Question 1. An alkene (molecular formula : C₅H₁₀) on ozonolysis forms acetone as one of the products. The alkene is—

1. 2-methyl-1-butene
2. 3-methyl-1-butene
3. 2-methyl-2-butene
4. Cyclopentane

Answer: 2.  3-methyl-1-butene

Question 2. Which of the following compounds can be used to prepare both ethylene and acetylene—

1. CH₃CH₂OH
2. BrCH₂CH₂Br
3. CH ₃CH₂Br
4. BrCH₂CH₂OH

Answer: 2. BrCH₂CH₂Br

Question 3. Alkyl chloride on dehydrochlorination produces 2 alkenes (C₆H₁₂) which on ozonolysis form four compounds—

1. CH₃CHO
2. CH₃CH₂CHO,
3. CH₃COCH₃ and
4. (CH₃)₂CHCHO.

The alkenes are—

1. 4-methylpent-2-ene and 2-methylpent-2-ene
2. 2-methyl pent-2-ene and 2,3-dimethyl but-2-ene
3. 4-methylpent-2-en§ and hex-3-ene
4. 2-methylpent-2-ene and hex-3-ene

Answer: 1. 4-methyl pent-2-ene and 2-methyl pent-2-ene

Question 4. The compound that exhibits geometrical isomerism is—

1. C₂H₅Br
2. (CH)₂(COOH)₂
3. CH₃CHO
4. (CH₂)₂(COOH)₂

Answer: 2. (CH)₂(COOH)₂

Question 5

1. X: cis-2-butene and Y: frans-2-butene
2. X: trans-2-butene and Y :cis-2-butene
3. X, Y both are cis-2-butene
4. X, Y both are trans-2-butene

Answer: 1. X: cis-2-butene and Y: frans-2-butene

Question 6. An alkene may be formed from a carbocation if—

1. One H- ion gets eliminated
2. One H+ ion gets added
3. One H+ ion gets eliminated
4. One H- ion gets added

Answer: 3. One H+ ion gets eliminated

Question 7. The number of moles of water produced when one mole acetylene undergoes complete combustion is—

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

Answer: 1.  1 mol

Question 8.   , A, B, C in the above reaction are respectively—

1. CH₃COCH₃, CH₃CHO, CO₂
2. CH₃COCOOH, CH₃COOH, CO₂
3. CH₃CH₂COOH, CH₃CHO, CO₂
4. CH₃COCH₃, CHgCOOH, CO₂

Answer: 4. CH₃COCH₃, CHgCOOH, CO₂

Question 9. The position of the double bond in an alkene can be determined by—

1. Hydrogenation
2. Ozonolysis
3. Hydroxylation
4. Hydroboration

Answer: 2. Ozonolysis

Question 20. Heavy water reacts with calcium carbide to form—

1. CaD₂
2. C₂D₂
3. Ca₂D₂O
4. CD₂

Answer: 2. C₂D₂

Question 21. In case of trisubstituted benzene, if the substituents are different, then the number of isomers will be

1. 5
2. 8
3. 6
4. 10

Answer: 4. 10

Question 22. The chemical formula ofCetane is—

1. C₆H₁₂
2. (CH₃)₃C(CH₂)₁₁CH₃
3. CH₃(CH₂)₁₄CH₃
4. (C₂H₅)₄C

Answer: 3. CH₃(CH₂)₁₄CH₃

Question 23. Which of the following gets converted into an explosive when it is turned into liquid by applying high pressure—

1. Propane
2. n-butane
3. Isobutane
4. Acetylene

Answer: 4. Acetylene

Question 24. The product which is not obtained when ethylene reacts with K3 mixed with Br₂/H₂O is—

1. BrCHCH₂Br
2. BrCH₂CH₂OH
3. HOCH₂CH₂OH
4. BrCH₂CH₂I

Answer: 3. HOCH₂CH₂OH

Question 25. Which of the following does not form a sooty flame—

1. Toluene
2. Benzene
3. Mesitylene
4. Butane

Answer: 4. Butane

Question 26. Which of the following statements is incorrect—

1. Delocalisation of electrons occur between two n bonds in a propadiene molecule
2. Delocalisation of electrons occur between two n bonds in a molecule of 1, 3-butadiene
3. Cumulated polyenes with odd number of double bonds exhibit geometrical isomerism if their terminal groups are different
4. Cumulated polyenes with even number of double bonds exhibit optical isomerism if their terminal groups are different

Answer: 1.  Delocalisation of electrons occur between two n bonds in a propadiene molecule

Question 27. Which of the given is a benzenoid aromatic compound—

1. Anthracene
2. Pyrrole
3. Pyridine
4. Cyclopentadienyl anion

Answer: 1. Anthracene

Question 28. Gas used in Hawker’s lamp for emitting bright light is—

1. Acetylene
2. Ethylene
3. Methane
4. Propane

Answer: 1. Acetylene

Question 29. The compounds which exist as liquids are—

1. C₅H₁₂
2. C₃H₈
3. C₂H₆
4. C₇H₁₆

Answer: 1,4

Question 30. Which of the given can be prepared by Wurtz reaction–

1. 2-methylpropane
2. 2,3-dimethyl butane
3. Hexane
4. All of them

Answer: 2,3

Question 31. Which of the following compounds do not produce acetylene on hydrolysis—

1. CaC₂
2. Al₄C₃
3. Be₂C
4. Zn(CH₄)₂

Answer: 2,3,4

Question 32. Which of the following options are correct with respect to Friedel-Crafts reaction —

1. Alkylation Reagent: CH₂=C₆H₅Cl
2. Solvent: C₆H₅NO₂, CS₂
3. Catalyst: AlCl₃ , H₂SO₄
4. All Of the Above

Answer: 2,3

Question 33. Lewisite and its antidote are—

1. Lewisite ClCH=CHAsC12
2. Antidote 1,1-dimercapto-l-propanol
3. Lewisite CH₂=CHAsCl₂
4. Antidote 2,3-dimercapto-l-propanol

Answer: 1,4

Question 34. Halogenation ofan alkene is a or an—

1. Substitution reaction
2. Elimination reaction
3. Addition reaction
4. Oxidation reaction

Answer: 1,4

Question 35. During detection of unsaturation in an unknown organic compound disappearance of the violet colour of dilute and cold KMn04 solution indicate—

1. Presence of ethylenic unsaturation in the compound
2. The presence of a group in the compound which gets easily oxidised by kmn04
3. Presence of only single covalent bond in the compound
4. All of the above are true

Answer: 1,2

Question 36. Which of the following options is correct—

1. Ortho- or para-orienting: — NR₂, —NHCOCH₃
2. Mete-orienting: —NO₃, —Cl
3. Ortho- or para-orienting: — CF₃, —SO₃H
4. Mete-orienting: —CHO, —COR

Answer: 1,4

Question 37. Which of the following statements are true for Kolbe’s electrolytic method—

1. It is an effective method for preparing symmetrical alkanes
2. Reduction of carboxylate ion occurs at the anode
3. Platinum electrodes are used in this method
4. Methane cannot be prepared by this method

Answer: 1,3,4

Question 38.   In this reaction, X and Y are-

1. X = CH₃COOH
2. X = HCOOH
3. F = CH₃COONa
4. Y = C₂H₅COONa

Answer: 1,3

Question 1. The energy of the state S1 in units of the hydrogen atom ground state energy is-

1. 0.75
2. 1.50
3. 2.25
4. 4.50

Answer: 3. 2.25

Question 2. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005% certainty with which the position ofthe electron can be located is{h = 6.6 × 10^-34 kg-m² s^-1, the mass of the electron, mg = 9.1 × 10^-31 kg).

1. 3.84 × 10^-3m
2. 1.52 × 10^-4 m
3. 5.10 ×10^-3m
4. 1.92 × 10^-3m

Answer: 2. 1.52 ×10^-4m

Question 3. The ionization enthalpy of a hydrogen atom is 1.312 × 10⁶J-mol^-1 The energy required to excite the electron in the atom from n – 1 to n = 2 is

1. 7.56 × 10⁵ J-mol^-1
2. 9.84 × 10⁵ J-mol^-1
3. 8.51 × 10⁵ J-mol^-1
4. 6.56 × 10⁵ J-mol^-1

Answer: 2. 9.84 ×10⁵ J-mol^-1

Question 4. If we apply potential difference so that an electron is accelerated continuously in a vacuum tube such that a decrease of 10% occurs in its de-Broglie wavelength. In such a case the change observed in the kinetic energy of the election will be approximately—

1. A Decrease Of 11%
2. An increase of 11.1%
3. An Increase Of 10%
4. An increase of 23.4%

Answer: 4. An increase of 23.4%

Question 5. The approximate wavelength of matter wave associated with an electron, that is accelerated by applying 100 V of potential difference in a discharge tube, will be-

1. 123pm
2. 12.3pm
3. 1.23 pm
4. 0.123pm

Answer: 1. 123pm

Question 6. If two particles are associated with the same kinetic energy, then the de Broglie’s wavelength (A) of these particles are—

1. Directly proportional to the velocity
2. Inversely proportional to the velocity
3. Independent of mass and velocity
4. Cannot be predicted.

