JEE Chemistry Chemical Bonding And Molecular Structure MCQs Set 48

Answer: (b) NO⁺, C₂²⁻, CN⁻, N₂

Explanation:
1. NO⁺, C₂²⁻, CN⁻ and N₂ each have 14 electrons.
2. In the other groups the electron counts differ (for example O₂⁻ has 17, NO has 15, O₂²⁻ has 18).

Teacher's Note:
1. Count total electrons, not only valence electrons.
2. The 14-electron group is the N₂ family.

Answer: (a) O₂²⁻

Explanation:
1. O₂²⁻ has 18 electrons and all MOs including π* are filled, so it is diamagnetic.
2. O₂⁺ (one unpaired), O₂ (two unpaired) and NO (one unpaired) are paramagnetic.

Teacher's Note:
1. Peroxide ion has no unpaired electron.
2. Add two electrons to O₂ to fill the π* orbitals.

Answer: (b) NO → NO⁺

Explanation:
1. NO → NO⁺: bond order 2.5 → 3 and paramagnetic → diamagnetic; the electron is removed from the antibonding π* orbital.
2. C₂ → C₂⁺: BO decreases (2 → 1.5). O₂ → O₂⁺: BO increases but both are paramagnetic. N₂ → N₂⁺: BO decreases.

Teacher's Note:
1. Removing an antibonding electron increases bond order.
2. Check the magnetic character of both reactant and product.

Answer: (b) F–H···F

Explanation:
1. Hydrogen bond strength increases with the electronegativity of the atom bonded to hydrogen and of the acceptor atom.
2. F is the most electronegative element, and in HF the H atom carries the largest partial positive charge, so F–H···F is the strongest.

Teacher's Note:
1. Strength order: F–H···F > O–H···O > N–H···N.
2. Both donor and acceptor matter.

Answer: (a) SF₄

Explanation:
1. SiF₄ and BF₄⁻ are regular tetrahedral, and XeF₄ is square planar, so all their bonds are equal.
2. SF₄ is a see-saw (distorted trigonal bipyramidal) molecule with two axial and two equatorial S–F bonds of different lengths.

Teacher's Note:
1. Lone pairs occupy equatorial positions in trigonal bipyramidal geometry.
2. Axial and equatorial bonds differ.

Answer: (a) O₂²⁻

Explanation:
1. O₂²⁻ has 18 electrons and all MOs are completely filled.
2. B₂ has two unpaired electrons, N₂⁺ has one, and O₂ has two.

Teacher's Note:
1. Fill the MOs and check for singly occupied orbitals.
2. The peroxide ion is diamagnetic.

Answer: (a) charge on the ion and size of the ion

Explanation:
1. Lattice energy ∝ q₁q₂/(r₊ + r₋).
2. It increases with higher ionic charges and smaller ionic sizes.

Teacher's Note:
1. Both factors must be considered together.
2. Higher lattice energy generally means higher melting point.

Answer: (a) different with 1, 0 and 2 lone pairs of electrons on the central atom, respectively

Explanation:
1. SF₄ is see-saw (sp³d, one lone pair), CF₄ is tetrahedral (sp³, no lone pair) and XeF₄ is square planar (sp³d², two lone pairs).
2. The shapes are different with 1, 0 and 2 lone pairs respectively.

Teacher's Note:
1. Count lone pairs and bond pairs to get the shape.
2. Same formula type does not mean the same shape.

Answer: (c) H₂

Explanation:
1. H₂ has two electrons in σ1s, both paired, so it is diamagnetic.
2. H₂⁻ (3 e⁻), H₂⁺ (1 e⁻) and He₂⁺ (3 e⁻) each have an unpaired electron.

Teacher's Note:
1. Count the electrons and place them in the MO diagram.
2. An odd number of electrons always means paramagnetism.

Answer: (c) BF₄⁻

Explanation:
1. A regular tetrahedron needs sp³ hybridisation with no lone pair on the central atom.
2. XeF₄ is square planar, SF₄ is see-saw and [Ni(CN)₄]²⁻ is square planar; BF₄⁻ is regular tetrahedral.

