}=\frac{2.303 \times 0.6020}{t}\) or \(t=138.6 \mathrm{sec}\)

Question 38. The rate of the first-order reaction is 0.04 mol L^-1 s^-1 at 10 seconds and 0.03 mol L^-1 s^-1 at 20 seconds after the initiation of the reaction. The half-life period of the reaction is

1. 44.1 s
2. 54.1s
3. 24.1 s
4. 34.1 s

Answer: 3. 24.1 s

For a first-order reaction, A → products and for the concentration of the reactant at two different times,

k = \(\frac{2.303}{t_2-t_1} \log \frac{[A]_1}{[A]_2}\)

k = \(\frac{2.303}{t_2-t_1} \log \frac{(\text { rate })_1}{(\text { rate })_2}\) (because rate ∝ [A])

k = \(\frac{2.303}{(20-10)} \log \left(\frac{0.04}{0.03}\right)=0.0287 \mathrm{sec}^{-1}\)

∴ \(t_{1 / 2}= \frac{0.693}{k}=\frac{0.693}{0.0287 \mathrm{sec}^{-1}}=24.14 \mathrm{sec}\)

Question 39. When the initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is

1. Second
2. More than zero but less than the first
3. Zero
4. First.

Answer: 4. First

The half-life period of a first-order reaction is independent of initial concentration, \(t_{1 / 2}=\frac{0.693}{k}\)

Question 40. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of the reaction is

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

Answer: 1. 1

As t_75% = 2 x t_50%, the order of the reaction is one.

Question 41. The half-life of a substance in a certain enzyme-catalyzed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L^-1 to 0.04 mg L^-1 is

1. 414 s
2. 552 s
3. 690 s
4. 276 s

Answer: 3. 690 s

A fall of concentration from 1.28 mg L^-1 to 0.04 mg L^-1 requires 5 half-lives.

∴ Time required = 5 x t_1/2= 5 x 138 = 690 s

Question 42. The half-life period of a first-order reaction is 1386 seconds. The specific rate constant of the reaction is

1. \(0.5 \times 10^{-2} \mathrm{~s}^{-1}\)
2. \(0.5 \times 10^{-3} \mathrm{~s}^{-1}\)
3. \(5.0 \times 10^{-2} \mathrm{~s}^{-1}\)
4. \(5.0 \times 10^{-3} \mathrm{~s}^{-1}\)

Answer: 2. \(0.5 \times 10^{-3} \mathrm{~s}^{-1}\)

Given, \(t_{1 / 2}=1386 \mathrm{~s}\)

For a first-order reaction, \(t_{1 / 2}=\frac{0.693}{k}\) (k = rate constant)

1386 = \(\frac{0.693}{k} \Rightarrow k=5 \times 10^{-4} \mathrm{~s}^{-1}=0.5 \times 10^{-3} \mathrm{~s}^{-1}\)

Question 43. If 60% of a first-order reaction was completed in 60 minutes, 50% of the same reaction would  be completed in approximately (log 4 = 0.60, log 5 = 0.69)

1. 45 minutes
2. 60 minutes
3. 40 minutes
4. 50 minutes

Answer: 1. 45 minutes

For a first-order reaction, \(k=\frac{2.303}{t} \log \frac{a}{a-x}\)

k = \(\frac{2.303}{60} \log \frac{100}{40}=\frac{2.303}{60} \times \log 2.5=0.0153\)

Again, \(t_{1 / 2}=\frac{2.303}{k} \log \frac{100}{50}=\frac{2.303}{0.0153} \times \log 2=45.31 \mathrm{~min}\)

Question 44. In a first-order reaction, A → B, if k is rate constant and the initial concentration of the reactant A is 0.5 M, then the half-life is

1. \(\frac{\log 2}{k}\)
2. \(\frac{\log 2}{k \sqrt{0.5}}\)
3. \(\frac{\ln 2}{k}\)
4. \(\frac{0.693}{0.5 k}\)

Answer: 3. \(\frac{\ln 2}{k}\)

For a 1st t order reaction, \(k=\frac{2.303}{t} \log _{10} \frac{a}{a-x}\)

At \(t_{1 / 2}, k=\frac{2.303}{t_{1 / 2}} \log _{10} \frac{a}{a-\frac{a}{2}}\) or \(t_{1 / 2}=\frac{2.303}{k} \log _{10} 2=\frac{\ln 2}{k}\)

Question 45. For a first-order reaction A → B the reaction rate at a reactant concentration of 0.01 M is found to be 2.0 x 10^-5 mol L^-1 s^-1. The half-life period of the reaction is

1. 30 s
2. 220 s
3. 300 s
4. 347 s

Answer: 4. 347 s

A → B

Rate of reaction = 2 x 10^-5 mol L^-1 s^-1

⇒ order of reaction is n = 1, rate = k [A]ⁿ = k[A]

⇒ k is the rate constant.

[A] = 0.01M

⇒ k = \(\frac{2 \times 10^{-5}}{0.01}=2 \times 10^{-3}, k=\frac{0.693}{t_{1 / 2}}\)

\(t_{1 / 2}=\frac{0.693}{2 \times 10^{-3}}=346.5 \approx 347 \mathrm{~s}\)

Question 46. The rate of a first-order reaction is 1.5 x 10^-2 mol L^-1 min^-1 at 0.5 M concentration of the reactant. The half-life of the reaction is

1. 0.383 min
2. 23.1 min
3. 8.73 min
4. 7.53 min

Answer: 2. 23.1 min

Rate \(\left(\frac{d x}{d t}\right)=k C\)

i.e., \(1.5 \times 10^{-2}=k \times 0.5$ or, $k=\frac{1.5 \times 10^{-2}}{0.5}\)

For first order reaction, \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693 \times 0.5}{1.5 \times 10^{-2}}=23.1 \mathrm{~min} \)

Question 47. The reaction A → B follows first-order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0. 9 mole of A to produce 0.675 mole of B?

1. 1 hour
2. 0.5 hour
3. 0.25 hour
4. 2 hours

Answer: 1. 1 hour

The time taken for the completion of the same fraction of change is independent of initial concentration.

Question 48. For a first-order reaction, the half-life period is independent of

1. The first power of final concentration
2. Cube root of initial concentration
3. Initial concentration
4. The square root of the final concentration.

Answer: 3. Initial concentration

For the first order reaction, the rate constant is given by, \(k_1=\frac{1}{t} \ln \frac{a}{a-x}\)

a = initial concentration,(a-x)= concentration at t time

At t = \(t_{1 / 2}, x=a /2\)

⇒ \(k_1=\frac{1}{t_{1 / 2}} \ln \frac{a}{a-a / 2} \Rightarrow k_1=\frac{1}{t_{1 / 2}} \ln 2 \Rightarrow k_1=\frac{0.693}{t_{1 / 2}}\)

Therefore, t_1/2 is independent of initial concentration.

Question 49. Given below are two statements: One is labeled as Assertion A and the other is labeled as Reason R:

Assertion A: A reaction can have zero activation energy.

Reason R: The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy.

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

Answer: 2. A is false but R is true.

If \(E_a=0\), then according to Arrhenius equation, \(k=A e^{-E_0 / R T} \Rightarrow k=A e^{0 / R T}=A\)

This implies that every collision results in a chemical reaction which cannot be true. So, a reaction cannot have zero activation energy.

Question 50. For reaction A → B, the enthalpy of the reaction is -4.2 kJ mol^-1 and the enthalpy of activation is 9.6 kJ mol^-1. The correct potential energy profile for the reaction is shown in the option.

Answer: 3

As the enthalpy of the reaction is negative, hence it is an exothermic reaction.

Question 51. The slope of Arrhenoius (In k vs 1/T) of a first-order reaction is -5 x 10³ K. The value of E_a of the reaction is [Given: R = 8.314 J K^-1 mol^-1)

1. \(-83 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(41.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(83.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(166 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 2. \(41.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

ln k = \(-\frac{E_a}{R T}+\ln A\)

–\(\frac{E_a}{R}=-5 \times 10^3=-5000\)

⇒ \(E_a=5000 \times 8.314=41570 \mathrm{~J} \mathrm{~mol}^{-1}\)

= \(41.57 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 52. For a reaction, activation energy E_a = 0, and the rate constant at 200 K is 1.6 x 10⁶ s^-1. The rate constant at 400 K will be [Given that gas constant R = 8.314 J K^-1 mol^-1]

1. \(3.2 \times 10^4 \mathrm{~s}^{-1}\)
2. \(1.6 \times 10^6 \mathrm{~s}^{-1}\)
3. \(1.6 \times 10^3 \mathrm{~s}^{-1}\)
4. \(3.2 \times 10^6 \mathrm{~s}^{-1}\)

Answer: 2. \(1.6 \times 10^6 \mathrm{~s}^{-1}\)

According to Arrhenius equation, \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_2 T_1}\right]\)

⇒ \(\log \frac{k_2}{1.6 \times 10^6}=0 ; \frac{k_2}{1.6 \times 10^6}=1\)

⇒ \(k_2=1.6 \times 10^6 \mathrm{~s}^{-1}\)

Question 53. The addition of a catalyst during a chemical reaction alters which of the following quantities?

1. Enthalpy
2. Activation energy
3. Entropy
4. Internal energy

Answer: 2. Activation energy

A catalyst provides an alternate path to the reaction which has lower activation energy.

Question 54. The activation energy of a reaction can be determined from the slope of which of the following graphs?

1. \(\ln k v s \frac{1}{T}\)
2. \(\frac{T}{\ln k} v s \frac{1}{T}\)
3. \(\ln k v s T\)
4. \(\frac{\ln k}{T} v s T\)

Answer: 1. \(\ln k v s \frac{1}{T}\)

According to Arrhenius equation, \(k=A e^{-E_a / R T}\)

ln k = ln A – \(\frac{E_a}{R T}\)

Hence, if ln k is plotted

against 1/T, slope of the line will be \(-\frac{E_a}{R}\)

Question 55. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20 °C to 35 °C? (R = 8.314 J mol^-1 K^-1)

1. \(34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(15.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(342 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(269 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 1. \(34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)

⇒  \(k_2=2 k_1, T_1=20+273=293 \mathrm{~K}\) or \(T_2=35+273=308 \mathrm{~K}\)

R = \(8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

⇒  \(\log 2=\frac{E_a}{2.303 \times 8.314}\left(\frac{1}{293}-\frac{1}{308}\right)\)

0.3010 = \(\frac{E_a}{19.147} \times \frac{15}{293 \times 308}\)

⇒  \(E_a=34673 \mathrm{~J} \mathrm{~mol}^{-1}=34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 56. In a zero-order reaction, for every 10 °C rise in temperature, the rate is doubled. If the temperature is increased from 10 °C to 100 °C, the rate of the reaction will become

1. 256 times
2. 512 times
3. 64 times
4. 128 times.

