1. Which of the following is correct representation of hydrogen bonds in \( H – F \)?

(various bond angles for \( H-F…H \))

• (a) Zig-zag with \( 109.5^{\circ} \)
• (b) Zig-zag with \( 140^{\circ} \)
• (c) Linear with \( 180^{\circ} \)
• (d) Bent with \( 120^{\circ} \)

Correct Answer: (b)

Explanation: In the solid state, hydrogen fluoride exists as zig-zag chains due to hydrogen bonding. The \( H-F…H \) bond angle is approximately \( 140^{\circ} \).

2. \( (CH_{3})_{4}N^{+}OH^{-} \) is stronger base than \( (CH_{3})_{3}N^{+}HOH^{-} \) due to:

• (a) intramolecular hydrogen bonding in \( (CH_{3})_{3}N^{+}HOH^{-} \)
• (b) intermolecular hydrogen bonding in \( (CH_{3})_{4}N^{+}OH^{-} \)
• (c) intermolecular hydrogen bonding \( (CH_{3})_{3}N^{+}HOH^{-} \) with water
• (d) none of these.

Correct Answer: (a)

Explanation: In trimethylammonium hydroxide, intramolecular hydrogen bonding is possible between the \( H \) attached to \( N \) and the \( OH^{-} \) group. This stabilization reduces the availability of the hydroxide ion, thereby decreasing the basic strength compared to the tetramethylammonium ion where no such H-bonding is possible.

3. Which is the correct order of bond length?

• (a) \( C\equiv N > C\equiv C > N\equiv N \)
• (b) \( N\equiv N > C\equiv N > C\equiv C \)
• (c) \( C\equiv C > C\equiv N > N\equiv N \)
• (d) \( C\equiv C > N\equiv N > C\equiv N \)

Correct Answer: (c)

Explanation: Nitrogen atoms contain lone pairs of electrons which cause repulsion with adjacent atoms. Consequently, the bond length order for these triple bonds is \( C\equiv C > C\equiv N > N\equiv N \).

4. The energy of bonding molecular orbital is less than combining atomic orbitals because:

• (a) molecular orbitals are more symmetrical than atomic orbitals
• (b) molecular orbitals are near to nucleus than atomic orbitals
• (c) probability of finding the electrons in bonding molecular orbital is less than atomic orbitals
• (d) probability of finding the electrons in bonding molecular orbitals is more than atomic orbitals.

Correct Answer: (d)

Explanation: A bonding molecular orbital is formed by the additive overlap of atomic wave functions (\( \Psi_A + \Psi_B \)). The electron probability density \( \Psi^2 = (\Psi_A + \Psi_B)^2 = \Psi_A^2 + \Psi_B^2 + 2\Psi_A\Psi_B \) is greater than the sum of the individual atomic densities, leading to lower energy and greater stability.

5. The compound with non-dipole moment (zero dipole moment) is:

• (a) methyl chloride
• (b) carbon tetrachloride
• (c) methylene chloride
• (d) chloroform

Correct Answer: (b)

Explanation: Carbon tetrachloride (\( CCl_4 \)) has a symmetrical tetrahedral structure. The individual \( C-Cl \) bond dipoles cancel each other out perfectly, resulting in a net dipole moment of zero.

6. A molecule (X) has (i) four sigma bonds formed by the overlapping of \( sp^{2} \) and s-orbital. (ii) one sigma bond formed by \( sp^{2} \) and \( sp^{2} \) orbitals and (iii) one \( \pi \) bond is formed by \( p_{x} \) and \( p_{x} \) orbitals. Which of the following is X?

• (a) \( C_{2}H_{6} \)
• (b) \( C_{2}H_{3}Cl \)
• (c) \( C_{2}H_{2}Cl_{2} \)
• (d) \( C_{2}H_{4} \)

Correct Answer: (d)

Explanation: In ethene (\( C_2H_4 \)), each carbon is \( sp^2 \) hybridised. There is one \( \sigma(sp^2-sp^2) \) bond between carbons, four \( \sigma(sp^2-s) \) bonds with hydrogens, and one \( \pi(p-p) \) bond.

