Chemical Bonding and Molecular Structure — Practice

Question 1

The hybridisation of the carbon atom in methane (CH₄) is:
(a) sp³ (b) sp² (c) sp (d) sp³d

Accepted Answer

(a) sp³.

Carbon has 4 σ bonds and no lone pairs (steric number 4) → sp³ hybridised, tetrahedral geometry.

Question 2

The shape of the water molecule (H₂O) is:
(a) bent (b) linear (c) trigonal planar (d) tetrahedral

Accepted Answer

(a) bent.

Oxygen has 2 bond pairs + 2 lone pairs = 4 pairs (tetrahedral electron geometry); the two lone pairs bend it to a V-shape with a 104.5° angle.

Question 3

The correct order of bond angles is:
(a) CH₄ > NH₃ > H₂O (b) H₂O > NH₃ > CH₄ (c) NH₃ > CH₄ > H₂O (d) CH₄ > H₂O > NH₃

Accepted Answer

(a) CH₄ > NH₃ > H₂O.

Angles are 109.5° > 107° > 104.5°. Each added lone pair (0, 1, 2) increases repulsion and squeezes the bond angle.

Question 4

The bond order of the nitrogen molecule N₂ is:
(a) 3 (b) 2 (c) 1 (d) 2.5

Accepted Answer

(a) 3.

N₂ has 14 electrons: 10 bonding and 4 antibonding, so bond order = (10 − 4)/2 = 3, a triple bond — which is why N₂ is so stable.

Question 5

Which molecule is paramagnetic?
(a) O₂ (b) N₂ (c) F₂ (d) H₂

Accepted Answer

(a) O₂.

O₂ has two unpaired electrons in its degenerate π* orbitals (by Hund's rule), making it paramagnetic. N₂, F₂ and H₂ have all electrons paired (diamagnetic).

Question 6

Which molecule has a net dipole moment of zero?
(a) CO₂ (b) H₂O (c) NH₃ (d) SO₂

Accepted Answer

(a) CO₂.

CO₂ is linear, so its two equal C=O bond dipoles point opposite and cancel (μ = 0). H₂O, NH₃ and SO₂ are bent/pyramidal and polar.

Question 7

The hybridisation of carbon in ethyne (C₂H₂) is:
(a) sp (b) sp² (c) sp³ (d) sp³d

Accepted Answer

(a) sp.

Each carbon has 2 σ bonds (to H and to C) and no lone pairs → steric number 2 → sp hybridised, linear. The triple bond is 1σ + 2π.

Question 8

The hybridisation of sulphur in SF₆ is:
(a) sp³d² (b) sp³d (c) sp³ (d) sp²

Accepted Answer

(a) sp³d².

Six σ bonds and no lone pairs → steric number 6 → sp³d² hybridisation, octahedral geometry, using two of sulphur's vacant 3d orbitals.

Question 9

According to Fajans' rules, which compound has the greatest covalent character?
(a) AlCl₃ (b) NaCl (c) KCl (d) CsCl

Accepted Answer

(a) AlCl₃.

Al³⁺ is small and highly charged, so it strongly polarises the chloride ion, giving AlCl₃ the most covalent character. The alkali chlorides (large, singly charged cations) are the most ionic.

Question 10

Which ionic solid has the highest lattice enthalpy?
(a) MgO (b) NaCl (c) KCl (d) NaF

Accepted Answer

(a) MgO.

Lattice enthalpy ∝ q₊q₋/(r₊+r₋). MgO has doubly charged, small ions (2+/2−), giving it a far higher lattice enthalpy and melting point than the singly charged halides.

Question 11

Which species has the highest bond order?
(a) O₂⁺ (b) O₂ (c) O₂⁻ (d) O₂²⁻

Accepted Answer

(a) O₂⁺.

Removing an electron from an antibonding π* orbital raises the bond order to 2.5. O₂ is 2, O₂⁻ is 1.5 and O₂²⁻ is 1, so O₂⁺ has the shortest, strongest bond.

Question 12

NH₃ has a larger dipole moment than NF₃ because:
(a) in NH₃ the bond and lone-pair dipoles add, while in NF₃ they oppose (b) NH₃ is planar (c) fluorine is less electronegative than hydrogen (d) NF₃ has no lone pair

Accepted Answer

(a) in NH₃ the bond and lone-pair dipoles add, while in NF₃ they oppose.

Both are pyramidal, but the direction of the N–H bond dipoles reinforces the lone-pair dipole in NH₃ and cancels part of it in NF₃, so NH₃ has the much larger net dipole moment.

Question 13

Hydrogen bonding is NOT possible in:
(a) CH₄ (b) H₂O (c) NH₃ (d) HF

Accepted Answer

(a) CH₄.

Hydrogen bonding requires H bonded to N, O or F. In CH₄ the hydrogen is bonded to carbon (low electronegativity), so no hydrogen bonding occurs.

Question 14

All three carbon–oxygen bonds in the carbonate ion (CO₃²⁻) are equal in length because of:
(a) resonance (delocalisation) (b) hydrogen bonding (c) sp hybridisation (d) an expanded octet

Accepted Answer

(a) resonance (delocalisation).

The true structure is a resonance hybrid in which the double-bond character is spread equally over all three C–O bonds, so they are identical in length — intermediate between a single and double bond.

Question 15

The shape of XeF₄ is:
(a) square planar (b) tetrahedral (c) octahedral (d) see-saw

Accepted Answer

(a) square planar.

Xe has 4 bond pairs + 2 lone pairs = 6 electron pairs (octahedral geometry); the two lone pairs occupy opposite axial positions, leaving the four F atoms in a square planar arrangement.

Practice — Chemical Bonding and Molecular Structure

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