Answer: 2. 512 times

At 10°C rise, the rate increases by 2

⇒ \(\frac{r_{100^{\circ} \mathrm{C}}}{r_{10^{\circ} \mathrm{C}}}=2^{\left(\frac{100-10}{10}\right)}=2^9=512 \text { times }\)

Question 57. The activation energy (E_a) and rate constants (k₁ and k₂) of a chemical reaction at two different temperatures (T₁) and T₂) are related by

1. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)
2. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\)
3. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}+\frac{1}{T_1}\right)\)
4. \(\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)

Answer: 2. \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) and 4. \(\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\)

⇒ \(k_1=A e^{-E_a / R T_1}, k_2=A e^{-E_a / R T_2}\)

ln \(k_1=\ln A-E_a / R T_1\)…(1)

ln \(k_2=\ln A-E_a / R T_2\)…(2)

From eq.(1) and (2), we have ln \(k_2-\ln k_1=\ln A-\frac{E_a}{R T_2}-\ln A+\frac{E_a}{R T_1}\)

ln \(\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \Rightarrow \ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\)

Question 58. The rate of the reaction, 2NO + Cl₂ → 2NOCl is given by the rate equation, rate = k[NO²[Cl₂]. The value of the rate constant can be increased by

1. Increasing the temperature
2. Increasing the concentration of NO
3. Increasing the concentration of the Cl₂
4. Doing all of these.

Answer: 1. Increasing the temperature

The rate constant is independent of the initial concentration of the reactants. It has a constant value at a fixed temperature. According to the Arrhenius equation, the value of the rate constant can be increased by increasing the temperature.

Question 59. The rate constants k₁ and k₂ for two different reactions are \(10^{16} \cdot e^{-2000 / T}\) and \(10^{15^2} \cdot e^{-1000 / T}\), respectively. Tire temperature at which k₁ = k₂ is

1. 2000 K
2. \(\frac{1000}{2.303}\) K
3. 1000 K
4. \(\frac{2000}{2.303}\) K

Answer: 2. \(\frac{1000}{2.303}\) K

⇒ \(k_1=10^{16} e^{-2000 / T}, k_2=10^{15} e^{-1000 / T}\)

When, \(k_1=k_2, 10^{16} e^{-2000 / T}=10^{15} e^{-1000 / T}\)

or \(10 e^{-2000 / T}=e^{-1000 / T}\)

Taking the natural logarithm of both sides, we get ln \(10-\frac{2000}{T}=\frac{-1000}{T}\) or \(2.303-\frac{2000}{T}=\frac{-1000}{T}\)

or, \(\frac{1000}{T}=2.303 \text { or } T=\frac{1000}{2.303} \mathrm{~K}\)

Question 60. The temperature dependence of the rate constant (k) of a chemical reaction is written in terms of the Arrhenius equation, k = A x e^-E*/RT. The activation energy (E*) of the reaction can be calculated by plotting

1. \(k v s T\)
2. \(k v s \frac{1}{\log T}\)
3. \(\log k v s \frac{1}{T}\)
4. \(\log k v s \frac{1}{\log T}\)

Answer: 3. \(\log k v s \frac{1}{T}\)

On plotting log k vs l/T, we get a straight line, the slope indicates the value of activation energy.

Question 61. The activation energy for a simple chemical reaction A \(\rightleftharpoons\) B is E_a in the forward direction. The activation energy for the reverse reaction

1. Is negative of E_a
2. Is always less than E_a
3. Can be less than or more than E_a
4. Is always double of E_a

Answer: 3. Can be less than or more than E_a

Activation energy is the minimum amount of energy required to convert reactant into product. The activation energy for reverse reaction can be less than or more than E_a depending on whether the reaction is exothermic or endothermic.

Question 62. When a biochemical reaction is carried out in the laboratory, outside the human body in the absence of an enzyme, then the rate of reaction obtained is 10^-6 times, and the activation energy of the reaction in the presence of an enzyme is

1. 6/RT
2. P is required
3. Different from E_a obtained in the laboratory
4. Can’t say anything.

Answer: 3. Different from Ea obtained in the laboratory

According to k = \(A e^{-E a / R T}\) (Arrhenius equation), the activation energy of a reaction in the presence of an enzyme is different from E_a obtained in the laboratory.

Question 63. How do enzymes increase the rate of reactions?

1. By lowering the activation energy
2. By increasing activation energy
3. By changing the equilibrium constant
4. By forming enzyme-substrate complex

Answer: 1. By lowering activation energy

Enzymes act like catalysts in biochemical reactions. The presence of an enzyme increases the rate of reaction by lowering the activation energy of the reactant.

Question 64. The activation energy of a chemical reaction can be determined by

1. Evaluating rate constants at two different temperatures
2. Evaluating velocities of reaction at two different temperatures
3. Evaluating rate constant at standard temperature
4. Changing concentration of reactants.

Answer: 1. Evaluating rate constants at two different temperatures

According to Arrhenius equation: \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_2 T_1}\right]\)

where E_a = activation energy

R = gas constant = 8.314 J K^-1 mol^-1

k₁ and k₂ are rate constants of the reaction at two different temperatures T₁ and T₂ respectively.

Question 65. By the action of enzymes, the rate of biochemical reaction

1. Does not change
2. Increases
3. Decreases
4. Either (1) or (3).

Answer: 2. Increases

Since the enzymes are regarded as biological catalysts, therefore their action increases the rate of biological reaction.

Question 66. An increase in the concentration of the reactants of a reaction leads to a change in

1. Activation energy
2. Heat of reaction
3. Threshold energy
4. Collision frequency.

Answer: 4. Collision frequency.

Collision frequency ∝ number of reacting molecules or atoms

The higher the concentration of reactant molecules, the higher is the probability of collision and so the collision frequency.

 

Surface Chemistry

Question 1. The correct option representing a Freundlich adsorption isotherm is

1. \(\frac{x}{m}=k p^{0.3}\)
2. \(\frac{x}{m}=k p^{2.5}\)
3. \(\frac{x}{m}=k p^{-0.5}\)
4. \(\frac{x}{m}=k p^{-1}\)

Answer: 1. \(\frac{x}{m}=k p^{0.3}\)

Freundlich adsorption isotherm equation is \(\frac{x}{m}=k p^{\frac{1}{n}}\left(1 \geq \frac{1}{n} \geq 0\right)\)

Question 2. Which one of the following characteristics is associated with adsorption?

1. ΔG and ΔH are negative but ΔS is positive.
2. ΔG and ΔS are negative but ΔH is positive.
3. ΔG is negative but ΔH and ΔS are positive.
4. ΔG, ΔH and ΔS all are negative.

Answer: 4. ΔG, ΔH and ΔS all are negative.

As the molecules of the adsorbate are held on the surface of the solid adsorbent, entropy decreases i.e., ΔS = -ve.

As ΔG=ΔH- TΔS

For the adsorption to occur, ΔG = -ve and it is possible only if ΔH= -ve

Question 3. In Freundlich adsorption isotherm, the value of 1/n is

1. Between 0 and 1 in all cases
2. Between 2 and 4 in all cases
3. 1 In case of physical absorption
4. 1 In case of chemisorption.

Answer: 1. Between 0 and 1 in all cases

Freundlich adsorption isotherm: \(\frac{x}{m}=k \cdot p^{1 / n} ; \quad 0 \leq \frac{1}{n} \leq 1\)

Read and Learn More NEET MCQs with Answers

Question 4. If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to the adsorption process?

1. x/m =f(p) at constant T
2. x/m =f(T) at constant p
3. p =f(T) at constant (x/m)
4. \(\frac{x}{m}\) = p x T

Answer: 4. \(\frac{x}{m}\) = p x T

⇒ \(\frac{x}{m}=p \times T\) is the incorrect relation.

The correct relation is amount of absorption \(\frac{x}{m} \propto \frac{p}{T}\)

Question 5. The Langmuir adsorption isotherm is deduced using the assumption

1. The adsorption sites are equivalent in their ability to adsorb the particles
2. The heat of adsorption varies with coverage
3. The adsorbed molecules interact with each other
4. The adsorption takes place in multilayers.

Answer: 1. The adsorption sites are equivalent in their ability to adsorb the particles

Langmuir adsorption isotherm is based on the assumption that every adsorption site is equivalent and that the ability of a particle to bind there is independent of whether nearby sites are occupied or not.

Question 6. A plot of log (x/m) versus log P for the adsorption of a gas on a solid gives a straight line with a slope equal to

1. log k
2. -log k
3. n
4. 1/n

Answer: 4. 1/n

This is according to Freundlich adsorption isotherm

Question 7. Which is not correct regarding the adsorption of a gas on the surface of a solid?

1. On increasing temperature, adsorption increases continuously.
2. Enthalpy and entropy change is negative.
3. Adsorption is more for some specific substance.
4. It is a reversible reaction.

Answer: 1. On increasing temperature adsorption increases continuously.

Adsorption is the ability of a substance to concentrate or hold gases, liquids or dissolved substances upon its surface. Solids absorb greater amounts of substances at lower temperatures. In general, adsorption decreases with an increase in temperature.

Question 8. Which one is an example of heterogeneous catalysis?

1. Decomposition of ozone in the presence of nitrogen monoxide,
2. Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron.
3. Oxidation of sulphur dioxide into sulphur trioxide in the presence of oxides of nitrogen.
4. Hydrolysis of sugar catalysed by H⁺ ions.

Answer: 2. Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron.

Question 9. The incorrect statement regarding enzymes is

1. Enzymes are biocatalysts
2. Like chemical catalysts enzymes reduce the activation energy of bioprocesses
3. Enzymes are polysaccharides
4. Enzymes are very specific for a particular reaction and substrate.

Answer: 3. Enzymes are polysaccharides

Enzymes are protein molecules of high molecular mass and form colloidal solutions in water.

Question 10. Which one of the following statements is not correct?

1. The value of the equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium.
2. Enzymes catalyse mainly biochemical reactions.
3. Coenzymes increase the catalytic activity of enzymes.
4. The catalyst does not initiate any reaction.

Answer: 1. The value of the equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium.

Catalysts do not change the value of equilibrium constant as they affect forward as well as backward reactions equa1ly.

Question 11. Which one of the following statements is incorrect about enzyme catalysis?

1. Enzymes are mostly proteinous in nature.
2. Enzyme action is specific.
3. Enzymes are denatured by ultraviolet rays and at high temperatures.
4. Enzymes are least reactive at optimum temperature.