7. Which one of the following is a planar molecule?

• (a) \( NH_{3} \)
• (b) \( H_{3}O^{+} \)
• (c) \( BCl_{3} \)
• (d) \( PCl_{3} \)

Correct Answer: (c)

Explanation: In \( BCl_3 \), the boron atom is \( sp^2 \) hybridised with no lone pairs, resulting in a trigonal planar geometry. The others (\( NH_3, H_3O^+, PCl_3 \)) are pyramidal due to the presence of a lone pair.

8. The bond length of HCl molecule is 1.275 \( \dot{A} \) and its dipole moment is 1.03 D. The ionic character of the molecule (in percent) is:

• (a) 100
• (b) 67.3
• (c) 33.6
• (d) 16.83

Correct Answer: (d)

Explanation: Theoretical dipole moment \( \mu = e \times d = (4.8 \times 10^{-10} \text{ esu}) \times (1.275 \times 10^{-8} \text{ cm}) = 6.12 \text{ D} \).
% ionic character \( = (\mu_{\text{exp}} / \mu_{\text{theo}}) \times 100 = (1.03 / 6.12) \times 100 = 16.83\% \).

9. The bond energies in (kJ \( mol^{-1} \)) of \( P-H, As-H \) and \( N-H \) respectively are:

• (a) 247, 318 and 389
• (b) 247, 389 and 318
• (c) 318, 389 and 247
• (d) 318, 247 and 389

Correct Answer: (d)

Explanation: Bond stability and energy decrease as the size of the central atom increases down the group. The order is \( N-H (389) > P-H (318) > As-H (247) \).

10. Which of the following statements is correct?

• (a) The number of electrons present in the valency shell of S in \( SF_{6} \) is 12.
• (b) The rate of ionic reactions are very slow.
• (c) According to VSEPR theory, \( SnCl_{2} \) is the linear molecule.
• (d) The correct order of ability of form ionic compounds is \( Al^{3+} > Mg^{2+} > Na^{+} \).

Correct Answer: (a)

Explanation: In \( SF_6 \), sulfur forms 6 covalent bonds with fluorine, resulting in 12 electrons in its valence shell (expanded octet).

11. Match the following geometry:

P. \( BCl_{3} \) – (ii) Planar triangular; Q. \( PdBr_{4}^{2-} \) – (v) Square planar; R. \( SF_{6} \) – (iv) Octahedral; S. \( I_{3}^{-} \) – (i) Linear

• (a) P-ii, Q-iii, R-iv, S-i
• (b) P-v, Q-iii, R-ii, S-i
• (c) P-ii, Q-v, R-iv, S-i
• (d) P-v, Q-iv, R-iii, S-ii

Correct Answer: (c)

Explanation: Correct geometry matches: \( BCl_3 \) is trigonal planar; \( PdBr_4^{2-} \) is square planar (\( dsp^2 \)); \( SF_6 \) is octahedral; \( I_3^- \) is linear.

12. Which one of the following sets of ions represents the collection of isoelectronic species?

• (a) \( K^{+}, Cl^{-}, Mg^{2+}, Sc^{3+} \)
• (b) \( Na^{+}, Ca^{2+}, Sc^{3+}, F^{-} \)
• (c) \( K^{+}, Ca^{2+}, Sc^{3+}, Cl^{-} \)
• (d) \( Na^{+}, Mg^{2+}, Al^{3+}, Cl^{-} \)

Correct Answer: (c)

Explanation: All ions in set (c) (\( K^+, Ca^{2+}, Sc^{3+}, Cl^- \)) have 18 electrons, making them isoelectronic.

13. According to Fajan’s rule polarization is more when:

• (a) small cation and large anion
• (b) small cation and small anion
• (c) large cation and large anion
• (d) large cation and small anion.

Correct Answer: (a)

Explanation: Polarization (covalent character) increases with high charge density on the cation (small size) and high polarizability of the anion (large size).