Answer: 1. Directly proportional to the velocity

Question 7. The ratio of magnetic moments of Fe (3) and CO (2) is

1. \(\sqrt{3}: \sqrt{7}\)
2. \(\sqrt{3}: \sqrt{7}\)
3. 7:3
4. 3:7

Answer: 2. \(\sqrt{3}: \sqrt{7}\)

Question 8. If it were possible for a hydrogen atom to exist with a position as the extra-nuclear particle, then the energy of position in the first excited state would be—

1. 13.6v
2. 3.4ev
3. -3.4ev
4. 6.8ev

Answer: 2. 3.4ev

Question 9. Identify the orbitals for which n = 4 and l = 1 

1. 4p_y
2. 4p_x
3. 4d_xy
4. 4d_x²-y²

Answer: 1. 4py

Question 10. The orbitals which have the same number of nodes are

1. 2s,3p
2. 3p,3d
3. 2s,2p
4. 3s,4d

Answer: 2. 3p,3d

Question 11. In the ground state, an element has 13 electrons in its Afshell. The element is

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

Answer: 1. Mn

Question 12. The angular momentum of an electron may have the values—

1. \(0.5 \frac{h}{\pi}\)
2. \(\frac{h}{\pi}\)
3. \(0.2 \frac{h}{\pi}\)
4. \(2.5 \frac{h}{2 \pi}\)

Answer: 1. \(0.5 \frac{h}{\pi}\)

Question 13. The line spectrum is noticed during the transition of an electron from a higher excited state to a lower one in the H-atom only when it falls from

1. 2s→ls
2. 2p→ls
3. 3s→2p
4. 4p→2p

Answer: 3. 3s→2p

Question 14. Select the pair of atoms having the same no. of electrons in their outermost shell —

1. Na, Ca
2. Mg, Fe
3. As, Bi
4. Rb, Sb

Answer: 3. Rb, Sb

Question 15. Which consists of particles of matter—

1. α -rays
2. β -rays
3. γ -rays
4. X-rays

Answer: 1.  -rays

Question 16. Which have two radial nodes-

1. 2p
2. 3s
3. 4p
4. 3p

Answer: 2. 3s

Question 17. The ratio of λα to λβ for the Balmer series of hydrogen spectra is given by

1. \(\frac{108}{80}\)
2. \(\frac{108}{90}\)
3. \(\frac{40}{54}\)
4. \(\frac{20}{27}\)

Answer: 1. \(\frac{108}{90}\)

Question 18. Indicate the conditions under which the ratio of Broglie wavelengths of a -particle and a proton will be—

1. When the ratio of their velocities is 4: 1
2. When the ratio of their velocities is 1: 8
3. When the ratio of their energies is 128: 1
4. When the ratio of their velocities is 1: 16

Answer: 2. When the ratio of their velocities is 1: 8

Question 19. Which ofthe following ions is paramagnetic—

1. Zn²⁺
2. Cu²⁺
3. He²⁺
4. ₂O²⁺

Answer: 2. Cu²⁺

Question 20. The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV. Then, which of the following statement(s) is/ are correct for He⁺ —

1. The energy of an electron in the second Bohr orbit is -13.6 ev
2. The kinetic energy of the electron in the first orbit is 54.46 ev
3. The kinetic energy of an electron in the second orbit is 13.6 ev
4. The speed of an electron in the second orbit is 2.19 × 10⁶ m.s^-1

Answer: 1. Energy electron in the second Bohr orbit is -13.6 ev

Question 21. For which of the following species, the expression for the energy of an electron in nth orbit, En = \(-\frac{13.6 Z^2}{n^2}\) eV. Atom-1. has the validity—

1. He²⁺
2. Li²⁺
3. Deuterium
4. Li²⁺

Answer: 2. He²⁺

Question 22. According to Bohr’s atomic theory, which of the following relations is correct—

1. The kinetic energy of electron oc z²/ n²
2. The product of the velocity of the electron and the principal quantum number α z²/n²
3. Frequency of revolution of the electron in an orbit z²/n³
4. Coulombic force of attraction on electron oc z³/n

Answer: 1. Kinetic energy of electron α z³/ n⁴

Question 23. Which is correct in the case of p -p-orbitals—

1. They are spherically symmetrical
2. They have strong directional character
3. They are three-fold degenerate
4. Their charge density along the x, y, and z -axes is zero

Answer: 2. They have strong directional character

Question 24. An isotone of \(\mathrm{f}_{32}^{76} \mathrm{Ge}\) is –

1. \({ }_{32}^{77} \mathrm{Ge}\)
2. \({ }_{33}^{77} \mathrm{As}\)
3. \({ }_{34}^{77} \mathrm{Se}\)
4. \({ }_{34}^{77} \mathrm{Se}\)

Answer: 2. \({ }_{33}^{77} \mathrm{As}\)

Question 25. Which of the following is correct—

1. Only the Lyman series is observed in both emission and absorption spectrum
2. The continuum in the line spectrum is noticed after a certain value of n
3. the wavelength of the mth line of Balmer Series is \(\frac{1}{\lambda}=R_H Z^2\left[\frac{1}{2^2}-\frac{1}{m^2}\right]\)
4. The number of spectral lines given when an electron drops from the 5th to the 2nd shell is six.

Answer: 1. Only the Lyman series is observed in both the emission and absorption spectrum.

Question 26. In Bohr’s models of atoms the lowest angular momentum, that an electron may have is

1. h
2. 0
3. \(\frac{h}{2 \pi}\)
4. \(\frac{h}{\pi}\)

Answer: 3. \(\frac{h}{2 \pi}\)

Question 27. Which one of the following is correct among the following sets of quantum number

1. \(n=1, l=2, m=1, s=\frac{1}{2}\)
2. \(n=3, l=1, m=0, s=\frac{1}{2}\)
3. \(n=5, l=1, m=2, s=\frac{1}{2}\)
4. \(n=4, l=1, m=-2, s=\frac{1}{2}\)

Answer: 2. \(n=3, l=1, m=0, s=\frac{1}{2}\)

Question 28. The number of total electrons in the n-th orbit of an atom is—

1. n
2. n2
3. 2n²
4. n-1

Answer: 3. 2n²

Question 29. Which of the following is the ground state electronic configuration of Cr (The atomic number of Cr is 24)—

1. ls²2s²2p⁶3s²3p⁶3d⁴4s²
2. ls²2s²2p⁶3s²z3p⁶3d⁵4s¹
3. ls²2s²2p⁶3s²3p⁶3d⁶
4. ls²2s²2p⁶3s²3p⁶3d³4s²4p¹

Answer:  2. ls²2s²2p⁶3s²z3p⁶3d⁵4s¹

Due to the half-filled electronic configuration of rf-subshell, Cr gets stabilized to a greater extent.

Question 30. The maximum how many number of electrons of the Cl atom follows the relation n + l = 3 —

1. 10
2. 8
3. 10
4. 16

Answer: 2.

n + Z = 3 represents 3s and 2p orbital. Electronic configuration of Cl: ls²2s²2p⁶3s²3p⁵

Question 31. Bohr’s theory does not apply to which of the following species?

1. H,
2. He²⁺
3.  Be³⁺
4. B⁴⁺

Answer:  2. He²⁺

Bohr’s theory does not apply to He²⁺ because it does not contain any electrons.

Question 1. The atomic number of the inert gas in which the total number of d -electrons is equal to the difference in the numbers of total p -and s -electrons is-

1. 18
2. 36
3. 54
4. 86

Answer: 2. 36

Question 2. The element with atomic number 118 has been discovered recently. Which of the following is not expected for this element

1. It is a radioactive and unstable element
2. It is a solid at room temperature
3. Its ionization enthalpy is minimal in the group
4. It has a 7s²7p⁶ outer shell configuration

Answer: 2. It is a solid at room temperature

Question 3. Which occupy the peak of the atomic volume curve

1. Transition Elements
2. Halogens
3. Alkali Metals
4. Alkaline earth metals

Answer: 3. Alkali Metals

Question 4. The ionization energy will be the maximum for the process

1. Ba → Ba²⁺
2. Be → Be²⁺
3. Cs → Cs⁺
4. Li → Li⁺

Answer: 2. Be → Be²⁺

Question 5. The correct order of second ionization potential of carbon, nitrogen, oxygen, and fluorine is

1. C>N>O>F
2. O>N>F>C
3. O>F>N>C^–
4. F>O>N>C

Answer: 3. O>F>N>C^–

Question 6. The most reactive metal is

1. Na
2. Fe
3. Hg
4. Cs

Answer: 4. Cs

Question 7. You are given Avogadro’s number of atoms of a gas X. If the atoms are converted into X+(g) by energy AH, then the ionization enthalpy of X is

1. \(\frac{2 \Delta H}{N_A}\)
2. \(\frac{2 N_A}{\Delta H}\)
3. \(\frac{\Delta H}{2 N_A}\)
4. \(\frac{N_A}{\Delta H}\)

Answer: 1. \(\frac{2 \Delta H}{N_A}\)

Question 8. The second electron affinity of oxygen is +744 kj.mol-1 then the second electron affinity of sulphur is

1. -200kj mol^-1
2. +450 kj – mol^-1
3. +800 kj – mol^-1
4. 1200 kl – mol^-1

Answer: 2. +450 kj – mol^-1

Question 9. What would be the atomic number of the next halogen if discovered in the future—

1. 116
2. 117
3. 118
4. 119

Answer: 2. 117

Question 10. An element belongs to group 17 and the fourth period of the periodic table. Which of the following properties will be shown by the element?