Answer: 4. Enzymes are least reactive at optimum temperature.

The enzyme activity rises rapidly with temperature and becomes maximum at a definite temperature, called optimum temperature.

Question 12. The enzyme which hydrolyses triglycerides to fatty acids and glycerol is called

1. Maltase
2. Lipase
3. Zymase
4. Pepsin.

Answer: 2. Lipase

Question 13. According to the adsorption theory of catalysis, the speed of the reaction increases because

1. The concentration of reactant molecules at the active centres of the catalyst becomes high due to adsorption
2. In the process of adsorption, the activation energy of the molecules becomes large
3. Adsorption produces heat which increases the speed of the reaction
4. Adsorption lowers the activation energy of the reaction.

Answer: 4. Adsorption lowers the activation energy of the reaction.

Adsorption causes a decrease in surface energy which appears as heat. thus, adsorption is an exothermic process and hence lowers the activation energy of the reaction.

Question 14. A colloidal system has particles of which of the following sizes?

1. 10^-9 m to 10^-12 m
2. 10^-6 m to 10^-9 m
3. 10^-4 m to 10^-10 m
4. 10^-5 m to 10^-7 m

Answer: 2. 10^-6 m to 10^-9 m

The particle size of colloids lies in the range of 10^-6 m to 10^-9 m. Particles themselves are invisible even under the most powerful microscope.

Question 15. Pumice stone is an example of

1. Solid sol
2. Foam
3. Sol
4. Gel.

Answer: 1. Solid sol

Pumice stone is an example of a solid sol in which the dispersed phase is gas and the dispersion medium is solid.

Question 16. Given below are two statements:

Statement-1: In the coagulation of a negative sol, the flocculating power of the three given ions is in the order: \(\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^{+}\)

Statement-2: In the coagulation of a positive sol, the flocculating power of the three given salts is in the order: \(\mathrm{NaCl}>\mathrm{Na}_2 \mathrm{SO}_4>\mathrm{Na}_3 \mathrm{PO}_4\)

In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both statement-1 and statement-2 are correct.
2. Both statement-1 and statement-2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but statement 2 is correct.

Answer: 3. Statement 1 is correct but statement 2 is incorrect.

In the coagulation of a positive sol, the flocculating power is in the order: \(\mathrm{PO}_4^{3-}>\mathrm{SO}_4^{2-}>\mathrm{Cl}^{-}\) or, \(\mathrm{NaCl}<\mathrm{Na}_2 \mathrm{SO}_4<\mathrm{Na}_3 \mathrm{PO}_4\)

Question 17. The right option for the statement “Tyndall effect is exhibited by” is

1. Urea solution
2. NaCl solution
3. Glucose solution
4. Starch solution.

Answer: 4. Starch solution.

As the starch solution is a colloidal solution hence it exhibits the Tyndall effect. NaCl, urea and glucose form a true solution.

Question 18. Measuring zeta potential is useful in determining which property of colloidal solution?

1. Viscosity
2. Solubility
3. Stability of the colloidal particles
4. Size of the colloidal particles

Answer: 3. Stability of the colloidal particles

Measuring, zeta potential is useful in determining the stability of the colloidal particles.

Question 19. Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI]I^-1 sol?

1. 50 mL of 0.1 M AgNO₃ + 50 mL of 0.1 M KI
2. 50 mL of 1 M AgNO₃ + 50 mL of 1.5 M KI
3. 50 mL of 1 M AgNO₃ + 50 mL of 2 M KI
4. 50 mL of 2 M AgNO₃ + 50 mL of 1.5 M KI

Answer: 2. 50 mL of 1 M AgNO₃ + 50 mL of 1.5 M KI

If the colloidal sol of AgI is prepared by adding KI solution to AgNO₃ till KI is in slight excess, iodide ion (I^–) will be adsorbed on the surface of AgI thereby, giving a negative charge to the sol.

⇒ \(\mathrm{AgI}+\underset{(\text { From } \mathrm{KI})}{\mathrm{I}^{-}} \longrightarrow \underset{\text { Negative sol }}{\mathrm{AgI}: \mathrm{I}^{-}}\)

Question 20. On which of the following properties does the coagulating power of an ion depend?

1. The magnitude of the charge on the ion alone
2. The size of the ion alone
3. Both the magnitude and sign of the charge on the ion
4. The sign of charge on the ion alone

Answer: 3. Both the magnitude and sign of the charge on the ion

According to the Hardy-Schulze rule, the coagulating power of an electrolyte depends on both the magnitude and sign of the charge of the effective ion or electrolyte.

Question 21. The coagulation values in millimoles per litre of the electrolytes used for the coagulation of As₂S₃ are given below:

1. (NaCl) = 52,
2. (BaCl₂) = 0.69,
3. (MgSO₄) = 0.22

The correct order of their coagulating power is

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

Answer: 3. 3 > 2 > 1

Coagulating power \(\propto \frac{1}{\text { Coagulation value }}\)

The lower the coagulation value, the higher is the coagulating power so, the correct order is: \(\underset{3} {\mathrm{MgSO}_4}>\underset{2} {\mathrm{BaCl}_2}>\underset{1}{\mathrm{NaCl}}\)

Question 22. Fog is a colloidal solution of

1. Solid in gas
2. Gas in gas
3. Liquid in gas
4. Gas in liquid.

Answer: 3. Liquid in gas

Fog is an example of aerosol in which the dispersed phase is liquid and the dispersion medium is gas.

Question 23. Which property of colloidal solution is independent of charge on the colloidal particles?

1. Electroosmosis
2. Tyndall effect
3. Coagulation
4. Electrophoresis

Answer: 2. Coagulation

Tyndall effect is the scattering of light by colloidal particles which is independent of the charge on them.

Question 24. The protecting power of lyophilic colloidal sol is expressed in terms of

1. Coagulation Value
2. Gold Number
3. Critical Micelle Concentration
4. Oxidation Number.

Answer: 2. Gold Number

Question 25. Which one of the following forms micelles in an aqueous solution above a certain concentration?

1. Dodecyl trimethyl ammonium chloride
2. Glucose
3. Urea
4. Pyridinium chloride

Answer: 1. Dodecyl trimethyl ammonium chloride

Question 26. Position of non-polar and polar parts in micelle

1. Polar at the outer surface but non-polar at the inner surface
2. Polar at the inner surface non-polar at the outer surface
3. Distributed over all the surface
4. Are present in the surface only.

Answer: 1. Polar at the outer surface but non-polar at the inner surface

Micelles are the clusters or aggregates formed in solution by association of colloids. Usually, such molecules have a lyophobic group and a lyophilic group. The long hydrocarbon is the lyophobic portion which tries to recede away from the solvent water and the ionisable lyophitic group which tends to go into water resulting in ions.

As the concentration is increased the lyophobic parts receding away from the solvent approach each other and form a cluster. Thus, the lyophobic ends are in the interior and lyophilic groups projecting outward in contact with the solvent.

Question 27. Which one of the following methods is commonly used method for the destruction of colloids?

1. Dialysis
2. Condensation
3. Filtration by animal membrane
4. By adding electrolyte

Answer: 4. By adding electrolyte

By adding electrolytes the colloidal particles are precipitated. The electrolytes neutralise the charge of colloids leading to their coagulation and thus, destroying the colloid.

Question 28. At the critical micelle concentration (CMC) the surfactant molecules

1. Associate
2. Dissociate
3. Decompose
4. Become Completely Soluble.

Answer: 1. Associate

The soap concentration at which micelles (spherical colloid molecules) first appear is called critical micelle concentration (CMC). At this condition, the surfactant molecules associate with each other

Question 29. The ability of an anion, to bring about coagulation of a given colloid, depends upon

1. The magnitude of the charge
2. Both magnitude and charge
3. Its charge only
4. Sign of the charge alone.

Answer: 2. Both magnitude and charge

Both magnitudes of charge and the nature of charge affect the coagulation of a given colloid. The greater the magnitude of the charge, the quicker will be the coagulation.

Question 30. When a few typical solutes are separated by a particular selective membrane such as protein particles, or blood corpuscles, this process is called

1. Transpiration
2. Endosmosis
3. Dialysis
4. Diffusion.

Answer: 3. Dialysis

Dialysis is the process of separating the particles of colloids from the particles of crystalloids by means of diffusion through a selective membrane placed in water.

General Principles And Processes Of Isolation Of Elements

Question 1. Match List 1 and List 2

Choose the correct answer from the options given below:

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

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

Haematite \(-\mathrm{Fe}_2 \mathrm{O}_3\)

Magnetite –\(\mathrm{Fe}_3 \mathrm{O}_4\)

Calamine –\(\mathrm{ZnCO}_3\)

Kaolinite –\(\mathrm{Al}_2 \mathrm{O}_3 \cdot 2 \mathrm{SiO}_2 \cdot 2 \mathrm{H}_2 \mathrm{O}\)

Question 2. Which one is malachite from the following?

1. \(\mathrm{CuCO}_3 \cdot \mathrm{Cu}(\mathrm{OH})_2\)
2. \(\mathrm{CuFeS}_2\)
3. \(\mathrm{Cu}(\mathrm{OH})_2\)
4. \(\mathrm{Fe}_3 \mathrm{O}_4\)

Answer: 1. \(\mathrm{CuCO}_3 \cdot \mathrm{Cu}(\mathrm{OH})_2\)

Malachite: CuCO₃ · Cu(OH)₂

Question 3. Identify the incorrect statement.

1. The scientific and technological process used for isolation of the metal from its ore is known as metallurgy.
2. Minerals are naturally occurring chemical substances in the earth’s crust.
3. Ores are minerals that may contain a metal.
4. Gangue is an ore contaminated with undesired materials.

Answer: 4. Gangue is an ore contaminated with undesired materials.

An ore rarely contains only a desired substance. it is usually contaminated with earthly or undesired materials known as gangue.

Question 4. “Metals are usually not found as nitrates in their ores.” Out of the following two (1 and 2) reasons which is/are true for the above observation?

1. Metal nitrates are highly unstable.
2. Metal nitrates are highly soluble in water.

1. 1 is false but 2 is true.
2. 1 is true but 2 is false.
3. 1 and 2 are true.
4. 1 and 2 are false

Answer: 1. 1 is false but 2 is true.

All nitrates are soluble in water and are quite stable as they do not decompose easily on heating.

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Question 5. Which one of the following is a mineral of iron?

1. Malachite
2. Cassiterite
3. Pyrolusite
4. Magnetite

Answer: 4. Magnetite

Magnetite is Fe₃O₄ and contains up to 70% of iron metal.