14. The correct order of hybridization of the central atom in \( NH_{3}, [PtCl_{4}]^{2-}, PCl_{5} \) and \( BCl_{3} \) is:

• (a) \( dsp^{2}, dsp^{3}, sp^{2}, sp^{3} \)
• (b) \( sp^{3}, dsp^{2}, sp^{3}d, sp^{2} \)
• (c) \( dsp^{2}, sp^{2}, sp^{3}, dsp^{3} \)
• (d) \( dsp^{2}, sp^{3}, sp^{2}, dsp^{3} \)

Correct Answer: (b)

Explanation: Hybridizations: \( NH_3 \) is \( sp^3 \); \( [PtCl_4]^{2-} \) is \( dsp^2 \); \( PCl_5 \) is \( sp^3d \); \( BCl_3 \) is \( sp^2 \).

15. The formal charge on the central oxygen atom in \( O_{3} \) molecule is:

• (a) 0
• (b) +1
• (c) -1
• (d) -2

Correct Answer: (b)

Explanation: In the Lewis structure of ozone, the central oxygen atom is bonded to two other oxygens and has one lone pair. Formal charge \( = 6 – 2 – (1/2 \times 6) = +1 \).

16. The electronegativity of H and Cl are 2.1 and 3.0 respectively. The correct statement about the nature of HCl is:

• (a) 17% ionic
• (b) 83% ionic
• (c) 50% ionic
• (d) 100% ionic

Correct Answer: (a)

Explanation: Using the Hannay-Smith equation: % ionic character \( = 16(\Delta \chi) + 3.5(\Delta \chi)^2 \). With \( \Delta \chi = 0.9 \), the value is approximately \( 17.2\% \).

17. Which of the following have identical bond order? (I) \( CN^{-} \), (II) \( O_{2}^{-} \), (III) \( NO^{+} \), (IV) \( CN^{+} \)

• (a) I, III
• (b) I, II
• (c) II, IV
• (d) I, II, III

Correct Answer: (a)

Explanation: Species with the same number of electrons (isoelectronic) have the same bond order. Both \( CN^- \) and \( NO^+ \) have 14 electrons and a bond order of 3.

18. Which of the following species is diamagnetic in nature?

• (a) \( He_{2}^{+} \)
• (b) \( H_{2} \)
• (c) \( H_{2}^{+} \)
• (d) \( H_{2}^{-} \)

Correct Answer: (b)

Explanation: \( H_2 \) has two electrons, both paired in the \( \sigma 1s \) orbital, making it diamagnetic. The others have unpaired electrons and are paramagnetic.

19. The formal charge of the O-atoms in the \( [:\dot{O}=N=\dot{O}:]^{+} \) ion is:

• (a) -2
• (b) -1
• (c) 0
• (d) +1

Correct Answer: (c)

Explanation: For each oxygen atom in the nitronium ion: Valence electrons = 6, lone pair electrons = 4, bonding electrons = 4. Formal charge \( = 6 – 4 – (1/2 \times 4) = 0 \).

20. Identify the correct sequence of increasing number of \( \pi \)-bonds: I. \( H_{2}S_{2}O_{6} \), II. \( H_{2}SO_{3} \), III. \( H_{2}S_{2}O_{5} \)

• (a) I, II, III
• (b) II, III, I
• (c) II, I, III
• (d) I, III, II

Correct Answer: (b)

Explanation: Number of \( \pi \)-bonds: \( H_2SO_3 = 1 \); \( H_2S_2O_5 = 3 \); \( H_2S_2O_6 = 4 \). Order is II < III < I.

21. Which of the following does not have a linear structure?

• (a) \( HgCl_{2} \)
• (b) \( SnCl_{2} \)
• (c) \( C_{2}H_{2} \)
• (d) \( CS_{2} \)

Correct Answer: (b)

Explanation: \( SnCl_2 \) has a bent (V-shaped) structure due to the presence of a lone pair on the tin atom. The others are linear.

22. Which of the following has a shape different from others?

• (a) \( PCl_{3} \)
• (b) \( NH_{3} \)
• (c) \( PH_{3} \)
• (d) \( BF_{3} \)

Correct Answer: (d)

Explanation: \( BF_3 \) is trigonal planar. \( PCl_3, NH_3, \) and \( PH_3 \) are all trigonal pyramidal due to one lone pair.