1. Gaseous, non-metallic
2. Liquid, non-metallic
3. Solid, non-metallic
4. Solid, metallic

Answer: 2. Liquid, non-metallic

Question 11. Five successive ionization enthalpies of an element are 800, 2427, 3638, 25024 & 32824kj. mol^-1 respectively. The number of valence electrons ofthe element is

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

Answer: 1. 3

Question 12. Choose the correct set of ionic radii among the following

1. Ti⁴⁺<Mn⁷⁺
2. ³⁵Cl^–<³⁷Cl^–
3. P³⁺>P⁵⁺
4. K⁺>Cl^–

Answer: 3. p³⁺>p⁵⁺

Question 13. Electron affinity( kj.mol^-1 ) of three halogens X, Y, and Z are -349, -333, and -325. X, Y, and Z are respectively.

1. F₂, Cl₂ and Br2
2. Cl₂, F₂ and Br₂
3. Br₂, Cl₂ and F₂
4. Br₂, F₂ and Cl₂

Answer: 1. F₂, Cl₂ and Br₂

Question 14. The correct order of ionic radii is

1. N^3- > O^2- >F^– > Na⁺
2. N^3- > Na⁺> O^2- >F^–
3. Na³⁺ > O^2- > N^3- >F^–
4. O^2- > F^–> Na²⁺ > N^3-

Answer: 1. N^3- > O^2- > F^– > Na⁺

Question 15. The ionization enthalpy of lithium is 520 kj. mol^-1. The amount required to convert 210 mg of -atoms in a gaseous state into Li+ ions is

1. -520kj
2. -173.3 kj
3. 2.47 kj
4. 15.60 kj

Answer: 4. 15.60 kj

Question 16. Electronic configuration ls²2s²2p⁶3s²3p⁶3d⁹ indicates

1. Mentalization
2. Non-metal atom
3. Non-mental anion
4. Mental atom

Answer: 1. Mentalcation

Question 17. In the periodic table metals used as catalysts belong to

1. F-block
2. d- block
3. p-block
4. s-block

Answer: 2. d- block

Question 18. The oxide of an element having configuration [Ne]3s¹ is

1. Amphoteric
2. Basic
3. Acidic
4. Neutral

Answer: 2. Basic

Question 19. Which one of the following orders is not by the property stated against it

1. F > Cl > Br >I : Oxidisingpower
2. F >Cl > Br >l : Electronegativity
3. F-F > Cl-Cl > Br-Br > I-I: Bond dissociation energy
4. HI > HBr > HCl > HF: Acidic property in water

Answer: 3. F-F > Cl-Cl > Br-Br > I-I: Bond dissociation energy

Question 20. In which of the following process energy is absorbed

1. F→F^–
2. H→h^–
3. Cl→cl^–
4. O→O^2-

Answer: 4. O→O^2-

Question 21. In which of the following pairs do both the metals have identical values of metallic radii

1. Ag and Au
2. Cu And Ag
3. Cu and Zn
4. Zn and Hg

Answer: 1. Ag and Au

Question 22. In which of the following pairs the difference between the covalent radii ofthe two metals is maximum

1. K, Ca
2. Mn, Fe
3. Co, Ni
4. Cr, Mn

Answer: 1. K, Ca

Question 23. The increasing order of the first ionization enthalpies of the elements B, P, S, and F is—

1. F<S<P<B
2. P<S<B<F
3. B<P<S<F
4. B<S<P<F

Answer: 4. B<S<P<F

Question 24. Which group of atoms have nearly the same atomic radius

1. Na, K, Rb, Cs
2. Li, Be, B, C
3. Fe, Co, Ni, Cu
4. F, Cl, Br, I

Answer: 3. Fe, Co, Ni, Cu

Question 25. The lattice energy is highest for

1. CsF
2. RbF
3. NaF
4. KF

Answer: 3. NaF

Question 26. The correct order of lattice energy for lithium halides is-

1. Li > LiBr > LiCl > LiF
2. LiCl > LiF > LiBr > Lil
3. LiBr > LiCl > Lil > LiF
4. LiF > LiCl > LiBr > Lil

Answer: 4. LiF > LiCl > LiBr > Lil

Question 27. An element with atomic number 20 is most likely to combine with the element whose atomic number is

1. 11
2. 16
3. 18
4. 10

Answer: 2. 16

Question 28. K⁺, Cl^–, Ca²⁺, and S^2- ions are isoelectronic. The decreasing order of their size is

1. S^2- > Cl- > K⁺> Ca²⁺
2. Ca²⁺ > K⁺> Cl^– > S^2-
3. K⁺  >Cl^–> Ca²⁺ >S^2-
4. Cl^– > S^2->Ca²⁺ >K⁺

Answer: 1. S^2-> Cl^– > K⁺ > Ca²⁺

Question 29. Which ofthe following oxides of Cr is amphoteric

1. CrO
2. CrO₃
3. Cr₂O₃
4. None of the theses

Answer: 3. Cr₂O₃

Question 30. The element with atomic number 118 has been discovered recently. In the periodic table, the element will occupy its position in the family of

1. Halogens
2. Insert gases
3. Chalcogens
4. Alkaline earth metals

Answer: 2. Insert gases

Question 31. In the relation, Electronegativity \(=\frac{0.359 Z_{\text {eff }}}{r^2}+0.744, r \text { is }\)

1. Ionic radius
2. Metallic radius
3. Van der Waals radius
4. Covalent radius

Answer: 4. Covalent radius

Question 32. Select the endothermic processes

1. H→H^–
2. O→ O^2-
3. Cr³⁺
4. Ar→Ar+

Answer: 2. O→O^2-

Question 33. Which of the following sequences contain the atomic number of only representative elements

1. 3,33,53,87
2. 7,17,25,37,48
3. 2,10,22,86
4. 9,35,51,88

Answer: 1. 3,33,53,87

Question 34. Which of the given elements will gain one electron more readily in comparison to other elements of their group

1. S
2. Na
3. O
4. Cl

Answer: 1. S

Question 35. MgSO₄ is soluble in water but BaSO₄ is not because

1. The hydration enthalpy of MgSO₄ is more than its lattice enthalpy
2. The lattice enthalpy of MgSO₄ is greater than its hydration enthalpy
3. The lattice enthalpy of BaSO₄ is greater than its hydration enthalpy
4. The hydration enthalpy of BaSO₄ is greater., than its lattice enthalpy

Answer: 1. The hydration enthalpy of MgSO₄ is more than its lattice enthalpy

Question 36. Consider the following ionization steps

• M(g)→M⁺(g) + e^–; ΔH = 100 eV
• M(g)→M²⁺(g) + 2e^–; ΔH = 250 eV

Select the correct statements

1. Δ_iH₁ofM(g) =100eV
2. Δ_iH₁ of M+(g) =150 eV
3. Δ_iH₂ of M(g)=250eV
4. Δ_iH₂ of M(g)=150 eV

Answer: 1. 1,2,4

Question 37. In which of the following sets of atomic numbers, all elements are in the same group—

1. 8,16,24
2. 3,11,37
3. 12,38,56
4. 10,18,54

Answer: 2. 3,11,37

Question 38. Which of the following elements are artificially made and do not exist in nature

1. Bi
2. Ge
3. Tc
4. At

Answer: 3. Tc

Question 39. Choose the pairs in which the IEX of the first element is greater than the IEl of the second element but in the case of IE₂ order is reversed

1. P, S
2. F, O
3. Mg, Al
4. N, O

Answer: 1. P, S

Question 40. Which ofthe following ions can form complexes

1. Al³⁺
2. Cu²⁺
3. Cr³⁺
4. Fe²⁺

Answer: 1. Al³⁺

Question 41. Which ofthe following pairs contain metalloid

1. In, Tl
2. Ge, Ga
3. As, Te
4. I, Bi

Answer: 2. Ge, Ga

Question 42. Which of the following elements exist in a liquid state at room temperature