Question 6. Cassiterite is an ore of

1. Sb
2. Ni
3. Mn
4. Sn

Answer: 4. Sn

Cassiterite is also known as tin stone (SnO₂), an ore of tin (Sn).

Question 7. Sulphide ores of metals are usually concentrated by the froth floatation process. Which one of the following sulphide ores offers an exception and is concentrated by chemical leaching?

1. Galena
2. Copper pyrite
3. Sphalerite
4. Argentite

Answer: 4. Argentite

The leaching process involves the treatment of the ore with a suitable reagent as to make it soluble while impurities remain insoluble. The ore is recovered from the solution by a suitable chemical method.

Argentite or silver glance, Ag₂S is an ore of silver. Silver is extracted from argentite by the MacArthur and Forest cyanide process (leaching process).

⇒ \(\mathrm{Ag}_2 \mathrm{~S}+4 \mathrm{NaCN} \longrightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{Na}_2 \mathrm{~S}\)

⇒ \(2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{Zn} \longrightarrow \mathrm{Na}_2\left[\mathrm{Zn}(\mathrm{CN})_4\right]+2 \mathrm{Ag}\)

Question 8. The roasting of sulphides gives the gas X as a byproduct. This is a colourless gas with a choking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic, acts as a reducing agent and its acid has never been isolated. The gas X is

1. \(\mathrm{CO}_2\)
2. \(\mathrm{SO}_3\)
3. \(\mathrm{H}_2 \mathrm{~S}\)
4. \(\mathrm{SO}_2\)

Answer: 4. \(\mathrm{SO}_2\)

Question 9. The reaction that does not take place in a blast furnace between 900 K to 1500 K temperature range during extraction of iron is

1. \(\mathrm{C}+\mathrm{CO}_2 \rightarrow 2 \mathrm{CO}\)
2. \(\mathrm{CaO}+\mathrm{SiO}_2 \rightarrow \mathrm{CaSiO}_3\)
3. \(\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{FeO}+\mathrm{CO}_2\)
4. \(\mathrm{FeO}+\mathrm{CO} \rightarrow \mathrm{Fe}+\mathrm{CO}_2\)

Answer: 3. \(\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{FeO}+\mathrm{CO}_2\)

Question 10. The maximum temperature that can be achieved in a blast furnace is

1. Upto 5000 K
2. Upto 1200 K
3. Upto 2200 K
4. Upto 1900 K.

Answer: 3. Upto 2200 K

It can withstand up to approximately 2000°C or 2200K.

Question 11. Considering the Ellingham diagram, which of the following metals can be used to reduce alumina?

1. Fe
2. Zn
3. Mg
4. Cu

Answer: 3. Mg

Any metal oxide with lower value of ΔG° is more stable than a metal oxide with higher ΔG°. This implies that the metal oxide placed higher in the Eliingham diagram can be reduced by the metal involved in the formation of the oxide placed lower in the diagram.

The relative tendency of the various metals to act as reducing agents is: Ca>Mg>AI>Zn>Fe>Cu

Thus, Mg being more reducing in nature, can reduce aluminium oxide (alumina) to aluminium.

Question 12. In the extraction of copper from its sulphide ore, the metal is finally obtained by the reduction of cuprous oxide with

1. Carbon monoxide
2. Copper (1) sulphide
3. Sulphur dioxide
4. Iron (2) sulphide.

Answer: 2. Copper (1) sulphide

It is an example of auto reduction. \(2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO}_2\)

Question 13. The metal oxide which cannot be reduced to metal by carbon is

1. Al₂O₃
2. PbO
3. ZnO
4. Fe₂O₃

Answer: 1. Al₂O₃

Oxides of less reactive metals (like PbO, ZnO, Fe₂O₃) can be reduced by carbon, while oxides of very reactive metals (like Al₂O₃) can be reduced only by the electrolytic method.

Question 14. Which of the following elements is present as the impurity to the maximum extent in the pig iron?

1. Manganese
2. Carbon
3. Silicon
4. Phosphorus

Answer: 2. Carbon

Pig iron contains about 470 carbon and many impurities such as S, Mn, P, Si, etc’ in smaller amounts.

Question 15. The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.

1. \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(b)}+3 \mathrm{CO}_{2(g)}\)
2. \(\mathrm{CaCO}_{3(s)} \rightarrow \mathrm{CaO}_{(s)}+\mathrm{CO}_{2(g)}\)
3. \(\mathrm{CaO}_{(s)}+\mathrm{SiO}_{2(s)} \rightarrow \mathrm{CaSiO}_{3(s)}\)
4. \(2 \mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{(g)}\)

Answer: 3. \(\mathrm{CaO}_{(s)}+\mathrm{SiO}_{2(s)} \rightarrow \mathrm{CaSiO}_{3(s)}\)

Slag is formed by the reaction \(\mathrm{CaO}+\mathrm{SiO}_2 \rightarrow \mathrm{CaSiO}_3\)

Question 16. Which of the following statements, about the advantage of roasting of sulphide ore before reduction is not true?

1. The ΔG_f° of the sulphide is greater than those for CS₂ and H₂S.
2. The ΔG_f° is negative for roasting of sulphide ore to oxide.
3. Roasting of the sulphide to the oxide is thermodynamically feasible.
4. Carbon and hydrogen are suitable reducing agents for metal sulphides.

Answer: 4. Carbon and hydrogen are suitable reducing agents for metal sulphides.

The standard free energies of formation (ΔG_f°) of most of the sulphides are greater than those of CS₂ and H₂S. Hence, neither carbon nor hydrogen can reduce metal sulphides to metal. The standard free energies of the formation of oxides are much lower than those of SO₂ Therefore, oxidation of metal sulphides to metal oxides is thermodynamically favourable. Hence, sulphide ore is roasted to the oxide before reduction.

Question 17. Nitriding is the process of surface hardening of steel by treating it in an atmosphere of

1. NH₃
2. O₃
3. N₂
4. H₂S

Answer: 1. NH₃

When steel is heated in the Presence of NH₃, iron nitride on the surface of the steel is formed which imparts a hard coating. This process is called nitriding.

Question 18. Aluminium is extracted from alumina (Al₂O₃) by electrolysis of a molten mixture of

1. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{HF}+\mathrm{NaAlF}_4\)
2. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{CaF}_2+\mathrm{NaAlF}_4\)
3. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{Na}_3 \mathrm{AlF}_6+\mathrm{CaF}_2\)
4. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{KF}+\mathrm{Na}_3 \mathrm{AlF}_6\)

Answer: 3. \(\mathrm{Al}_2 \mathrm{O}_3+\mathrm{Na}_3 \mathrm{AlF}_6+\mathrm{CaF}_2\)

The electrolytic mixture contains alumina(Al₂O₃), cryolite(Na₃AlF₆) and fluorspar(CaF₂) in the ratio of 20:40:20. Due to the presence of these, the conductivity of alumina increases and fusion temperature decreases from 2000°C to 900°c

Question 19. The purification of aluminium, by electrolytic refining, is known as

1. Hoopes process
2. Baeyers process
3. Hall’s process
4. Serpecks process.

Answer: 1. Hoopes process

Aluminium metal obtained from the Hoopet electrolytic refining process is about 99.9% pure. The cell used for this process consists of three layers. The upper layer is pure Al, which acts as the cathode, and the middle layer is a mixture of fluorides of Al and Ba, which acts as an electrolyte. The lowest layer is impure Ali which acts as anode. On electrolysis, pure Al is transferred from the bottom to the top layer, through the middle layer.

Question 20. Calcium is obtained by

1. Reduction of calcium chloride with carbon
2. Electrolysis of molten anhydrous calcium chloride
3. Roasting of limestone
4. Electrolysis of solution of calcium chloride in H₂O

Answer: 2. Electrolysis of molten anhydrous calcium chloride

Calcium is obtained by the electrolysis of a fused mixture of anhydrous CaCl₂ and CaF₂ in a graphite-lined tank which serves as anode. The cathode is a hollow movable iron rod which is kept cool. During electrolysis, calcium is deposited at the cathode while Cl₂ is liberated at the anode.

Question 21. Extraction of gold and silver involves leaching with CN^– ion. Silver is later recovered by

1. Distillation
2. Zone refining
3. Displacement with Zn
4. Liquation.

Answer: 3. Displacement with Zn

Extraction of gold and silver involves leaching the metal with CN^– and the metals silver and gold are later recovered by displacement method.

⇒ \(4 M_{(s)}+8 \mathrm{CN}_{(a q)}^{-}+2 \mathrm{H}_2 \mathrm{O}_{(a q)}+\mathrm{O}_{2(g)} \rightarrow 4\left[M(\mathrm{CN})_2\right]_{(a q)}^{-}+4 \mathrm{OH}_{(a q)}^{-}\)

⇒ \(2\left[\mathrm{M}(\mathrm{CN})_2\right]_{(\mathrm{miq})}^{-}+\mathrm{Zn}_{(s)} \rightarrow 2 \mathrm{M}_{(x)}+\left[\mathrm{Zn}(\mathrm{CN})_4\right]^{2-}{ }_{(\omega q)}\)

Question 22. Which one of the following methods can be used to obtain highly pure metal which is liquid at room temperature?

1. Zone refining
2. Electrolysis
3. Chromatography
4. Distillation

Answer: 4. Distillation

Question 23. Identify the correct statement from the following

1. Wrought iron is impure iron with 4% carbon.
2. Blister copper has a blistered appearance due to the evolution of CO₂.
3. Vapour phase refining is carried out for nickel by the van Arkel method.
4. Pig iron can be moulded into a variety of shapes.

Answer: 4. Wrought iron is impure iron with 4% carbon.

1. Pig iron is impure iron with 4% carbon
2. Blister copper has a blistered appearance due to the evolution of SO₂.
3. Vapour phase refining is carried out for the nickel bv Mondt process.
4. Pig iron can be moulded into a variety of shapes

Question 24. Match items of Column 1 with the items of Column 2 and assign the correct code:

Answer: 3

Question 25. Which of the following pairs of metals is purified by the van Arkel method?

1. Ga and In
2. Zr and Ti
3. Ag and Au
4. Ni and Fe

Answer: 2. Zr and Ti

van Arkel method is used for the purification of Zr and Ti.

Question 26. The method of zone refining of metals is based on the principle of

1. Greater mobility of the pure metal than that of the impurity
2. Higher melting point of the impurity than that of the pure metal
3. The greater noble character of the solid metal than that of the impurity
4. Greater solubility of the impurity in the molten state than in the solid.

Answer: 4. Greater solubility of the impurity in the molten state than in the solid.