23. Which of the following structures is expected to have three bond pairs and one lone pair?

• (a) Tetrahedral
• (b) Trigonal planar
• (c) Pyramidal
• (d) Octahedral

Correct Answer: (c)

Explanation: A pyramidal geometry arises when a central atom has three bond pairs and one lone pair in its valence shell.

24. The \( sp^{3}d^{2} \) hybridization of central atom of a molecule would lead to:

• (a) square planar geometry
• (b) tetrahedral geometry
• (c) trigonal bipyramidal geometry
• (d) octahedral geometry

Correct Answer: (d)

Explanation: \( sp^3d^2 \) hybridization corresponds to an octahedral spatial arrangement of electron pairs.

25. In the electron-dot structure \( \dot{N}=N=\dot{N} \), calculate the formal charge from left to right nitrogen atom:

• (a) -1, -1, +1
• (b) -1, +1, -1
• (c) +1, -1, -1
• (d) +1, -1, +1

Correct Answer: (b)

Explanation: For terminal nitrogens: \( 5 – 4 – 2 = -1 \). For central nitrogen: \( 5 – 0 – 4 = +1 \). The sequence is -1, +1, -1.

26. Hybridization of \( C_{2} \) and \( C_{3} \) of \( H_{3}C-CH=C=CH-CH_{3} \) are:

• (a) sp, \( sp^{3} \)
• (b) \( sp^{2} \), sp
• (c) \( sp^{2}, sp^{2} \)
• (d) sp, sp

Correct Answer: (b)

Explanation: \( C_2 \) is part of a double bond (\( sp^2 \)). \( C_3 \) is bonded to two double bonds (\( =C= \)), which is sp hybridised.

27. \( XeF_{2} \) is isostructural with:

• (a) \( SbCl_{3} \)
• (b) \( BaCl_{2} \)
• (c) \( TeF_{2} \)
• (d) \( ICl_{2}^{-} \)

Correct Answer: (d)

Explanation: Both \( XeF_2 \) and \( ICl_2^- \) are linear molecules with \( sp^3d \) hybridization on the central atom (2 bond pairs and 3 lone pairs).

28. Which of the following is a polar molecule?

• (a) \( SiF_{4} \)
• (b) \( XeF_{4} \)
• (c) \( BF_{3} \)
• (d) \( SF_{4} \)

Correct Answer: (d)

Explanation: \( SF_4 \) has a “see-saw” shape due to the presence of one lone pair. Its asymmetric geometry results in a non-zero net dipole moment. The others are symmetrical and non-polar.

29. Which of the following is paramagnetic?

• (a) \( CN^{-} \)
• (b) \( NO^{+} \)
• (c) \( CO \)
• (d) \( O_{2}^{-} \)

Correct Answer: (d)

Explanation: The superoxide ion (\( O_2^- \)) has 17 electrons. According to Molecular Orbital theory, it has one unpaired electron in its antibonding orbitals.

30. Which one of the following molecules contains no \( \pi \) bond?

• (a) \( SO_{2} \)
• (b) \( NO_{2} \)
• (c) \( CO_{2} \)
• (d) \( H_{2}O \)

Correct Answer: (d)

Explanation: Water (\( H_2O \)) contains only single sigma bonds between oxygen and hydrogen. The others all contain double bonds.

31. Which one of the following is not correct in respect to hybridization of orbitals?

• (a) The orbitals present in the valence shell only are hybridized.
• (b) The orbitals undergoing hybridization have almost equal energy.
• (c) Promotion of electron is not essential condition for hybridization.
• (d) Pure atomic orbitals are more effective in forming stable bonds than hybrid orbitals.

Correct Answer: (d)

Explanation: Hybrid orbitals are more effective than pure atomic orbitals at forming stable bonds because they have better directional characteristics and greater overlap.