1. Selenium
2. Germanium
3. Cesium
4. Gallium

Answer: 2. Germanium

Question 43. Which of the following elements can form two or more chlorides

1. Cu
2. Vs
3. Hg
4. Cs

Answer: 1. Cu

Question 44. Which of the following hydrides rarely show basic properties

1. PH₂
2. SbH₃
3. AsH₃
4. BiH₃

Answer: 2. SbH₃

Question 45. The correct sequences of increasing electropositive character are

1. Fe<Cu<Mg
2. Cu < Fe < Mg
3. Cs < Ca < Sr
4. Ca<Sr<Cs

Answer: 2. Cu < Fe < Mg

Question 46. Select the correct sequence of electron-gain enthalpy

1. S > Se > O
2. Cl > F >I
3. F > Cl >I
4. S > O > Se

Answer: 3. F > Cl >I

Question 47. The correct orders of electronegativity are

1. C < Si < P < N
2. Si < P < C < N
3. P < Se < S < N
4. S < P < Se < N

Answer: 3. P < Se < S < N

Question 48. Among which of the following pairs of elements, the first has lesser ionization energy than the second

1. Na, K
2. N, O
3. B, Be
4. Br, I

Answer: 3. B, Be

Question 49. Which of the given elements has the lowest electron affinity

1. C
2. P
3. O
4. S

Answer: 2. P

Question 50. Which of the following atoms has the highest first ionization energy

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

Answer: 4. Sc

Question 1. Which of die following substances undergo disproportionation reactions in the basic medium

1. F₂
2. P₄
3. S₂
4. Br₂

Answer: 2,3,4

Question 2. In which of the following compounds, the oxidation number of oxygen is fractional

1. B₄O₁₀
2. B₂H₆
3. CSO₂
4. KO₃

Answer: 1,3,4

Question 3. Wlien Cl₂ is passed through NaOH in the cold, the oxidation number of changes from

1. 0 to -1
2. 0 to +2
3. 0 to -2
4. 0 to -1

Answer: 1,2

Question 4. In which of the following cases equivalent mass of a reductant is equal to its molecular mass

1. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}\)

2. \(\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\)

3. \(2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{NaI}\)

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

Answer: 2,3

Question 5. Identify the redox reactions

1. 2CuSO₄ + 4KI→2CuI + I₂ + 2K₂SO₄
2. BaCl₂ + Na₂SO₄→BaSO₄ + 2NaGl
3. 3I₂ + 6NaOH→NaIO3 + 5Nal + 3H₂O
4. CuSO₄ + 4NH₃→[Cu(NH₃)₄]SO₄

Answer: 1,3

Question 6. When ammonium nitrate is heated, the oxidation numbers of the N-atoms present it change from

1. -3 to +1
2. -3 to 0
3. -2 to +4
4. +5 to 0

Answer: 2,4

Question 7. For the reaction, \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}-\)

1. \(\mathrm{S}_2 \mathrm{O}_3^{2-} \text { gets oxidised to } \mathrm{S}_4 \mathrm{O}_6^{2-}\)

2. \(\mathrm{S}_2 \mathrm{O}_3^{2-} \text { gets reduced to } \mathrm{S}_4 \mathrm{O}_6^{2-}\)

3. I₂ gets oxidized to I¯

4. I₂ gets reduced to I¯

Answer: 1,4

Question 8. Which of the following statements about the following reaction is wrong

2Cu₂O + Cu₂s→6cu + SO₂

1. Both Cu₂O and cu₂s are reduced
2. Only Cu₂s are reduced
3. Cu₂s is the oxidation
4. Only cu₂O is reduced

Answer: 2,3,4

Question 9. Which of the following orders represents the correct descending order of oxidation numbers

1. HNO₂ > NO > NH₄Cl > N₂
2. HNO₃ > NO > N₂ > NH₂Cl
3. H₂S₂O₇ > Na₂S₂O₃ > Na₂S4O₆ > S₈
4. H₂SO₅ > H₂SO₃ > SCl₂ > H₂S

Answer: 2,4

Question 10. Which ofthe following reactions are not reactions

1. SO₂(g) +H₂O(f) H₂SO₃(aq)

2. Ca(s) + H₂(g) → CaH₂(s)

3. 2H₂S(aq) + SO₂(g)→2H20(l) + 3S(s)

4. \(\begin{aligned}
2 \mathrm{PCl}_5(g)+ & \mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow \\
& 2 \mathrm{POCl}_3(a q)+2 \mathrm{HCl}(a q)+\mathrm{SO}_2 \mathrm{Cl}_2(g)
\end{aligned}\)

Answer: 1,4

Question 11. In which compounds do Cr exist +6 oxidation state

1. CrO₂Cl₂
2. Na₂[Cr(CN)₆]
3. CrO₅
4. K₂Cr₂O₇

Answer: 1,3,4

Question 12. When ammonium nitrite (NH₂NO₂) is heated

1. Oxidation of nitrogen takes place
2. Reduction of nitrogen takes place
3. The overall reaction is a disproportionation reaction
4. The overall reaction is a double decomposition reaction

Answer: 1,2,3

Question 13. In which compounds does an atom exist in two different oxidation states

1. H₂SO₅
2. NH₄NO₃
3. Fe₂O₃
4. H₂O₂

Answer: 1,2

Question 14. In the balanced equation for the reaction—

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+x \mathrm{HI} \rightarrow \mathrm{H}_2 \mathrm{~S}+y \mathrm{I}_2+z \mathrm{H}_2 \mathrm{O}\)

1. x=y-z
2. y=z
3. x=2y
4. z=2x

Answer: 2,3

Question 15. In the reaction,

⇒ \(\mathrm{KMnO}_4+\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{MnO}_2+\mathrm{SO}_4^{2-}+\mathrm{OH}^{-}\)

(Assume formula masses of KMnO₂ and Na₂S₂O₃ M₁ and M₂ respectively)—

1. The equivalent mass of KMnO₂ = M₁/3
2. The equivalent mass of Na₂S₂O₃ = M₂
3. The equivalent mass of KMnO₄ = M₁/5
4. The equivalent mass of Na₂S₂O₂ = M₂/8

Answer: 1,4

Question 16. In the balanced equation for the reaction,

⇒ \(\mathrm{UO}^{2+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{UO}_2^{2+}+\mathrm{Cr}^{3+}+\mathrm{H}_2 \mathrm{O}\) the coefficient of-

1. UO²⁺
2. \(\mathrm{UO}_2^{2+} \text { is } 3\)
3. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} \text { is } 1\)
4. H₂O is 7

Answer: 2,3,4

Question 17. The disproportionation of 1 mol of \(\mathrm{MnO}_4^{2-}\) ions in a neutral aqueous solution results in

1. 1/3 mol of MnO¯₄
2. 2/3 mol of MnO₂
3. 2/3 mol of MnO₄
4. 1/3 mol of MnO₂

Answer: 3,4

Question 18. In the reaction the oxidation number of marked with (*)-

1. Increases by 2 units
2. Increases by 1 unit
3. Decreases by 2 units
4. Decreases by 3 units

Answer: 1,3

Question 19. For the reaction: SO₂ + 2H₂S→3S + 2H₂O 

1. The equivalent mass of the oxidant is 64
2. Equivalent mass ofoxidantis 16
3. The number of electrons accepted oxidant is 4
4. The number of electrons lost by reductant is 6

Answer: 2,3

Question 20. The species that cannot be reducing agents are

1. SO₃
2. \(\mathrm{SO}_3^{2-}\)
3. H₂SO₄
4. S^2-

Answer: 1,3

Question 21. Which are conserved all redox reactions

1. Charge
2. Mass
3. Either charger or Mass
4. Neither charge nor mass

Answer: 1,2

Question 22. The equivalent weight of K₂Cr₂O₇ in an acidic medium is expressed in terms of its molecular weight (M) as

1. \(\frac{M}{3}\)
2. \(\frac{M}{4}\)
3. \(\frac{M}{6}\)
4. \(\frac{M}{7}\)

Answer: 3. \(\frac{M}{6}\)

In an acidic medium, K₂Cr₂O₇ undergoes reduction, forming a Cr³⁺ ion.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

In this reaction, the equivalent weight of K₂Cr₂O₇

⇒ \(=\frac{\text { Molecular weight of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{\begin{array}{c}
\text { No of electrons gained by a molecule of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \\
\text { in its reduction }
\end{array}}=\frac{M}{6}\)

Question 23. If Cl₂ is passed through hot aqueous NaOH, the products formed have Cl in different oxidation states. These are indicated as

1. -1 and +1
2. -1 and +5
3. -1 and +5
4. -1 and +3

Answer: 2. -1 and +5

Reaction: Cl₂ + 6NaOH→NaCl + 5NaC1O₃ + 3H₂O The oxidation number of Cl in NaCl is -1 and that in NaCIO₃ is +5.