Elements which are used as semiconductors such as Si, Ge, Ga, etc. are refined by this method, which is based on the difference in solubility of impurities in the molten and solid state of the metal.

 

p Block Elements

Question 1. In which of the following compounds, nitrogen exhibits the highest oxidation state?

1. \(\mathrm{N}_2 \mathrm{H}_4\)
2. \(\mathrm{NH}_3\)
3. \(\mathrm{N}_3 \mathrm{H}\)
4. \(\mathrm{NH}_2 \mathrm{OH}\)

Answer: 3. \(\mathrm{N}_3 \mathrm{H}\)

N₂H₂ ⇒ 2x + 4(+1) = 0

⇒ 2x +4=0 ⇒ x = -2

N₃H ⇒ x + 3(+1) = 0 ⇒ x =-3

N₃H ⇒ 3x + 1(+1) = 0

⇒ 3x + 1 = 0 ⇒ x = -1/3

NH₂OH ⇒ x + 2 + 1(-2) + 1 = 0

⇒ x + 1 = 0 ⇒ x = -1

Thus, the highest oxidation state is -1/3

Question 2. Nitrogen forms N₂, but phosphorus does not form P₂, however, it converts P₄, reason is

1. Triple bond present between phosphorus atom
2. pπ – pπ bonding is weak
3. pπ – pπ bonding is strong
4. Multiple bonds form easily.

Answer: 2. pπ – pπ bonding is weak

For strong n-bonding, pπ – pπ bonding should be strong. In the case of P, due to its larger size as compared to N-atom, pπ- Pπ bonding is not so strong.

Question 3. Which of the following oxides is most acidic?

1. \(\mathrm{As}_2 \mathrm{O}_5\)
2. \(\mathrm{P}_2 \mathrm{O}_5\)
3. \(\mathrm{N}_2 \mathrm{O}_3\)
4. \(\mathrm{Sb}_2 \mathrm{O}_5\)

Answer: 3. \(\mathrm{N}_2 \mathrm{O}_3\)

Among N, P, As and Sb, the former has the highest electronegativity (EN) so its oxide is the most acidic.

As the electronegativity value of the element increases, the acidic character of the oxide also increases.

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Question 4. Which of the following has the highest dipole moment?

1. \(\mathrm{SbH}_3\)
2. \(\mathrm{AsH}_3\)
3. \(\mathrm{NH}_3\)
4. \(\mathrm{PH}_3\)

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

Due to the greater electronegativity of nitrogen, the dipole moment for NH₃ is greater.

Question 5. The basic character of hydrides of the V group elements decreases in the order

1. \(\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3\)
2. \(\mathrm{SbH}_3>\mathrm{AsH}_3>\mathrm{PH}_3>\mathrm{NH}_3\)
3. \(\mathrm{SbH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{NH}_3\)
4. \(\mathrm{NH}_3>\mathrm{SbH}_3>\mathrm{PH}_3>\mathrm{AsH}_3\)

Answer: 1. \(\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3\)

AI1 the hydrides of group V elements have one lone pair of electrons on their central atom. Therefore, they can act as Lewis bases. The basic character of these hydrides decreases down the group.

Question 6. Among the following oxides, the lowest acidic is

1. \(\mathrm{As}_4 \mathrm{O}_6\)
2. \(\mathrm{As}_4 \mathrm{O}_{10}\)
3. \(\mathrm{P}_4 \mathrm{O}_6\)
4. \(\mathrm{P}_4 \mathrm{O}_{10}\)

Answer: 1. \(\mathrm{As}_4 \mathrm{O}_6\)

The acidic character of the oxides decreases with the decrease in the oxidation state and also decreases down the group

Question 7. Which of the following fluorides does not exist?

1. NF₅
2. PF₅
3. ASF₅
4. SbF₅

Answer: 1. NF₅

Nitrogen cannot form pentahalides because it cannot expand its octet due to non-availability of d-orbitals.

Question 8. Which one has the lowest boiling point?

1. \(\mathrm{NH}_3\)
2. \(\mathrm{PH}_3\)
3. \(\mathrm{AsH}_3\)
4. \(\mathrm{SbH}_3\)

Answer: 2. \(\mathrm{PH}_3\)

The boiling point of hydrides increases with an increase in atomic number but ammonia has an exceptionally high boiling point due to hydrogen bonding. Thus, the correct order of boiling point is, BiH₃ > SbH₃ > NH₃ > AsH₃ > PH₃

Question 9. The number of electrons shared in the formation of nitrogen molecules is

1. 6
2. 10
3. 2
4. 8

Answer: 1. 6

Nitrogen molecule is diatomic containing a triple bond between two N atoms, \(\ddot{\mathrm{N}} \equiv \ddot{\mathrm{N}}\) therefore, nitrogen molecule is formed by sharing six electrons

Question 10. Nitrogen is a relatively inactive element because

1. Its atom has a stable electronic configuration
2. It has a low atomic radius
3. Its electronegativity is fairly high
4. The dissociation energy of its molecule is fairly high.

Answer: 4. Dissociation energy of its molecule is fairly high.

N₂ molecule contains a triple bond between N atoms having very high dissociation energy (946 kJ mol^-1) due to which it is relatively inactive.

Question 11. Pure nitrogen is prepared in the laboratory by heating a mixture of

1. \(\mathrm{NH}_4 \mathrm{OH}+\mathrm{NaCl}\)
2. \(\mathrm{NH}_4 \mathrm{NO}_3+\mathrm{NaCl}\)
3. \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NaOH}\)
4. \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NaNO}_2\)

Answer: 4. \(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NaNO}_2\)

Question 12. Which of the following statements is not correct for nitrogen?

1. Its electronegativity is very high.
2. d-orbitals are available for bonding.
3. It is a typical non-metal.
4. Its molecular size is small.

Answer: 2. d-orbitals are available for bonding.

In the case of nitrogen, d-orbitals are not available for bonding. N: 1s² 2s² 2p³

Question 13. Urea reacts with water to form A which will decompose to form B. B When passed through Cu_(aq)²⁺, deep blue colour solution C is formed. What is the formula of C from the following?

1. \(\mathrm{CuSO}_4\)
2. \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}\)
3. \(\mathrm{Cu}(\mathrm{OH})_2\)
4. \(\mathrm{CuCO}_3 \cdot \mathrm{Cu}(\mathrm{OH})_2\)

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

Question 14. Aqueous solution of ammonia consists of

1. \(\mathrm{H}^{+}\)
2. \(\mathrm{OH}^{-}\)
3. \(\mathrm{NH}_4^{+}\)
4. \(\mathrm{NH}_4^{+}\) and \(\mathrm{OH}^{-}\)

Answer: 4. \(\mathrm{NH}_4^{+}\) and \(\mathrm{OH}^{-}\)

Aqueous solution of ammonia contains \(\mathrm{NH}_4^{+}\) and OH- ions.

Question 15. Which of the following oxides of nitrogen is paramagnetic?

1. \(\mathrm{NO}_2\)
2. \(\mathrm{N}_2 \mathrm{O}_3\)
3. \(\mathrm{N}_2 \mathrm{O}\)
4. \(\mathrm{N}_2 \mathrm{O}_5\)

Answer: 1. \(\mathrm{NO}_2\)

NO₂ is paramagnetic due to the presence of one unpaired electron.

Question 16. Which of the following is a nitric acid anhydride?

1. \(\mathrm{NO}\)
2. \(\mathrm{NO}_2\)
3. \(\mathrm{N}_2 \mathrm{O}_5\)
4. \(\mathrm{N}_2 \mathrm{O}_3\)

Answer: 3. \(\mathrm{N}_2 \mathrm{O}_5\)

When two molecules of nitric acid undergo heating, lose a water molecule to form an anhydride.

Thus, N₂O₅ is nitric acid anhydride.

Question 17. When copper is heated with a cone. HNO₃ it produces

1. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2,\mathrm{NO}\) and \(\mathrm{NO}_2\)
2. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{N}_2 \mathrm{O}\)
3. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{NO}_2\)
4. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{NO}\)

Answer: 3. \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\) and \(\mathrm{NO}_2\)

Question 18. Zn gives H₂ gas with H₂SO₄ and HCl but not with HNO₃ because

1. Zn acts as an oxidising agent when reacting with HNO₃
2. HNO₃ is a weaker acid than H₂SO₄ and HCl
3. In the electrochemical series, Zn is above the hydrogen
4. NO₃ is reduced in preference to hydronium ion.

Answer: 4. NO₃ is reduced in preference to hydronium ion.

Zinc is in the top position of hydrogen in the electrochemical series. So, Zn displaces H₂ from dilute H₂SO₄ and HCI with liberation of H₂.

⇒ \(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{ZnSO}_4+\mathrm{H}_2\)

On the other hand, HNO₂ is one oxidising agent. Hydrogen obtained in the reaction is converted into H₂O.

⇒ \(\mathrm{Zn}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+2 \mathrm{H}\)

⇒ \(2 \mathrm{HNO}_3 \rightarrow \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}_2+\mathrm{O}\)

⇒ \(2 \mathrm{H}+\mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{O}\)

Question 19. Sugarcane in reaction with nitric acid gives

1. CO₂ and SO₂
2. (COOH)₂
3. 2HCOOH(two moles)
4. No reaction.

Answer: 2. (COOH)₂

⇒ \(\underset{\text { Cane sugar }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\underset{\text { From } \mathrm{HNO}_3}{18[\mathrm{O}]} \longrightarrow \underset{\text { Oxalic acid }}{6(\mathrm{COOH})_2}+5 \mathrm{H}_2 \mathrm{O}\)

Question 20. Which of the following phosphorus is the most reactive?

1. Scarlet phosphorus
2. White phosphorus
3. Red phosphorus
4. Violet phosphorus

Answer: 2. White phosphorus

White phosphorus has low ignition temperature so it is the most reactive among all the allotropes.

Question 21. Each of the following is true for white and red phosphorus except that they

1. Are both soluble in CS₂
2. Can be oxidised by heating in air
3. Consist of the same kind of atoms
4. Can be converted into one another.

Answer: 1. Are both soluble in CS₂

Red phosphorus is insoluble in CS₂ and only white P is soluble in C₂.

Question 22. A compound X upon reaction with H₂O produces a colourless gas ‘Y with rotten fish smell. Gas Y is absorbed in a solution of CuSO₄ to give Cu₃P₂ as one of the products. Predict the compound ‘X.