32. Allyl cyanide molecule contains:

• (a) 9 sigma bonds, 4 pi bonds and no lone pair
• (b) 9 sigma bonds, 3 pi bonds and one lone pair
• (c) 8 sigma bonds, 5 pi bonds and one lone pair
• (d) 8 sigma bonds, 3 pi bonds and two lone pairs

Correct Answer: (b)

Explanation: Structure: \( CH_2=CH-CH_2-C\equiv N \). It contains 9 sigma bonds, 3 pi bonds (1 from \( C=C \), 2 from \( C\equiv N \)), and one lone pair on nitrogen.

33. Bond angle is minimum for:

• (a) \( H_{2}O \)
• (b) \( H_{2}S \)
• (c) \( H_{2}Se \)
• (d) \( H_{2}Te \)

Correct Answer: (d)

Explanation: Bond angle decreases as the electronegativity of the central atom decreases down the group. The angle is minimum for \( H_2Te \) (approx. \( 90^{\circ} \)).

34. Which of the following has non-zero dipole moment?

• (a) \( PCl_{3}Br_{2} \)
• (b) \( PBr_{3}Cl_{2} \)
• (c) \( PCl_{5} \)
• (d) \( CO_{2} \)

Correct Answer: (a)

Explanation: \( PCl_3Br_2 \) is an asymmetric trigonal bipyramidal structure and thus has a non-zero dipole moment. \( PBr_3Cl_2, PCl_5, \) and \( CO_2 \) are symmetrical.

35. Which of the following statement is correct?

• (a) \( BCl_{3} \) and \( AlCl_{3} \) both are Lewis acids but \( BCl_{3} \) is stronger than \( AlCl_{3} \)
• (b) \( BCl_{3} \) and \( AlCl_{3} \) both are Lewis acids but \( AlCl_{3} \) is stronger than \( BCl_{3} \)
• (c) \( BCl_{3} \) and \( AlCl_{3} \) both are equally strong Lewis acids
• (d) both \( BCl_{3} \) and \( AlCl_{3} \) are not Lewis acids

Correct Answer: (a)

Explanation: Due to more effective \( p\pi-p\pi \) back-bonding in \( AlCl_3 \) (similar sizes of Cl and Al 3p orbitals) compared to \( BCl_3 \), boron remains more electron-deficient. Thus, \( BCl_3 \) is a stronger Lewis acid.

36. \( N_{2} \) and \( O_{2} \) are converted into monoanions \( N_{2}^{-} \) and \( O_{2}^{-} \) respectively. Which of the following statements is wrong?

• (a) In \( N_{2}^{+} \), N-N bond weakens.
• (b) In \( O_{2}^{+} \), O-O bond order increases.
• (c) In \( O_{2}^{+} \), paramagnetism decreases.
• (d) \( N_{2}^{+} \) becomes diamagnetic.

Correct Answer: (d)

Explanation: Statement (d) is wrong because when \( N_2 \) (14 electrons, diamagnetic) loses an electron to become \( N_2^+ \) (13 electrons), it acquires an unpaired electron and becomes paramagnetic.

37. Mark the incorrect statement in the following:

• (a) The bond order in the species \( O_{2}, O_{2}^{+} \) and \( O_{2}^{-} \) decreases as \( O_{2}^{+} > O_{2} > O_{2}^{-} \)
• (b) The bond energy in a diatomic molecule always increases when an electron is lost.
• (c) Electrons in antibonding M.O. contribute to repulsion between two atoms.
• (d) With increase in bond order, bond length decreases and bond strength increases.

Correct Answer: (b)

Explanation: Bond energy decreases if the electron is removed from a bonding orbital (e.g., \( N_2 \rightarrow N_2^+ \)) and increases only if removed from an antibonding orbital (e.g., \( O_2 \rightarrow O_2^+ \)).

38. Among the following species, identify the isostructural pairs: \( NF_{3}, NO_{3}^{-}, BF_{3}, H_{3}O^{+}, HN_{3} \)

• (a) \( [NF_{3}, NO_{3}^{-}] \) and \( [BF_{3}, H_{3}O^{+}] \)
• (b) \( [NF_{3}, HN_{3}] \) and \( [NO_{3}^{-}, BF_{3}] \)
• (c) \( [NF_{3}, H_{3}O^{+}] \) and \( [NO_{3}^{-}, BF_{3}] \)
• (d) \( [NF_{3}, H_{3}O^{+}] \) and \( [HN_{3}, BF_{3}] \)

Correct Answer: (c)

Explanation: \( NF_3 \) and \( H_3O^+ \) are both pyramidal (3 bond pairs + 1 lone pair). \( NO_3^- \) and \( BF_3 \) are both trigonal planar (3 bond pairs + 0 lone pairs).

