CBSE Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs Set 08

Answer: (c) BF₃

Explanation:
1. Boron has three valence electrons and forms three B–F bonds, so it has only 6 electrons around it in BF₃.
2. In H₂O, NH₃ and CH₄ the central atoms (O, N and C) each have 8 electrons.

Teacher's Note:
1. Count the electrons around the central atom after drawing the Lewis structure.
2. BF₃, BeCl₂ and AlCl₃ are the standard incomplete-octet examples.

Answer: (a) Ca²⁺ and O²⁻

Explanation:
1. Lattice enthalpy increases with the charges on the ions and decreases with their size.
2. Ca²⁺ and O²⁻ carry the largest charges (2+ and 2−) and are small ions, so the attraction between them is the greatest.

Teacher's Note:
1. Compare the product of the charges first.
2. Then compare ionic sizes if the charges are equal.

Answer: (d) KF > KCl > KBr > KI

Explanation:
1. All the compounds have the same cation K⁺ and singly charged anions.
2. The anion size increases in the order F⁻ < Cl⁻ < Br⁻ < I⁻, so the ions come closest in KF.
3. A smaller internuclear distance means a larger lattice enthalpy, so KF > KCl > KBr > KI.

Teacher's Note:
1. When charges are equal, lattice enthalpy depends on size.
2. Smaller ions pack closer and hold more strongly.

Answer: (b) A has a low ionisation enthalpy and B has a large negative electron gain enthalpy

Explanation:
1. A⁺ forms easily only if A loses an electron easily, which needs a low ionisation enthalpy.
2. B⁻ forms easily only if B gains an electron with the release of much energy, which needs a large negative electron gain enthalpy.

Teacher's Note:
1. Think of the cation and the anion separately.
2. The third factor is a large lattice enthalpy.

Answer: (d) +1

Explanation:
1. Formal charge = V − L − S/2. For the central O: V = 6, L = 2 (one lone pair), S = 6 (three shared pairs).
2. Formal charge = 6 − 2 − 3 = +1.
3. Check: the end O with the double bond is 0 and the end O with the single bond is −1; the sum is 0.

Teacher's Note:
1. Use the formula for each atom separately.
2. The sum of the formal charges equals the charge on the molecule.

Answer: (b) +1

Explanation:
1. For N: V = 5, L = 0 (no lone pair), S = 8 (four shared pairs).
2. Formal charge = 5 − 0 − 8/2 = +1.
3. The two single-bonded O atoms carry −1 each, so the total is +1 − 1 − 1 = −1, the charge on NO₃⁻.

Teacher's Note:
1. Nitrogen in NO₃⁻ has four bonds and no lone pair.
2. Check the total charge at the end.

Answer: (a) 6 and 1

Explanation:
1. C–H bonds: 3 in CH₃ and 1 in CHO, making 4 σ bonds.
2. The C–C bond is 1 σ bond, and the C=O double bond has 1 σ and 1 π bond.
3. So σ = 4 + 1 + 1 = 6 and π = 1.

Teacher's Note:
1. Count each single bond as σ.
2. Each double bond adds one π bond only.

Answer: (c) 8 σ and 1 π

Explanation:
1. C–H bonds: 3 + 1 + 2 = 6.
2. C–C bonds: one single C–C σ bond and one σ bond in C=C, making 2.
3. So σ = 6 + 2 = 8, and the C=C contributes 1 π bond.

Teacher's Note:
1. Count C–H bonds carbon by carbon.
2. Each double bond adds only one π bond.

Answer: (d) 4/3

Explanation:
1. Each of the three resonance structures has one N=O double bond and two N–O single bonds, so 4 bonds in all.
2. These 4 bonds are shared among 3 equivalent N–O positions, so the bond order is 4/3.

Teacher's Note:
1. Total number of bonds ÷ number of equivalent positions gives the resonance bond order.
2. The three N–O bonds are of equal length.

Answer: (c) HI

Explanation:
1. Bond length increases with the size of the halogen atom.
2. The sizes follow F < Cl < Br < I, and the bond lengths are 91, 127, 141 and 160 pm respectively.
3. So HI has the longest H–X bond.

Teacher's Note:
1. Compare the sizes of the halogens down group 17.
2. The longer the bond, the lower the bond enthalpy.