1. \(\mathrm{Ca}_3 \mathrm{P}_2\)
2. \(\mathrm{NH}_4 \mathrm{Cl}\)
3. \(\mathrm{As}_2 \mathrm{O}_3\)
4. \(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2\)

Answer: 1. \(\mathrm{Ca}_3 \mathrm{P}_2\)

⇒ \(\underset{\mathrm{Y}}{\mathrm{Ca}_3 \mathrm{P}_2}+6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_2+ \underset{\mathrm{Y}}{2 \mathrm{PH}_{3(g)}}\)

⇒ \(3 \mathrm{CuSO}_4+ \underset{\mathrm {Y}}{2 \mathrm{PH}_3} \rightarrow \underset{\text {Copper phosphide}}{\mathrm{Cu}_3 \mathrm{P}_2}+3 \mathrm{H}_2 \mathrm{SO}_4\)

Question 23. PH₄I + NaOH forms

1. \(\mathrm{PH}_3\)
2. \(\mathrm{NH}_3\)
3. \(\mathrm{P}_4 \mathrm{O}_6\)
4. \(\mathrm{P}_4 \mathrm{O}_{10}\)

Answer: 1. \(\mathrm{PH}_3\)

⇒ \(\mathrm{PH}_4 \mathrm{I}+\mathrm{NaOH} \rightarrow \mathrm{NaI}+\mathrm{PH}_3+\mathrm{H}_2 \mathrm{O}\)

Question 24. Identify the incorrect statement related to PCl₅ from the following:

1. PCl₅ molecule is non-reactive.
2. Three equatorial P – Cl bonds make an angle of 120° with each other.
3. Two axial P – Cl bonds make an angle of 180° with each other.
4. Axial P – Cl bonds are longer than equatorial P – Cl bonds.

Answer: 1. PCl₂ molecule is non-reactive.

It is a reactive gas as it easily provides Cl₂ gas.

Question 25. PCl₃ reacts with water to form

1. \(\mathrm{PH}_3\)
2. \(\mathrm{H}_3 \mathrm{PO}_3, \mathrm{HCl}\)
3. \(\mathrm{POCl}_3\)
4. \(\mathrm{H}_3 \mathrm{PO}_4\)

Answer: 2. \(\mathrm{H}_3 \mathrm{PO}_3, \mathrm{HCl}\)

⇒ \(\mathrm{PCl}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \mathrm{HCl}\)

Question 26. Which of the following oxoacids of phosphorus has the strongest reducing property?

1. \(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7\)
2. \(\mathrm{H}_3 \mathrm{PO}_3\)
3. \(\mathrm{H}_3 \mathrm{PO}_2\)
4. \(\mathrm{H}_3 \mathrm{PO}_4\)

Answer: 3. \(\mathrm{H}_3 \mathrm{PO}_2\)

Acids which contain P – H bonds have strong reducing properties. Among the given compounds, H₃PO₂ is the strongest reducing agent as it contains two P – H bonds.

Question 27. Which is the correct statement for the given acids?

1. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.
2. Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid.
3. Both are diprotic acids.
4. Both are triprotic acids.

Answer: 1. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.

Question 28. Strong reducing behaviour of H₃PO₂ is due to

1. High electron gain enthalpy of phosphorus
2. High oxidation state of phosphorus
3. Presence of two —OH groups and one P—H bond
4. Presence of one —OH group and two P—H bonds.

Answer: 4. Presence of one —OH group and two P—H bonds.

All oxyacids of phosphorus which have P-H bonds act as strong reducing agents. H₃PO₂ has two P-H bonds hence, it acts as a strong reducing agent.

Question 29. Which of the following statements is not valid for oxoacids of phosphorus?

1. Orthophosphoric acid is used in the manufacture of triple superphosphate.
2. Hypophosphorous acid is a diprotic acid.
3. All oxoacids contain tetrahedral four-coordinated phosphorus.
4. All oxoacids contain at least one P = O unit and one P—OH group.

Answer: 2. Hypophosphorous acid is a diprotic acid.

Hypophosphorous acid is a monoprotic acid

Question 30. Oxidation states of P in \(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5, \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6, \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7\) are respectively

1. +3, +5, +4
2. +5, +3, +4
3. +5, +4, +3
4. +3, +4, +5

Answer: 4. +3, +4, +5

The oxidation state can be calculated as:

H₄P₂O₅ : +4 + 2x+ 5(- 2) = 0 = 2x – 6 = 0 = x = +3

H₄P₂O₆ : +4 + 2x + 6(- 2) = 0 = 2x – 8 = 0 t x= +4

H₄P₂O₇ : +4 + 2x + 7(- 2) = 0 + 2x – 10 = 0 = x = +5

Question 31. How many bridging oxygen atoms are present in P₄O₁₀?

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

Answer: 1. 6

Question 32. The structural formula of hypophosphorous acid is

Answer: 3

The formula of hypophosphorous acid is H₃PO₂ as shown in (3). It is a monobasic acid

Question 33. H₃PO₂ is the molecular formula of an acid of phosphorus. Its name and basicity respectively are

1. Phosphorous acid and two
2. Hypophosphorous acid and two
3. Hypophosphorous acid and one
4. Hypophosphoric acid and two.

Answer: 3. Hypophosphorous acid and one

H₃PO₂ is named as hypophosphorous acid. As it contains only one P-OH group, its basicity is one.

Question 34. Which one of the following substances is used in the laboratory for fast drying of neutral gases?

1. Phosphorus pentoxide
2. Active charcoal
3. Anhydrous calcium chloride
4. Na₃PO₄

Answer: 1. Phosphorus pentoxide

P₂O₅ absorbs moisture much more readily than anhydrous CaCl₂

Question 35. P₂O₅ is heated with water to give

1. Hypophosphorous acid
2. Phosphorous acid
3. Hypophosphoric acid
4. Orthophosphoric acid.

Answer: 4. Orthophosphoric acid.

⇒ \(\mathrm{P}_2 \mathrm{O}_5+3 \mathrm{H}_2 \mathrm{O}\) \(\underrightarrow{^{\Delta}}\) \(2 \mathrm{H}_3 \mathrm{PO}_4\)

Question 36. The basicity of orthophosphoric acid is

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

Answer: 2. 3

Orthophosphoric acid, H₃PO₄ contains three P-OH groups and is, therefore, tribasic

Question 37. When orthophosphoric acid is heated to 600°C, the product formed is

1. \(\mathrm{PH}_3\)
2. \(\mathrm{P}_2 \mathrm{O}_5\)
3. \(\mathrm{H}_3 \mathrm{PO}_3\)
4. \(\mathrm{HPO}_3\)

Answer: 4. \(\mathrm{HPO}_3\)

On heating, it gives pyrophosphoric acid at 525 K and metaphosphoric acid at 875 K

Question 38. Given below are two statements:

Statement-1: The boiling points of the following hydrides of group 16 elements increase in the order: H₂O < H₂S < H₂Se < H₂Te.

Statement 2: The boiling points of these hydrides increase with the increase in molecular mass.

In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both statement-1 and statement-2 are correct.
2. Both statement-1 and statement-2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but statement 2 is correct.

Answer: 2. Both statement-1 and statement-2 are incorrect.

Question 39. Which is the correct thermal stability order for H₂E (E = O, S, Se, Te and Po)?

1. \(\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}<\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S}\)
2. \(\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}\)
3. \(\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}\)
4. \(\mathrm{H}_2 \mathrm{Po}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}\)

Answer: 4. \(\mathrm{H}_2 \mathrm{Po}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}\)

The thermal stability of hydrides decreases from H₂O to H₂Po. This is because as the size of atom E it H₂E increases, the bond H-E becomes weaker and thus, breaks on heating. Therefore, the correct order of thermal stability is H₂Po < H₂Te < H₂Se < H₂S < H₂O.

Question 40. The acidity of diprotic acids in aqueous solutions increases in the order

1. \(\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\)
2. \(\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Te}\)
3. \(\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}\)
4. \(\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{~S}\)

Answer: 1. \(\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\)

As the atomic size increases down the group, the bond length increases the bond strength decreases and the cleavage of the E-H bond becomes easier thus, more will be acidity. Thus, the correct order is: H₂S < H₂Se < H₂Te.

Question 41. Which of the following bonds has the highest energy?

1. S-S
2. O-O
3. Se-Se
4. Te-Te

Answer: 1. S-S

The bond energy of S – S is exceptionally high due to its catenation tendency.

Question 42. Which of the following does not give oxygen on heating?

1. \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\)
2. \(\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7\)
3. \(\mathrm{KClO}_3\)
4. \(\mathrm{Zn}\left(\mathrm{ClO}_3\right)_2\)

Answer:

⇒ \(\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{N}_2+\mathrm{Cr}_2 \mathrm{O}_3+4 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Zn}\left(\mathrm{ClO}_3\right)_2\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{ZnCl}_2+3 \mathrm{O}_2\)

⇒ \(\mathrm{KClO}_3\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{KCl}+3 / 2 \mathrm{O}_2\)

⇒ \(2 \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) \(\underrightarrow{^{\Delta}}\) \(2 \mathrm{~K}_2 \mathrm{CrO}_4+\mathrm{Cr}_2 \mathrm{O}_3+3 / 2 \mathrm{O}_2\)

Question 43. Which would quickly absorb oxygen?

1. Alkaline solution of pyrogallol
2. Cone. H₂SO₄
3. Lime water
4. Alkaline solution of CuSO₄

Answer: 1. Alkaline solution of pyrogallol

The alkaline solution of pyrogallol absorbs oxygen quickly.

Question 44. Oxygen will directly react with each of the following elements except

1. P
2. Cl
3. Na
4. S

Answer: 2. Cl

Chlorine does not react directly with oxygen.

Question 45. It is possible to obtain oxygen from air by fractional distillation because

1. Oxygen is in a different group of the periodic table from nitrogen
2. Oxygen is more reactive than nitrogen
3. Oxygen has higher b.pt. Than nitrogen
4. Oxygen has a lower density than nitrogen.

Answer: 3. Oxygen has higher b.pt. Than nitrogen

Air is liquefied by making use of the |oule Thomson effect (cooling by expansion of the gas). Water vapour and CO₂ are removed by solidification.

The remaining constituents of liquid air i.e., liquid oxygen and liquid nitrogen are separated by means of fractional distillation as fractional distillation is a process of separation of mixture based on the difference in their boiling points. (b.pt. of O₂ = – 183°C : b.pt. of N₂ = -195.8°C)

Question 46. Match the following: 

Which of the following is the correct option?

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

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

CO – neutral, BaO – basic,

Al₂O₃ – amphoteric and Cl₂O₇ – acidic.