Question 24. In an aqueous alkaline solution, two-electron reductions of HO‾₂ give—

1. HO‾
2. H₂O
3. O₂
4. O‾₂

Answer: 1. HO‾

Question 25. Consider the following reactions

⇒ \(x \mathrm{MnO}_4^{-}+y \mathrm{C}_2 \mathrm{O}_4^{2-}+z \mathrm{H}^{+} \rightarrow x \mathrm{Mn}^{2+}+2 y \mathrm{CO}_2+\frac{z}{2} \mathrm{H}_2 \mathrm{O}\)

The values of x, y, and z in the reaction are respectively

1. 5,2 And 8
2. 5,2 and 16
3. 2,5 and 8
4. 2,5 and 16

Answer: 4. 2,5 and 16

⇒ \({\left[\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2}\)

⇒ \(\frac{\left[\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2+2 e\right] \times 5}{2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}}\)

‍∴ x=2, y=5 and z=16

Question 26. In which of the following reactions, H₂O₂ acts as a reducing agent-

1. H₂O₂ + 2H+ + 2e→2H₂O
2. H₂O₂-2e→ O₂ + 2H⁺
3. H₂O₂ + 2e→2OH¯
4. H₂O₂ + 2OH¯-2e→O₂ + 2H₂O

Choose the correct option 

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

Answer: 1. 2,4

In the reaction, H₂O₂ → O₂ + 2H+ + 2e, electrons are lost by H₂O₂ and hence H₂O₂ acts as a reductant. In the reaction, H₂O₂+2OH¯→ O₂ + 2H₂O + 2e, electrons are lost by H₂O₂ and hence H₂O₂ acts as a reductant.

Question 27. The Pair in which phosphorus atoms have a formal oxidation state of +3 is-

1. Orthophosphoric and Pyrophosphoric acid
2. Pyrophosphorus and Hypophosphoric acid
3. Orthophosphoric and Hypophosphoric acid
4. Pyrophosphorus and Pyrophosphoric acid

Answer: 1. Orthophosphorus and pyrophosphoric acid

Orthophosphoric acid: \(\begin{aligned}
& +1+3-2 \\
& \mathrm{H}_3 \mathrm{PO}_3
\end{aligned}\)

Pyrophosphorus acid:

Let the oxidation number of P in pyrophosphoric acid be x.

So, 4(+1) + 2x + 5(-2) = 0

or, 2x = 6 or, x = +3

Question 28. Which of the following reactions is an example of a redox creation:

1. XeFG + H₂O→XeOF4 + 2HF
2. XeF₆ + 2H₂O→XeO₂F₂ + 4HF
3. XeF₄ +O₂F₄→XeF₆ + O₂
4. XeF₂ + PF₅→[XeF]+[PF₆]-

Answer: 3. XeF₄ + O₂F₄→XeF₆ + O₂

⇒ \(\stackrel{+6-1}{\mathrm{XeF}_6}+\mathrm{H}_2 \stackrel{-2}{\mathrm{O}} \rightarrow \mathrm{XeOF}_4^{-2-1}+2 \mathrm{HF}\)

⇒ \(\stackrel{+6}{\mathrm{XeF}_6-1}+2 \mathrm{H}_2 \stackrel{-2}{\mathrm{O}} \xrightarrow[\rightarrow]{+6} \mathrm{XeO}_2^{-2-1} \mathrm{~F}_2+4 \mathrm{HF}\)

⇒ \(\stackrel{+2}{\mathrm{XeF}_2^{-1}}+\stackrel{+5-1}{\mathrm{PF}_2^{-1}} \rightarrow\left[\stackrel{+2}{\mathrm{XeF}} \mathrm{F}^{-1}\right]\left[\mathrm{PF}_6^{-1}\right]\)

For these reactions, there is no change in the oxidation number of the respective elements. So these reactions are not redox reactions.

⇒ \(\stackrel{+4}{\mathrm{XeF}_4}+\stackrel{+4}{\mathrm{O}_2} \mathrm{~F}_4^{-1} \rightarrow \stackrel{+6}{\mathrm{X}} \mathrm{XeF}_6-\stackrel{0}{\mathrm{O}}_2\)

Question 29. The oxidation states of Cr in [Cr(H₂O) ]Cl₃ [Cr(C₆HG)₂] and K2[Cr(CN₂)(0)₂(O₂)(NH₃)] respectively are

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

Answer: 3. +3,0 and +6

⇒ \(\left[\stackrel{+3}{\mathrm{Cr}}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3,\left[\stackrel{0}{\mathrm{Cr}}\left(\mathrm{C}_6 \mathrm{H}_6\right)_2\right]\)

In K₂[Cr(CN)₂(O₂)(0)₂(NH₃)] compound, let, the oxidation number of Cr be x.

or, 2 + X-2-4-2 + 0

or, x = +6

Question 30. A mixture of potassium, Oxalic acid, and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number-

1. S
2. H
3. Cl
4. C

Answer: 3. Cl

Question 31. In which of the following compounds, nitrogen exhibits the highest oxidation state—

1. N₂H₄
2. NH₃
3. N₃H
4. NH₂OH

Answer: 3. N₃H

⇒ \(\stackrel{-2}{\mathrm{~N}_2} \mathrm{H}_4, \stackrel{-3}{\mathrm{~N}} \mathrm{H}_3, \stackrel{-1 / 3}{\mathrm{~N}} \mathrm{H}_3 \mathrm{H}, \stackrel{-1}{\mathrm{~N}} \mathrm{H}_2 \mathrm{OH}\)

Question 32. In acidic medium, H₂O₂ changes \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) to CrO₅ which has two (—0—0— ) bonds. The oxidation state of Cr in CrO₅ is-

1. +5
2. +3
3. +6
4. -10

Answer: 3. +6 let the oxidation number of Cr in CrO₅ be x

Question 33.

1. H₂O₂ + O₃ →H₂O + 2O₂
2. H₂O₂ + Ag₂O→ 2Ag + H₂O + O₂

The role of hydrogen peroxide in the above reactions is respectively—

1. Oxidisingin (1) and reducing (2)
2. Reducing (1) and oxidising (2)
3. Reducing (1) and (12)
4. Oxidisingin (1) and (2)

Answer: 3. Reducing (1) and (12)

⇒ \(\mathrm{H}_2 \stackrel{-1}{\mathrm{O}}_2+\stackrel{0}{\mathrm{O}}_3 \rightarrow \mathrm{H}_2 \mathrm{O}+2 \stackrel{0}{\mathrm{O}}_2\)

In this reaction, H₂O₂ undergoes oxidation and forms O₂. Hence, it acts as a reductant

Question 34. Assuming complete ionization, the same moles of which of the following compounds will require the least amount of acidified KMnO₄ for complete oxidation—

1. FeSO₄
2. FeSO₃
3. FeC₂O₄
4. Fe(NO₂)₂

Answer: 1. FeSO₄ will require the least amount of acidified KMnO₄ for complete oxidation.

Question 35. Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior-

1. Cu + 2H₂SO₄→CuSO₄ + SO₂ + 2H₂O
2. S + 2H₂SO₄→3SO₂ + 2H₂O
3. C + 2H₂SO₄→CO₂ + 2SO₂ + 2H₂O
4. CaF₂ + H₂SO₄→CaSO₄ + 2HF

Answer: 4. CaF₂ + H₂SO₄→CaSO₄ + 2HF

In this reaction, there is no change in the oxidation number of any elements, present. Thus, it is not a redox reaction.

Question 76. For the redox reaction

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

The correct coefficients of the reactants for the balanced equation are:

1. 16,5,2
2. 2,5,16
3. 2,16,5
4. 5,16,2

Answer: 2. 2,5,16

⇒ \({\left[\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2}\)

⇒ \({\left[\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2+2 e\right] \times 5}\)

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

Question 77. When KMnO₄ reacts with KBr in an alkaline medium and gives a bromate ion, the oxidation state of Mn changes from +7 to

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

Answer: 2. +4

⇒ \(2 \mathrm{MnO}_4^{-}+\mathrm{Br}^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \stackrel{+4}{2} \mathrm{MnO}_2+\mathrm{BrO}_3^{-}+2 \mathrm{OH}^{-}\)

Question 38. K₂Cr₂O₇ in an acidic medium converts into

1. Cr²⁺
2. Cr²⁺
3. Cr⁴⁺
4. Cr⁵⁺

Answer: 2. Cr²⁺

Question 39. The oxidation state of iron in hemoglobin is

1. 0
2. +2
3. -2
4. +3

Answer: 2. +2

Question 40. What is the oxidation number of Br in KBrO₂

1. +6
2. +7
3. +5
4. +8

Answer: 2. +7

Question 41. Substances that are oxidized and reduced in the following reaction are respectively—

⇒ \(\mathrm{N}_2 \mathrm{H}_4(l)+2 \mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{N}_2(g)+4 \mathrm{H}_2 \mathrm{O}(l)\)

1. N₂H₄H₂O
2. N₂H₄H₂O₂
3. N₂H₂O₂
4. H₂O₂N₂

Answer: 2. N₂H₄H₂O₂

Question 1. For a process at a particular temperature and pressure G = ΔH- TΔS. It can be written as: – What does the quantity within the bracket indicate?
Answer:

For a process at a particular temperature and pressure G = ΔH- TΔS.

In a process at a given pressure and a temperature of T K, the change in Gibbs free energy (AG) and the change in entropy ofthe universe is expressed by the relation

⇒ \(\begin{aligned}
& \Delta G=-T \Delta S_{\text {univ }} \text { or, }-\Delta G=T \Delta S_{\text {univ }} \\
& \text { Given equation: }-\Delta G=T\left[-\frac{\Delta H}{T}+\Delta S\right]
\end{aligned}\)

Comparing equations [1] and [2] we get that the quantity within the bracket in equation [2] is ΔS_univ.