Answer: (a) N₂

Explanation:
1. The bond enthalpies are H₂ 435.8, O₂ 498, N₂ 941.4 and F₂ 158.8 kJ mol⁻¹.
2. N₂ has a triple bond (bond order 3), while the others have single or double bonds.

Teacher's Note:
1. Higher bond order gives a stronger bond.
2. F₂ is weak because of lone-pair repulsion on the small F atoms.

Answer: (b) BF₃

Explanation:
1. BF₃ is trigonal planar and symmetrical, so the three B–F bond dipoles cancel each other.
2. HCl, H₂O and NH₃ are not symmetrical, so their bond dipoles do not cancel.

Teacher's Note:
1. Zero dipole moment needs both polar bonds and a symmetrical shape.
2. CO₂, BeF₂, CCl₄ and CH₄ are other examples.

Answer: (d) the N–H bond dipoles and the lone-pair dipole point the same way, while in NF₃ the N–F bond dipoles oppose the lone-pair dipole

Explanation:
1. Both NH₃ and NF₃ are pyramidal with one lone pair on N.
2. In NH₃ the nitrogen is the negative end of each N–H bond, so the bond dipoles add to the lone-pair dipole.
3. In NF₃ fluorine is the negative end, so the bond dipoles point away from N and partly cancel the lone-pair dipole.

Teacher's Note:
1. Draw the direction of each bond dipole and of the lone pair.
2. F is more electronegative than N, but H is less.

Answer: (b) O₃

Explanation:
1. Ozone has two equivalent Lewis structures, O=O⁺–O⁻ and ⁻O–O⁺=O, differing only in the position of the double bond.
2. NH₃, CH₄ and H₂O have only one Lewis structure each.

Teacher's Note:
1. Resonance needs more than one valid Lewis structure with the nuclei in the same positions.
2. Other examples are CO₃²⁻, NO₃⁻ and SO₃.

Answer: (a) CH₄

Explanation:
1. In NH₄⁺ the lone pair of N goes to H⁺, in H₃O⁺ the lone pair of O goes to H⁺, and in BF₄⁻ F⁻ donates a pair to BF₃.
2. CH₄ has four ordinary covalent bonds, each formed by one electron from C and one from H.

Teacher's Note:
1. A coordinate bond needs a lone pair donor and an empty orbital acceptor.
2. Count how the last bond is formed.

Answer: (c) KCN

Explanation:
1. KCN is made of K⁺ and CN⁻ ions held by an ionic bond.
2. Inside the CN⁻ ion, carbon and nitrogen are joined by a covalent triple bond.

Teacher's Note:
1. Look for a polyatomic ion combined with a metal ion.
2. NaOH, NH₄Cl and K₂SO₄ are similar.

Answer: (d) equally shared between them

Explanation:
1. Two identical atoms have the same electronegativity.
2. Neither can pull the shared pair, so it is shared equally and the bond is non-polar covalent, as in Cl₂ or O₂.

Teacher's Note:
1. Identical atoms have zero electronegativity difference.
2. Spins of paired electrons are opposite, not identical.

Answer: (a) SF₆

Explanation:
1. In SF₆ sulphur forms six bonds and has 12 electrons around it, using its vacant 3d orbitals.
2. N, C and C in NH₃, CO₂ and CCl₄ have 8 electrons only, as they have no d orbitals in the second period.

Teacher's Note:
1. Elements of the third period and beyond can expand the octet.
2. PF₅, SF₆ and IF₇ are common examples.

Answer: (b) NO

Explanation:
1. Nitric oxide has 5 + 6 = 11 valence electrons, an odd number.
2. All atoms cannot get an octet with an odd number of electrons.
3. CO₂, N₂ and H₂O all have an even number of electrons.

Teacher's Note:
1. Count the total valence electrons first.
2. NO₂ (17 electrons) is another example.

Answer: (c) 4.8 D

Explanation:
1. μ = Q × r = 1.602 × 10⁻¹⁹ C × 100 × 10⁻¹² m = 1.602 × 10⁻²⁹ C m.
2. In debye: 1.602 × 10⁻²⁹ ÷ 3.336 × 10⁻³⁰ = 4.80 D.

Teacher's Note:
1. Convert pm to m before multiplying.
2. Divide by 3.336 × 10⁻³⁰ to change C m into D.