Question 47. The angular shape of the ozone molecule (O₃) consists of

1. 1σ and 1π bond
2. 2σ and 1π bond
3. 1σ and 2π bonds
4. 2σ and 2π bonds

Answer: 2. 2σ and 1π bond

The angular shape of the ozone molecule (O₃)

Question 48. The gases respectively absorbed by alkaline pyrogallol and oil of cinnamon are

1. \(\mathrm{O}_3, \mathrm{CH}_4\)
2. \(\mathrm{O}_2, \mathrm{O}_3\)
3. \(\mathrm{SO}_2, \mathrm{CH}_4\)
4. \(\mathrm{N}_2 \mathrm{O}, \mathrm{O}_3\)

Answer: 2. \(\mathrm{O}_2, \mathrm{O}_3\)

Alkaline pyrogallol absorbs O₂ and oil of cinnamon absorbs O₃

Question 49. Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other?

1. Is soluble in water.
2. Is used as a food preservative.
3. Forms acid-rain.
4. Is a reducing agent.

Answer: 2. Is used as a food preservative.

NO₂ is not used as a food preservative.

Question 50. Sulphur trioxide can be obtained by which of the following reactions?

1. \(\mathrm{CaSO}_4+\mathrm{C}\) \(\underrightarrow{^{\Delta}}\)
2. \(\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \underrightarrow{^{\Delta}}\)
3. \(\mathrm{S}+\mathrm{H}_2 \mathrm{SO}_4 \underrightarrow{^{\Delta}}\)
4. \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{PCl}_5 \underrightarrow{^{\Delta}}\)

Answer: 2. \(\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \underrightarrow{^{\Delta}}\)

⇒ \(\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\) \(\underrightarrow{^{\Delta}}\) \(\mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{SO}_3\)

Question 51. Match List 1 with List 2

Choose the correct answer from the options given below:

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

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

Question 52. Which of the following oxoacids of sulphur has — O — O — linkage

1. H₂SO₃, sulphurous acid
2. H₂SO₄, sulphuric acid
3. H₂S₂O₈, peroxodisulphuric acid
4. H₂S₂O₇, pyrosulphuric acid

Answer: 3. H2S₂O₂, peroxodisulphuric acid

Peroxodisulphuric acid, H₂S₂O₈ has – O – O linkage.

Question 53. Identify the correct following:

1. \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\)
2. \(\mathrm{H}_2 \mathrm{SO}_3\)
3. \(\mathrm{H}_2 \mathrm{SO}_4\)
4. \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8\)

Answer: 1. \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\)

Question 54. In which pair of ions do both species contain an S — S bond?

1. \(\mathrm{S}_4 \mathrm{O}_6^{2-}, \mathrm{S}_2 \mathrm{O}_3^{2-}\)
2. \(\mathrm{S}_2 \mathrm{O}_7^{2-}, \mathrm{S}_2 \mathrm{O}_8^{2-}\)
3. \(\mathrm{S}_4 \mathrm{O}_6^{2-}, \mathrm{S}_2 \mathrm{O}_7^{2-}\)
4. \(\mathrm{S}_2 \mathrm{O}_7^{2-}, \mathrm{S}_2 \mathrm{O}_3^{2-}\)

Answer: 1. \(\mathrm{S}_4 \mathrm{O}_6^{2-}, \mathrm{S}_2 \mathrm{O}_3^{2-}\)

Question 55. Oleum is

1. Castor oil
2. Oil of vitriol
3. Fuming H₂SO₄
4. None of these.

Answer: 3. Fuming H₂SO₄

Pyrosulphuric acid or oleum (+6) is H₂S₂O₇ which is obtained by dissolving SO₃ and is called fuming sulphuric acid.

Question 56. Match List 1 (substances) with List 2 (processes) employed in the manufacture of the substances and select the correct option.

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

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

Question 57. Statement 1: Acid strength increases in the order given as HF « HCl« HBr « HI.

Statement 2: As the size of the elements F, Cl, Br, and I increases down the group, the bond strength of HF, HC1, HBr, and HI decreases and so the acid strength increases.

In light of the above statements, choose the correct answer from the options given below.

1. Statement 1 is incorrect but statement 2 is true.
2. Both statement 1 and statement 2 are true.
3. Both statement 1 and statement 2 are false.
4. Statement 1 is correct but statement 2 is false.

Answer: 2. Both statement 1 and statement 2 are true.

Question 58. In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?

1. \(\mathrm{CO}_2<\mathrm{SiO}_2 <\mathrm{SnO}_2<\mathrm{PbO}_2\): Increasing oxidizing power
2. \(\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}\) : Increasing acidic strength
3. \(\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S} <\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\) : Increasing pK values
4. \(\mathrm{NH}_3<\mathrm{PH}_3<\mathrm{AsH}_3<\mathrm{SbH}_3\) : Increasing acidic character

Answer: 3. \(\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{~S} <\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}\) : Increasing pK values

pK_a

H₂O: 14

H₂S: 7

H₂Se: 3.872

H₂Te: 2.6

Question 59. Which of the following statements is not true for halogens?

1. All form monobasic oxyacids.
2. All are oxidizing agents.
3. All but fluorine show positive oxidation states.
4. Chlorine has the highest electron-gain enthalpy.

Answer: 3. All but fluorine show positive oxidation states.

All halogens show both positive and negative oxidation states while fluorine shows only negative oxidation states except +1 in HOF.

Question 60. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?

1. \(\mathrm{Br}_2>\mathrm{I}_2>\mathrm{F}_2>\mathrm{Cl}_2\)
2. \(\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2\)
3. \(\mathrm{I}_2>\mathrm{Br}_2>\mathrm{Cl}_2>\mathrm{F}_2\)
4. \(\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2\)

Answer: 4. \(\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2\)

The order of bond dissociation enthalpy is

A reason for this anomaly is the relatively large electron-electron repulsion among the lone pairs in F₂ a molecule where they are much closer to each other than in the case of Cl₂

Question 61. The variation of the boiling points of the hydrogen halides is in the order HF > HI > HBr > HCl. What explains the higher boiling point of hydrogen fluoride?

1. There is strong hydrogen bonding between HF molecules.
2. The bond energy of HF molecules is greater than in other hydrogen halides.
3. The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule.
4. The electronegativity of fluorine is much higher than for other elements in the group.

Answer: 1. There is strong hydrogen bonding between HF molecules.

HF forms strong intermolecular H-bonding due to the high electronegativity of F. Hence, the boiling point of HF is abnormally high. Boiling points of other hydrogen halides gradually increase from HCl to HI due to an increase in the size of halogen atoms from Cl to I which further increases the magnitude of van der Waals forces.

Question 62. Among the following which is the strongest oxidising agent?

1. Br₂
2. I₂
3. Cl₂
4. F₂

Answer: 4. F₂

Standard reduction potentials of halogens are positive and decrease from fluorine to iodine. So, F₂ is the strongest oxidising agent.

Question 63. Which one of the following arrangements does not give the correct picture of the trends indicated against it?

1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy
2. F₂ > Cl₂ > Br₂ > I₂: Electronegativity
3. F₂ > Cl₂ > Br₂ > I₂: Oxidizing power
4. F₂ > Cl₂ > Br₂ > I₂: Electron gain enthalpy

Answer: 1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy and 4. F₂ > Cl₂ > Br₂ > I₂: Electron gain enthalpy

In the case of diatomic molecules (X₂) of halogens, the bond dissociation energy decreases in the order: Cl₂ > Br₂ > F₂> I₂.

This is due to the relatively large electron-electron repulsion among the lone pairs is F, then in the case of Cl₂.

The oxidising power, electronegativity and reactivity decrease in the order: F₂ > C₂ > Br₂ > I₂

Electron gain enthalpy of halogens follows the given order: Cl₂>F₂>Br₂>I₂

The low value of electron gain enthalpy of fluorine is probably due to the small size of the fluorine atom.

Question 64. Which one of the following orders is not in accordance with the property stated against it?

1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy
2. F2 > Cl2 > Br2 > I₂: Oxidising power
3. HI > HBr > HCl > HF: Acidic property in water
4. F₂ > Cl₂ > Br₂ > I₂: Electronegativity

Answer: 1. F₂ > Cl₂ > Br₂ > I₂: Bond dissociation energy

The lower value of bond dissociation energy of fluorine is due to the high inter-electronic repulsions between non-bonding electrons in the 2p-orbitals of fluorine. As a result, the F – F bond is weaker in comparison to the CI – Cl and Br – Br bonds.

Question 65. Which statement is wrong?

1. Bond energy of F₂ > Cl₂
2. Electronegativity of F > Cl
3. F is more oxidising than Cl
4. Electron affinity of Cl > F

Answer: 1. Bond energy of F₂ > Cl₂

Due to more repulsion in between non-bonding electron pairs (2p) of two fluorines (due to the small size of F-atom) in comparison to non-bonding electron pairs (3p) in chlorine, the bond energy of F, is less than Cl₂.

B.E(F₂) = 158.8 kJ/mole and

B.E. (Cl₂) =242.6 kJ/ mole

Question 66. Which of the following has the greatest electron affinity?

1. I
2. Br
3. F
4. Cl

Answer: 4. Cl

In general, the electron affinity decreases from top to bottom in a group. But in group 17, fluorine has lower electron affinity as compared to chlorine due to the very small size of the fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2s orbitals of fluorine and thus, the incoming electron does not experience much attraction

Question 67. Which of the following displaces Br₂ from an aqueous solution containing bromide ions?

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

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

Since chlorine is a stronger oxidising agent than bromine, therefore it will displace bromine from an aqueous solution containing bromide ions. \(\mathrm{Cl}_2+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_2\)

Question 68. Which of the following species has four lone pairs of electrons?

1. I
2. O
3. Cl^–
4. He

Answer: 3. Cl^–

Outer electronic configuration of \(\mathrm{Cl}=3 s^2 3 p_x^2 3 p_y^2 3 p_z^1\)

Outer electronic configuration of \(\mathrm{Cl}^{-}=3 s^2 3 p_x^2 3 p_y^2 3 p_z^2\),

i.e., 4 lone pair of electrons

Question 69. Match the following:

Which of the following is the correct option?

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

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

Question 70. When Cl₂ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

1. Zero to +1 and zero to -5
2. Zero to -1 and zero to +5
3. Zero to -1 and zero to +3
4. Zero to +1 and zero to -3

Answer: 2. Zero to -1 and zero to +5

⇒ \(\stackrel{0}{\mathrm{C}} \mathrm{l}_2+\underset{\text{(hot and conc.)}}{6 \mathrm{NaOH}} \rightarrow 5 \stackrel{-1}{\mathrm{NaCl}}+\mathrm{NaClO}_3+3 \mathrm{H}_2 \mathrm{O}\)

This is an example of a disproportionation reaction and oxidation state of chlorine changes from 0 to -1 and +5

Question 71. Which of the following is used in the preparation of chlorine?