Question 2. At a particular temperature and pressure, when will the changes in the entropy of the system (ΔS_sys) and the surroundings soundings (ΔS_surr) be equal but the opposite in sign in a chemical reaction? In this condition, what will the value of AG be? Will the reaction be spontaneous in this condition?
Answer:

At a particular temperature and pressure, a reaction will be at equilibrium in the reaction AS_univ = 0. Now, ΔS_univ = ΔS_sys + ΔS_surr or, ΔS_sys + ΔS_surr = 0 or, ΔS_surr =

So, at equilibrium ΔS_sys  = -ΔS_surr

We know, ΔG = -TΔS_univ

At equilibrium, ΔS_univ = 0, So, ΔG = 0.

Since the reaction is at equilibrium, neither the forward process nor the reverse process is spontaneous.

Question 3. At a given temperature and pressure, for a chemical reaction AH < 0 & AS <0. Is the spontaneity of this reaction dependent upon temperature? If it depends and if the reaction reaches equilibrium at a temperature of Tg, then will the reaction be spontaneous below or above Te?
Answer:

At a given temperature and pressure, a process will be spontaneous in this process ΔG < 0. Again, ΔG = ΔH- TΔS. So, at a given temperature and pressure, ΔH -TΔS < 0 for a spontaneous process. For the given physical change, ΔH < 0 and ΔS < 0. Therefore, for this process. ΔH- TΔS will be less than zero only when |ΔH| is greater than |7ΔS|. Now the value of |TΔS| depends on temperature.

Therefore, the spontaneity of this process will depend on temperature. Ifthe equilibrium temperature ofthe process is Tg, then at equilibrium ΔG = 0 = ΔH- TgΔS or, ΔH = Tg x AS A process occurs spontaneously ΔG < 0 . or, ΔH- TΔS < 0 [ v ΔG = ΔH- TΔS ] Therefore, ΔH- TΔS < 0 or, TgΔS- TΔS <0 or, (Te- T)ΔS < 0 For the given process, AS < 0. So, Tg-T> 0 or, Tg> T Therefore, the given process will be spontaneous at a temperature below the equilibrium temperature.

Question 4. A physical change at a fixed pressure attains equilibrium at 353 K. In this process, ΔH > 0 and AS > 0. What will be the direction (from left to right or right to left) of this process at 350 K and 355 K?
Answer:

For a spontaneous process at constant temperature and pressure, AG = AH- TAS and AG < 0. At equilibrium, ΔG = 0 or, ΔH- TΔS = 0 or, ΔH = TΔS [T’ = equilibrium temperature = 353 K]

ΔH = 353 x ΔS and ΔG = ΔH- TΔS = (353- DΔS For this physical change, ΔS > 0. Hence, AG will be negative if 353- T < 0 or, T > 353K. This indicates that the process will be spontaneous above 353K and it will be non-spontaneous when T < 353K (as at this condition ΔG > 0 ). Therefore, at 355K the direction ofthe process will be left to right and at 350 K the direction ofthe process will be from right to left.

Question 5. At a given temperature and pressure, for a chemical reaction AH > 0 and AS < 0. The reaction is non-spontaneous at all temperatures. But the reverse reaction is spontaneous at all temperatures—Give reason.
Answer:

At constant temperature and pressure, a process will be spontaneous if ΔG < 0 for the process Again, ΔG = ΔH- TΔS

For the given reaction, ΔH > 0 and ΔS < 0.

Hence, from equation [1] we get, AG is positive for this process. So it is non-spontaneous. But for the same temperature and pressure, the AG value of the reverse reaction must be negative. Therefore, the reverse reaction is spontaneous.

Question 6. The results obtained by applying the mathematical form of the first law of thermodynamics for different processes are given below. Identify the processes (m = P-V work only):ΔU = 0, q + w = 0, q = 0, ΔU < 0. w < 0 q= O,ΔV>O , ΔU =w=0
Answer:

Given

The results obtained by applying the mathematical form of the first law of thermodynamics for different processes are given below.

According to the first law of thermodynamics, ΔU=q +w…..[1]

In the process, ΔU = 0. U is a state function. So, in a cyclic process, Its change is zero. So, from equation (1], q + w = 0. Thus, the given conditions indicate a cyclic process.

Again in the isothermal process, the change in internal energy for an ideal gas is zero. Thus according to equation [1], for isothermal change of an ideal gas, q + w = 0. Thus, the given conditions indicate an isothermal change.

In the process, q = 0. So, Δ = w or -ΔU = -w (from equation [1]).

As q = 0 , the process is adiabatic. In such a process, the work done by the system (Δv < 0) is equal to a decrease in internal energy (Δt< 0). So, the given conditions indicate that the process is adiabatic and work is done by the system.

As q = 0 and if (P-V work) = 0, from equation [1], At)1 also becomes zero, q = 0 indicates that the process is adiabatic. Since w =’ 0 and ΔV> 0, the volume of the system increases in process but no work is done by the system. This happens in case of an expansion against zero pressure. Thus, the given conditions indicate the adiabatic expansion against zero pressure

Question 7. The entropy of the system decreases on condensation of a vapour though it occurs spontaneously. Explain.
Answer:

The entropy of the system decreases on condensation of a vapour though it occurs spontaneously.

On condensation of vapour, the entropy change of the system is, ΔS = S_liquid-S_vapour. ΔS= S_liquid– S_vapour, the entropy of the system decreases in the process. Condensation of a vapor is an exothermic process. Hence, the entropy of the surroundings increases. However, the increase in entropy of the surroundings is greater than the decrease in entropy of the system. So, Δ S_toatal is always positive in this process.

ΔS_total = ΔS_sytem +ΔS_surroundings

So, condensation of a vapor is a spontaneous process.

Question 8. The formation of ice from water is exothermic, but water does not convert into ice spontaneously at ordinary temperature and pressure. Why?
Answer:

The formation of ice from water is exothermic, but water does not convert into ice spontaneously at ordinary temperature and pressure.

At a given temperature and pressure, a process will be spontaneous if it occurs with a decrease in free energy.

Again, at a fixed temperature and pressure, the free energy change of a system in a process is, ΔG = ΔH- TΔS; where ΔH and ΔS are enthalpy and entropy change of the system respectively.

During the conversion of water Into Ice, the entropy of the system decreases because the entropy of a liquid Is greater than that of a solid. So, for the conversion of water Into Ice, AS Is -ve.

Again, All Is also- ve as It Is an exothermic process. Thus, both ΔH and TΔS are -ve. But at ordinary temperature and pressure, |TΔS| Is greater than|TΔS]. Hence, the value of ΔG becomes |ΔH|. So, the conversion of water Into Ice at ordinary temperature and pressure Is not spontaneous.

Question 9. For the reaction \(\mathrm{CH}_4(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) ;\) AH0 = negative at 25°C. If the standard enthalpies of combustion of CH₄(g) and CH₃OH(g) at 25°C are -x kj mol-1 & -y kl.mol-1 respectively, then will the value of x be less than, greater than, or equal to the value of y?
Answer:

Given

For the reaction \(\mathrm{CH}_4(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) ;\) AH0 = negative at 25°C. If the standard enthalpies of combustion of CH₄(g) and CH₃OH(g) at 25°C are -x kj mol-1 & -y kl.mol-1 respectively,

⇒  \(\begin{array}{r}
\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-x \mathrm{~kJ} \\
\mathrm{CH}_3 \mathrm{OH}(\mathrm{g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H^0=-y \mathrm{~kJ}
\end{array}\)

Subtracting equation [2] from equation [l], we obtain

According to the question, (y-x) = negative. Thus, x > y.

Question 10. At a certain temperature and pressure, ΔH = 0 for the reaction: X + X = X₂. Show that the reaction is non-spontaneous at this temperature and pressure.
Answer:

At a certain temperature and pressure, for a chemical reaction, ΔG = AH- TAS.

Now, AH = 0 for given the reaction. So, AG = -TAS In the given reaction, X₂  is produced from two atoms of X. So, the entropy of the system decreases. So, AS < 0.

AG =-T x (negative quantity) = positive quantity The positive value of AG indicates that the reaction is non-spontaneous.

Question 11. The magnitude of work done by a gas in a reversible expansion is greater than that in an irreversible expansion provided the initial and final states are identical in both processes —explain
Answer:

The magnitude of work done by a gas in a reversible expansion is greater than that in an irreversible expansion provided the initial and final states are identical in both processes

Suppose, an ideal gas expands isothermally from its initial state P1 V1 to its final state P₂ V₂.

A reversible process consists of an infinite number of small steps, and at each step, the driving force, is infinitesimally greater than die opposing force. Suppose, in a reversible expansion of an ideal gas, die external pressure of the gas at a step of the process is decreased by a very small amount of dP, causing the increase in volume of the gas by dV. The work done by the gas will be, dw = ~{P_ex~dP)dV = -PdV [v Pv- dP = P (pressure of the gas) and dP x dV is very small as in the P-V diagram the points, m, and n are very close and pressure in these two points are not much different]. In the above diagram, PdV is given by the area of the strip, mnqp. The total work done in the reversible process will be equal to the sum of the die areas of all such strips. Thus, in such a process, work is done.