Teacher's Note:
1. Lone pairs distort the ideal geometry.
2. [Ni(CN)₄]²⁻ is dsp² (square planar) because CN⁻ is a strong-field ligand.

Answer: (c) H₂S < NH₃ < SiH₄ < BF₃

Explanation:
1. H₂S has two lone pairs and a bond angle of about 92°; NH₃ has one lone pair (107°).
2. SiH₄ is regular tetrahedral (109°28′) and BF₃ is trigonal planar (120°).
3. The order is H₂S < NH₃ < SiH₄ < BF₃.

Teacher's Note:
1. More lone pairs mean a smaller angle.
2. sp² gives 120°, sp³ gives about 109°.

Answer: (b) sp² and sp³

Explanation:
1. Boron forms three σ bonds with no lone pair, so it is sp² (trigonal planar).
2. Each oxygen has two σ bonds and two lone pairs, so it is sp³.

Teacher's Note:
1. Boric acid is a planar molecule linked by hydrogen bonds.
2. Steric number 3 gives sp² and steric number 4 gives sp³.

Answer: (b) Bond length in NO is greater than in NO⁺

Explanation:
1. Bond length is inversely proportional to bond order.
2. NO⁺ has the higher bond order (3), so it has the shorter bond; NO has the longer bond.

Teacher's Note:
1. Higher bond order means shorter and stronger bond.
2. NO⁺ is isoelectronic with N₂.

Answer: (c) SH₂

Explanation:
1. Bond angles: SO₂ 119.5°, H₂O 104.5°, NH₃ 107° and H₂S about 92°.
2. The central atom S is larger and less electronegative than O, so bond pairs are far from it and the angle shrinks to about 90°.

Teacher's Note:
1. Hydrides of group 16 and 15 show decreasing angles down the group.
2. H₂S has two lone pairs, like H₂O, but a much smaller angle.

Answer: (c) NO₂ and O₃

Explanation:
1. NO₂ and O₃ are both angular (bent), so their bond dipoles do not cancel and they have permanent dipoles.
2. SiF₄ (tetrahedral) and CO₂ (linear) are symmetrical and their dipoles cancel.

Teacher's Note:
1. Symmetric molecules have zero net dipole moment.
2. Bent molecules generally have a non-zero dipole moment.

Answer: (c) A bond is formed when attractive forces overcome repulsive forces

Explanation:
1. A covalent bond involves sharing, not transfer, of electrons, so (a) is wrong.
2. Pure water contains H₃O⁺ and OH⁻ ions from autoionisation, so (b) is wrong, and HF is more polar than HBr, so (d) is wrong.
3. A bond forms when the net force of attraction between nuclei and electrons exceeds the repulsion, which is (c).

Teacher's Note:
1. Bond formation lowers the potential energy of the system.
2. Polarity increases with the electronegativity difference.

Answer: (a) 3, 2, 1

Explanation:
1. Lone pairs = (8 − number of bonded F atoms)/2.
2. XeF₂: (8 − 2)/2 = 3; XeF₄: (8 − 4)/2 = 2; XeF₆: (8 − 6)/2 = 1.

Teacher's Note:
1. Xenon has 8 valence electrons.
2. Each F atom uses one electron of Xe.

Answer: (a) Al in AlH₃ changes to AlH₄⁻

Explanation:
1. AlH₃ is sp² (three bonds, planar) and AlH₄⁻ is sp³ (tetrahedral), so the hybridisation changes.
2. O in H₂O and in H₃O⁺ is sp³ and N in NH₃ and NH₄⁺ is sp³; only the shape changes there.

Teacher's Note:
1. Count the steric number before and after.
2. A change of shape does not necessarily mean a change of hybridisation.

Answer: (d) O₂²⁻ < O₂⁻ < O₂ < O₂⁺

Explanation:
1. Bond orders: O₂²⁻ = 1, O₂⁻ = 1.5, O₂ = 2, O₂⁺ = 2.5.
2. Bond strength increases with bond order: O₂²⁻ < O₂⁻ < O₂ < O₂⁺.

Teacher's Note:
1. Higher bond order means greater bond strength.
2. Adding electrons to π* lowers the bond order.