39. In which set of species, are all the species paramagnetic?

• (a) \( B_{2}, O_{2}, N_{2} \)
• (b) \( B_{2}, O_{2}, N_{2}^{+} \)
• (c) \( B_{2}, F_{2}, O_{2} \)
• (d) \( B_{2}, O_{2}, Li_{2} \)

Correct Answer: (b)

Explanation: \( B_2 \) and \( O_2 \) have two unpaired electrons each. \( N_2^+ \) has one unpaired electron. All are paramagnetic.

40. Which is non-polar, but contains polar bonds?

• (a) \( HCl \)
• (b) \( H_{2}O \)
• (c) \( SO_{3} \)
• (d) \( CO_{2} \)

Correct Answer: (d)

Explanation: \( CO_2 \) contains polar \( C=O \) bonds, but its linear symmetry causes the individual dipoles to cancel, making the molecule non-polar.

41. Likely bond angles of \( SF_{4} \) molecule are:

• (a) \( 89^{\circ}, 117^{\circ} \)
• (b) \( 120^{\circ}, 180^{\circ} \)
• (c) \( 45^{\circ}, 118^{\circ} \)
• (d) \( 117^{\circ}, 92^{\circ} \)

Correct Answer: (a)

Explanation: In the see-saw structure of \( SF_4 \), the lone pair causes distortion of the ideal \( 90^{\circ} \) and \( 120^{\circ} \) angles to approximately \( 89^{\circ} \) and \( 117^{\circ} \).

42. Bond order and magnetic moment of \( CO^{+} \) is:

• (a) 2.5 and paramagnetic moment
• (b) 3.5 and diamagnetic moment
• (c) 3.5 and paramagnetic moment
• (d) 2.5 and diamagnetic moment

Correct Answer: (c)

Explanation: The removal of an electron from the weakly antibonding orbital of \( CO \) leads to an increase in bond order to 3.5. The presence of one unpaired electron makes it paramagnetic.

43. Which of the following statements is correct?

• (a) \( sp^{3} \) hybrid orbitals have equal s and p character.
• (b) The bond angle decreases with the decrease of s character of hybridized orbital.
• (c) Resonance decreases the stability of a molecule.
• (d) Resonance is due to delocalization of sigma electrons.

Correct Answer: (b)

Explanation: Bond angle is directly proportional to s-character. As s-character decreases (from \( sp \) to \( sp^2 \) to \( sp^3 \)), the bond angle decreases.

44. Which of the following represents the zero overlap?

• (a) s and \( p_z \) on z-axis
• (b) \( p_x \) and \( p_x \) on z-axis (sidewise)
• (c) \( p_x \) and \( p_x \) on x-axis (head-on)
• (d) \( p_x \) and s on z-axis

Correct Answer: (d)

Explanation: Overlap between a \( p_x \) orbital and an s-orbital along the z-axis (the internuclear axis) is zero because the positive and negative parts of the overlap cancel each other.

45. The ground state electronic configuration of CO molecule is:

• (a) \( 1\sigma^{2} 2\sigma^{2} 1\pi^{4} 3\sigma^{2} \)
• (b) \( 1\sigma^{2} 2\sigma^{2} 3\sigma^{2} 1\pi^{2} 2\pi^{2} \)
• (c) \( 1\sigma^{2} 2\sigma^{2} 1\pi^{4} 3\sigma^{2} 2\pi^{2} \)
• (d) \( 1\sigma^{2} 1\pi^{4} 2\sigma^{2} 3\sigma^{2} \)

Correct Answer: (a)

Explanation: The valence electronic configuration for \( CO \) (14 electrons) following the energy order of molecular orbitals is \( 1\sigma^{2} 2\sigma^{2} 1\pi^{4} 3\sigma^{2} \).