Absolute value of work done,|w| = area ABNM Now, in isothermal irreversible expansion, suppose, the gas is expanded from volume to V2 against an external pressure which is equal to the final pressure of the gas. In this process work done,

⇒ \(w=\int_{V_1}^{V_2} P_2 d V=-P_2 \int_{V_1}^{V_2} d V\left[P_2=\text { constant during expansion }\right]\)

The absolute value of work done,|w|=P₂(V₂– V₁) =area ABM:

From the graphs, it is clear that work done in reversible isothermal expansion is greater than that in irreversible isothermal expansion since the area of ABNM is greater than that of ABNE. This also indicates that the work involved in a process depends upon the path (or nature) of the process.

Question 12. A gas cools down during adiabatic expansion and it becomes hot during adiabatic compression —explain
Answer:

A gas cools down during adiabatic expansion and it becomes hot during adiabatic compression

In an adiabatic process, heat is not exchanged between the system and its surroundings. In this process, if the gas is expanded, then the gas does work at the expense of its internal energy. As a result, the internal energy as well as the temperature of the gas decreases due to the adiabatic expansion of an ideal gas.

As no heat is exchanged between the system and its surroundings in an adiabatic process, any work done on the gas during adiabatic compression increases the internal energy of the gas. As a result, the temperature of the gas increases.

Question 13. Given: A(s) + B₂(g)→AB₂(g);ΔHº = -x kJ \(\mathrm{A}(\mathrm{s})+\frac{3}{2} \mathrm{~B}_2(\mathrm{~g}) \rightarrow \mathrm{AB}_3(\mathrm{~g}) ; \Delta H^0=-y \mathrm{~kJ}\)
Answer: Multiplying the equation by 2, we get,

2A(s) + 2B₂(g) 2AB₂(g); ΔH0 = -2xkJ………[3]

Writing the equation in a reverse manner and multiplying both sides by 2, we get,2AB₃(g) 2A(s) + 3B₂(g); AH0 = +2y kJ Adding equations [3] and [4], we get,

A(5) + 2B2(g) + 2AB3(g-2A(s) + 2AB2(g) + 3B2(g); AH0 = (2y-2jc)kJ

2AB₃(g)→2AB₂(g) + B₂(g); AH0 = +2ykJ

Therefore, the standard reaction enthalpy for the given reaction = 2(y- x) kJ.

Question 14. At 25°C the standard enthalpy of formation of freon gas (CHC1F₂) is -480.0 kj.mol-1. Write down the thermochemical equation representing the formation reaction of the compound.
Answer:

Given

At 25°C the standard enthalpy of formation of freon gas (CHC1F₂) is -480.0 kj.mol-1.

The constituent elements of the compound, are CHC1F₂ carbon, hydrogen, chlorine, and fluorine. The standard states of these elements are C(graphites), H₂(g), Cl₂(g), and F₂(g). Thus, the thermochemical equation for the formation reaction-

⇒ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{Cl}_2(g)+\mathrm{F}_2(g) \rightarrow \mathrm{CHClF}_2(g) ;\)

⇒ \(\Delta H_f^0\left[\mathrm{CHClF}_2(g)\right]=-480.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 15. For each of the given changes, state whether the final enthalpy Is greater or less than the initial enthalpy: H₂Q(r)→ H₂O(l), H₂O(g) H₂O(l)
Answer:

Change is a melting process. So, it is an endothermic process. Hence, in this process, ΔH = +ve.

ΔH=H[H₂0(l)]- H[H₂0(s)]. As ΔH > 0, H[H₂O(l)] > H[H₂O(S)], indicating greater enthalpy for the final state than the initial state.

Change is a condensation process and so, it is an exothermic process. Hence in this process AH = -ve. AH = H[H20(/)]-H[H20(g)] Since AH < 0, H[H20(/)] < H[H20(g)]. Thus enthalpy of the final state is less than that of the initial state.

Question 16. There occurs no exchange of heat between a system and its surroundings in an adiabatic process. So, the change in entropy of the system is zero in this process. Justify the statement.
Answer:

There occurs no exchange of heat between a system and its surroundings in an adiabatic process. So, the change in entropy of the system is zero in this process.

The statement is incorrect. It is applicable only for a reversible adiabatic change.

We know, the change in entropy of a system in a reversible process \(d S=\frac{\delta q_{r e V}}{T}\) (where 6qrev is heat lost or gained by the system in the process at 7’K). Now, for the reversible adiabatic process, our = 0.

Therefore, dS – 0. So, the entropy change Is zero in a reversible adiabatic process. In the case of an irreversible process, there is no relation between the entropy change and the heat lost or gained by the system in the process.

However, it can be shown that the entropy change in an irreversible process is \(d S>\frac{\delta q_{l r}}{T}\) (where oq[rr is the heat lost or gained by the system in the process at T K). Now, in an irreversible adiabatic process 6q(rr = 0. So, dS (or AS) > 0. Tilus the entropy of the system increases in an irreversible adiabatic process.

Question 17. Mg) + A(g)→A₂(g); these types of reactions are generally exothermic. Explain.
Answer:

Mg) + A(g)→A₂(g); these types of reactions are generally exothermic.

In the reaction, one molecule A₂(g) is formed from two atoms of A(g). Thus, the entropy of the system decreases. For any chemical reaction at fixed temperature and pressure, ΔG =ΔH- TΔS. For the given reaction, ΔS<0. The reaction will be spontaneous if ΔG < 0.

So, negative quantity = ΔH- (T X negative quantity) = ΔH + positive quantity

∴ ΔH = (-ve) quantity- (+ye) quantity ; So, ΔH < 0

∴ The given reaction is exothermic.

Question 18. At ordinary temperature and pressure, solid NH₄C1 dissolves in water to form NH4(aq) and Cl(aq) ions. The process is endothermic. Indicate the signs (+ or-) of ΔS_sys, ΔS_surr, ΔH And ΔG
Answer:

As the process is endothermic, AH > 0.

At ordinary temperature and pressure, the dissolution of solid NH₄C1 occurs spontaneously, so AG < 0.

NH+₄ and Cl- ions in the crystal lattice of NH₄C1 are held at fixed positions and consequently, they are not able to move. But in an aqueous solution of NH₄C1, NH, and Cl- ions are distributed throughout the solution. Naturally, the randomness of the ions in aqueous solution of NH₄C1 is greater than that in solid NH₄C1. Hence, in the dissolution process of NH4CI, ΔS_sys > 0.

As the process is endothermic, the system absorbs heat from the surroundings. Consequently, the surroundings lose heat. As a result, the entropy of the surroundings decreases (AS_surr < 0).

However, the increase in the entropy system is greater than the decrease in the entropy of the surroundings. So, the net entropy change of the system and its surroundings is always positive, and the process occurs spontaneously.

Question 19. Classify as an open, closed, or isolated system:

1. A cup of coffee placed on a table
2. Water in a beaker is boiled by heating
3. Lead nitrate is heated in a test tube
4. Solid NH₄Cl is heated in a closed vessel
5. Substances present in a soda-water bottle
6. Mercury enclosed in the thermometer

Answer:

1. Open system
2. Open system
3. Open system
4. Closed system
5. Closed system
6. Closed system

Question 20. Assuming experimental conditions are the same, compare (ΔH-ΔU) values for the given reactions

1. \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Answer: In case of reaction (1) \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Therefore , \(\Delta H-\Delta U=\Delta n R T=-\frac{3}{2} R T\)

In case of reaction \(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)

Therefore, \(\Delta H-\Delta U=\Delta n R T=-\frac{1}{2} R T\)

Hence, the magnitude of H-U is greater in the case of 2

In case of reaction \(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)

Hence, the magnitude of (AH- AU) is greater in the case of 2

Question 1. Write the name of the element which is diagonally related to the element beryllium. Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively. State the blocks in the periodic table to which elements A and C belong. Write the formulas of the compounds formed between B and C and A and C. State the nature ofthe bonds.
Answer:

Aluminium (Al)

Given

Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively.

11A: 1, 14B: ls²2s²2p⁶3s²3p²,

17C: ls²2s²2p⁶3s²3p⁵

From the electronic configuration is evident that A is an s -a block element whereas C is a p -a block element.

The compound formed between B and C has the formula BC4 and the nature of the bond is covalent. A compound formed by a combination of A and C is AC (electrovalent).

Question 2. Classify as basic, amphoteric or acidic: BeO, Al₂O₃, CaO, SiO₂ State the modem periodic law.
Answer:

Given

BeO, Al₂O₃, CaO, SiO₂

BeO, Al₂O₃: amphoteric; CaO: basic; Si02: acidic

Question 3. Write the name of the element which is diagonally related to the element beryllium.

Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively.