1. Both MnO₂ and KMnO₄
2. Only KMnO₄
3. Only MnO₂
4. Either MnO₂ or KMnO₄

Answer: 1. Both MnO₂ and KMnO₄

⇒ \(\mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \uparrow\) \(2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{MnCl}_2+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{Cl}_2 \uparrow\)

Question 72. Which of the following elements is extracted commercially by the electrolysis of an aqueous solution of its compound?

1. Cl
2. Br
3. Al
4. Na

Answer: 1. Cl

Chlorine is obtained by the electrolysis of brine (concentrated NaCl solution). Chlorine is liberated at the anode.

Question 73. When chlorine is passed over dry slaked lime at room temperature, the main reaction product is

1. Ca(ClO₂)₂
2. CaCl₂
3. CaOCl₂
4. Ca(OCl)₂

Answer: 3. CaOCl₂

Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 74. In the manufacture of bromine from seawater, the mother liquor containing bromides is treated with

1. Carbon dioxide
2. Chlorine
3. Iodine
4. Sulphur dioxide.

Answer: 2. Chlorine

Bromide in the mother liquor (containing MgBr₂) is oxidised to Br₂ by passing CI₂ which is a stronger oxidising agent.

⇒ \(2 \mathrm{Br}^{-}+\mathrm{Cl}_2 \rightarrow \mathrm{Br}_2+2 \mathrm{Cl}^{-}\)

Question 75. The bleaching action of chlorine is due to

1. Reduction
2. Hydrogenation
3. Chlorination
4. Oxidation.

Answer: 4. Oxidation.

The bleaching action of chlorine is due to oxidation in the presence of moisture. The bleaching effect is permanent.

⇒ \(\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}+[\mathrm{O}]\)

Colouring matter + [O] → Colourtress matter

Question 76. Bleaching powder reacts with a few drops of cone. HCl to give

1. Chlorine
2. Hypochlorous Acid
3. Calcium Oxide
4. Oxygen.

Answer: 1. Chlorine

⇒ \(\mathrm{CaOCl}_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2\)

Question 77. Among the following, the correct order of acidity is

1. \(\mathrm{HClO}_2<\mathrm{HClO}<\mathrm{HClO}_3<\mathrm{HClO}_4\)
2. \(\mathrm{HClO}_4<\mathrm{HClO}_2<\mathrm{HClO}<\mathrm{HClO}_3\)
3. \(\mathrm{HClO}_3<\mathrm{HClO}_4<\mathrm{HClO}_2<\mathrm{HClO}\)
4. \(\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4\)

Answer: 4. \(\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4\)

The acidic character of the oxoacids increases with an increase in the oxidation number of the halogen atom i.e., \(\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4\).

This can be explained on the basis of the relative stability of the anions Ieft after the removal of a proton. Since the stability of the anion decreases in the order : \(\mathrm{ClO}_4^{-}>\mathrm{ClO}_3^{-}>\mathrm{ClO}_2^{-}>\mathrm{ClO}^{-}\) acid strength also decreases in the same order.

Question 78. Which of the statements given below is incorrect?

1. O₃ molecule is bent.
2. ONF is isoelectronic with O₂N^–.
3. OF₂ is an oxide of fluorine.
4. Cl₂O₇ is an anhydride of perchloric acid.

Answer: 3. OF₂ is an oxide of fluorine.

OF₂ (oxygen difluoride) is a fluoride of oxygen because fluorine is more electronegative than oxygen.

Question 79. The correct order of increasing bond angles in the following species is

1. \(\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2<\mathrm{ClO}_2^{-}\)
2. \(\mathrm{ClO}_2<\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2^{-}\)
3. \(\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2^{-}<\mathrm{ClO}_2\)
4. \(\mathrm{ClO}_2^{-}<\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2\)

Answer: 4. \(\mathrm{ClO}_2^{-}<\mathrm{Cl}_2 \mathrm{O}<\mathrm{ClO}_2\)

Question 80. Which one of the following oxides is expected to exhibit paramagnetic behaviour?

1. CO₂
2. SiO₂
3. SO₂
4. ClO₂

Answer: 4. ClO₂

Question 81. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): ICi is more reactive than I₂.

Reason (R): The I — Cl bond is weaker than the I — I bond. In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both (A) and (R) are correct and (R) is the correct explanation of (A).
2. Both (A) and (R) are correct but (R) is not the correct explanation of (A).
3. (A) is correct but (R) is not correct.
4. (A) is not correct but (R) is correct.

Answer: 1. Both (A) and (R) are correct and (R) is the correct explanation of (A).

In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because the X-X’ (ICl) bond in interhalogens is weaker than the X-X (I-I) bond in halogens except for the F-F bond.

Question 82. Match the interhalogen compounds of column A with the geometry in column B and assign the correct code.

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

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

Question 83. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

Assertion A: Helium is used to dilute oxygen in the diving apparatus.

Reason R: Helium has a high solubility in O₂.

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: 1. A is true but R is false.

Helium is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

Question 84. Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.

1. Noble gases have large positive values of electron gain enthalpy.
2. Noble gases are sparingly soluble in water.
3. Noble gases have very high melting m and boiling points.
4. Noble gases have weak dispersion forces.

Answer: 3. Noble gases have very high melting m and boiling points.

Noble gases have very low melting and boiling points because the only type of interatomic interaction in these elements is weak dispersion forces.

Question 85. Match the Xenon compounds in Column A with their structure in Column B and assign the correct code.

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

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

Question 86. Identify the incorrect statement, regarding the molecule Xe04.

1. XeO₄ molecule is square planar.
2. There are four pπ – dπ bonds.
3. There are four sp³ – p, σ bonds.
4. XeO₄ molecule is tetrahedral.

Answer: 1. XeO₄ molecule is square planar.

Question 87. Which compound has a planar structure?

• \(\mathrm{XeF}_4\)
• \(\mathrm{XeOF}_2\)
• \(\mathrm{XeO}_2 \mathrm{~F}_2\)
• \(\mathrm{XeO}_4\)

Answer: 1. \(\mathrm{XeF}_4\)

In XeF₄ the ‘Xe’ atom is sp³d² hybridised, which contains tu,o lone pair orbitals and four bond pair orbitals. Therefore, the shape of XeF₄ molecule is square plar.rar, with one lone pair orbital over and the other below the plane.

 

Haloalkanes And Haloarenes

Question 1. The given compound

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

Answer: 3

Question 4. In the replacement reaction . 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 S_N1 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

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

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

Answer: 1

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. S_N¹ reaction yields a 1:1 mixture of both enantiomers
2. The product obtained by the S_N² 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

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)

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

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

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

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

Answer: 4

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

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_N² mechanism which is favoured by polar aprotic solvent i.e., N, N’-dimethylformamide (DMF),

Question 15. Which of the following biphenyls 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:

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.

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

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

Question 18. In an S_N1 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, S_N1 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 S_N1 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?

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

Answer: None

Due to chirality only compound (4) will undergo racemisation.

Hence, all the given options are incorrect.

Question 20. Given:

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.

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. S_N1 and S_N2
2. S_N1 and S_N1
3. S_N2 and S_N2
4. S_N2 and S_N1

Answer: 3. S_N2 and S_N2

If the reaction is S_N1, there will be the formation of carbocation and the rearrangement takes place. In these reactions there is no rearrangement hence both are S_N2 mechanisms.

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

Answer: 1

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

Question 24. Which one is most reactive towards S_N1 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. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{C}\left(\mathrm{CH}_3\right)\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{Br}\)

S_N1 reactions proceed via the formation of a carbocation intermediate.

The more stable is the carbocation more reactive the alkyl aryl halide towards S_N1.

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.

Question 25. The correct order of increasing reactivity of

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, 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. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3\)

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

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 S_N2 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?

Answer: 2

S_N2 mechanism is followed in the case of primary and secondary alkyl halides i.e. S_N2 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?

Answer: 2

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 S_N1 mechanism?

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

Answer: 4. Benzyl chloride

S_N1 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

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

Answer: 1. R

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.

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

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

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

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.

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. \(\mathrm{DCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}\)

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

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

Answer: 1

S_N2 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.

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.

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?

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

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

Answer: 2

It gives D.D.T

(p,p di chlorodiphenyltrichloroethane).

Question 53. Industrial preparation of chloroform employs acetone and

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

Answer: 3. Chlorine gas

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.

Coordination Compounds

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

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

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.

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

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 (EDTA^4-) 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]^_4–is

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

Answer: 2. 6

(C₂O₄^2-) → 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

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 (C₂O₄^2-) 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.

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

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.

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

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.

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.

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 NO^–₂ is linked through N or through O. Octahedral complexes of the type Ma₄b₂ exhibit 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.

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

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

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 :

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

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

Answer: 4. 2

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

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)

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

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

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

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.

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.

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.

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 e_g orbitals since they are directed towards the ligands and the energy gain is considerably greater. In the case of unevenly occupied t_2g, orbitals, the fan-teller distortion is very weak since the t_2g, set does not point directly at the ligands and therefore, the energy gain is much less.

High spin complexes:

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

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 t_2g and e_g orbitals respectively.

For [Fe(H₂O)₆]³⁺ complex ion,

Fe³⁺ (3d⁵) = t³_2g e²_g=-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

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 t_2g orbitals

y = no. of electrons occupying e_g orbitals

= (- 0.4 x 3 + 0.6 x 1)Δ₀ [High spin d⁴ =t_2g³e_g¹)

= (- 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

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. \(\frac{-2}{5} \Delta_o+P\)

(Δ₀ = 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 t_2g orbital

y = number of electrons occupying e_g 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.

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

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

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

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

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

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.

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

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.

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

The stronger the ligands larger the value of Δ_oct.

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

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

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

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

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

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

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 e_g set i.e. (n -1)d²_z and (n – 1)d_x²-y²  orbitals), one ns and three np(np_x np_y and np_z) 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

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

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

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 \(\left[\mathrm{K}_3 \mathrm{Cu}(\mathrm{CN})_4\right]\)

Question 87. Which of the following is considered to be an 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:

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

 

d And f Block Elements

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)d^1-10.

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:

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

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)

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 TiF₆^2- 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 VO₄^3- 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

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

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

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

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 MnO^–₄ = 5 moles of C₂O^2-₄–

20 mL of 0.1 M KMnO₄ = 2 mmol of KMnO₄

Also, 50 mL of 0.1 M C₂H₂O₄ = 5 mmol of C₂O^2-₄

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

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.