State the blocks in the periodic table to which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.

Symbol of one transition element — Fe.

Symbol of one inner-transition element — Ce

Ionic radius is highest for O₂– for the given case.

Electronegativity is lowest for Na for the given case.

Ionization energy is highest for Ar for the given case.

Question 4. State the group number in the modem periodic table where solid, liquid, and gaseous elements are present at room temperature. Identify solid, and liquid elements. Indicate the given elements as alkali metal, alkaline earth metal, coinage metal, or chalcogen: Li, Ca, S, Cu.
Answer:

Group-17 of modem periodic table contains solid (iodine, I2), liquid (bromine, Br2), and gaseous (chlorine, Cl2) elements at the same time

Group 17 of modern periodic table contains solid

(iodine, I2), liquid (bromine, Br2) and gaseous

(chlorine, Cl2) elements at the same time.

Question 5. Among which of the following pairs of elements, the first has lesser ionization energy than the second—

1. Na, K
2. N, O
3. B, Be
4. Br, I

Answer: 3. B, Be

Question 6. Name a pair of elements that exhibit a diagonal relationship.
Answer:

Li and Mg.

Question 7. Name a chalcogen and an alkaline earth metal.
Answer:

Oxygen (O) is an example of chalcogen and magnesium (Mg) is an alkaline earth metal

Question 10. Mention the names of the most electropositive and most electronegative stable elements in the periodic table. 2
Answer:  

Question 11. What is a metalloid? Give one example. 
Answer:

Metalloid

The most electropositive stable element of the periodic table is cesium (Cs) and the most electronegative stable element of the periodic table is fluorine (F).

Question 11. Which has greater ionization energy and why—S or P?
Answer:

Phas greater ionization energy than S. According to Hund’s rule the half-filled electronic configuration is highly stable and the outermost 3p -orbital of the P atom is half-filled.

To produce P+ ions by removing an electron from the 3p orbital ofPrequireshuge energyresulting a very high value of ionization energy.

On the other hand, the S atom has a partially filled 3p -orbital which attains a stable half-filled electronic configuration in the S+ ion.

Therefore, to produce S to S+ comparatively lower energy is required resulting in a lower value of ionization energy.

Question 13. Arrange as directed:

1. SiO₂, NO₂, Al₂O₃, ClO₂ (Increasing acidic property)
2. Na⁺, Al³⁺, F^–, Cl^– (Increasing ionic radius)

Arrange the following ions in the increasing order of their ionic radius. F-, Mg²⁺, Al³⁺, O^2-

Answer:

Al₂O₃ < SiO₂ < NO₂ < ClO₂

Al³⁺ < Na⁺ < F^– < Cl^–

Question 14. Arrange the following ions in the increasing order of their ionic radius. F^–, Mg²⁺, Al³⁺, O^2-
Answer:

Al³⁺ < Mg²⁺ < F^–< O^2-

Question 15. Which ofthe given elements has the lowest electron affinity

1. C
2. P
3. O
4. S

Answer: 2. P

Question 16. Arrange the following elements according to their decreasing oxidising ability: I, Br, F, Cl.
Answer:

F > Cl > Br >I

Question 17. Arrange as per instruction: Na20, B203, A1203, MgO (increasing order of basicity) Be, 0, Cl, I (increasing order of electro-negativity)
Answer:

B₂O₃ < Al₂O₂₃< MgO < Na₂O

Be <I < Cl <

Question 19. Two atoms X and Y have electronic configurations [He] 2sz2p3 and [Ne] 3s23p2 respectively. Which period and group of the periodic table do they belong to? The second electron affinity of an element is always endothermic. Explain.
Answer:

X —1 Period 2, Group-15. Y —1 Period 3, Group-14.

Question 20. The Electron affinity of Cl is greater than that of. Explain. Indicate the basic/acidic behavior of the hydrides of the following elements: F, C, O, and N Metallic properties of elements in the periodic table.
Answer:

HF: acidic, CH4: neutral, H20: neutral, NH3: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increase).

Question 21. Metallic property of elements down the group in the periodic table
Answer:

HF: acidic, CH4: neutral, H20: neutral, NH3: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increases).

Question 22. Arrange the following in increasing order of ionic radius: Na⁺, F^–, O^2-, Al³⁺, N^3-.
Answer:

Al³⁺ < Na⁺ < F^– < O^2-< N^3-

Question 23. Why does nitrogen have a higher ionization enthalpy than that of Oxygen? Arrange the following in increasing order of acidity: N02, A1203, Si02, C102
Answer:

Na < Al < Mg < Si

Question What will be the order of, Mg, A1, and Sn terms of the first ionisation enthalpy?
Answer:

Mg⁺ < Na⁺ < F^– < O^2-

Question 25. Arrange the following ions in order of increasing ionic radii: Na⁺, F^–, O^2-, Mg²⁺
Answer:

Mg²⁺ < Na⁺ < F^– < O^2-

Question 26. Why is the electron-gain enthalpy of oxygen less than that of sulfur? Arrange the following metal oxides in terms of ascending order of basicity: ZnO, MgO, CaO, CuO.
Answer:

Electron-gain enthalpy of O is less than that of S as the 2p -orbital of the outermost shell of O-atom is much smaller in size than the 3p -orbital of the outermost shell of the S-atom.

So, the additional electron-electron repulsive force produced due to the addition of one electron to the 2p -orbital of the oxygen atom from outside is more than the additional electron-electron repulsive force developed by the addition of one electron to the 3p -orbital of the S-atom.

So the electron-gain enthalpy of the O-atom becomes less than that S-atom.

Question 27. Why is the first ionization enthalpy of helium maximum among all the elements? Arrange the given compounds in terms of ascending order of oxidizing property: HC1, HBr, HI, HF
Answer:

Configuration of He is Is2, i.e., all electrons of He are presenting Is -orbital. These electrons are attracted very strongly by the nucleus.

Further, there are no inner electronic orbitals to shield these electrons from the pull ofthe nucleus.

So removal of an electron from the Is -orbital requires a large amount of energy. Therefore, the first ionization enthalpy ofHe is maximum.

Hydrogen halides do not show oxidizing properties. However, the reducing power of the hydrogen halides follows the sequence: HF <HCl < HBr < H

Question 28. Determine the position of an element in the long form of the periodic table if its electronic configuration is [18Ar]3d104s2.
Answer:

It is a d -block element. Hence period no. = principle quantum no. ofthe outermost shell = 4 and its group no. = total electronin 3d and 4s -orbital = 10 + 2 = 12.

Question 29. Mention the name and the position of two elements, one of which is the most electronegative and the most electropositive periodic table.
Answer:

Most electronegative element =F (group 17, period-2)

Most electropositive element = Cs (group 1, period-6)

Question 30. Which of the following two elements has a diagonal relationship? Li, Be, A1, and Si Between 2gCu and igK which one has higher ionization enthalpy and why do Be and Al
Answer:

1. Electronic configuration of 19K: ls²2s²2p⁶3s²3p⁶4s¹
2. Electronic configuration of 2gCu: ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹

The nucleus of the Cu-atom contains 10 more protons than that of the K-atom.

Additional nuclear pull on the outermost electron (4s¹) in Cu-atom is not counter-balanced by the shielding effect of ten 3d-electrons because orbitals have poor screening effects.

Thus the effective nuclear charge acting on the 4s-electron of copper is greater than that acting on the 4s-electron of

Question 31. Arrange the following ions in ascending order of radius: Na⁺, F^–, O^2-, Mg²⁺
Answer:

Mg²⁺< Na⁺ <F^– <O^2-

Question 32. Is the electronegativity of Sn²⁺ and Sn²⁴⁺ equal or different?
Answer:

Elements in a higher oxidation state have higher electronegativity compared to the elements in a lower oxidation state. Thus Sn4+ is more electronegative than Sn2

Question 33. What is the oxidation state of TI in the compound TlCl₃ ? Which is the stronger oxidizing agent between CO2 and PbO2 and why?
Answer:

+1

PbO₂. Due to the inert pair effect, Pb is very unstable in the +4 oxidation state. Thus Pb4+ can be easily reduced to Pb2.

Question 34. Which iw is more stable between BCl₃ and TlCl₃ and why? What is the oxidation state of Zn in Zn-Hg Arrange the following elements in the increasing order of their first ionisation enthalpy.IX Be. Na, Mg 56 Arrange the following elements in the decreasing order of their electro-negativity. Si N. F. Cl
Answer:

Because of the poor shielding effect by the inner d – and f – electrons, the inert pair effect is maximum for Tl.

Thus the most stable oxidation state of TL is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BCl₃ is stable.

Question 37. The outermost electronic configuration of the atom of an element is 3s-3p3. Mention the position of the element in the long periodic table. Why is electron gain enthalpy of oxygen less than that of sulfur?
Answer:

Because of the poor shielding effect by the inner d – and f- electrons, the inert pair effect is maximum for Tl. Thus the most stable oxidation state of TL is +1 and not +3. Therefore TICl₃ is unstable. B does not exhibit an inert pair effect and thus BC13 is stable.