Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure worksheet with answer key

Answer: (c) AlCl₃

Al has only three valence electrons; after three shared pairs with Cl it holds six electrons, an incomplete octet. P in PCl₃, S in H₂S and Si in SiF₄ each reach an octet.

Answer: (a) H₂SO₄

In H₂SO₄ sulphur, a third-period element, has an expanded octet of 12 electrons. S in SCl₂, N in NF₃ and C in CCl₄ have eight.

Answer: (d) 32

S gives 6 and each of the four O atoms gives 6: 6 + 24 = 30; the two negative charges add two electrons, so 32.

Answer: (b) 391 kJ mol⁻¹

The three N–H bonds break with different enthalpies because the chemical environment changes after each step, so the mean is used: (449 + 385 + 339)/3 = 1173/3 = 391 kJ mol⁻¹.

Answer: (c) 6 σ and 3 π

Every bond between two atoms has one σ bond: three C–H, C=C, C–C and C≡N give 6 σ. The double bond adds one π and the triple bond two π: 3 π.

K(g) → K⁺(g) needs +419 and Br(g) → Br⁻(g) releases 325: net 419 − 325 = +94 kJ mol⁻¹, so forming the gaseous ions costs energy. Lattice formation releases 671 kJ mol⁻¹, which more than compensates for the 94 kJ mol⁻¹, so the solid is stabilised by its lattice enthalpy.

• net +94 kJ mol⁻¹ to form the gaseous ions (1)
• lattice enthalpy released (671) more than compensates, so KBr(s) is stable (1)

(i) No: atomic orbitals combine only if their energies are the same or nearly the same, and the energy of 2s is appreciably higher than that of 1s. (ii) No: although their energies are equal, 2pz and 2px do not have the same symmetry about the molecular (z) axis.

• (i) no, energies too different (1)
• (ii) no, different symmetry about the z axis (1)

(i) ICl₂⁻: V = 7, M = 2, A = 1; H = ½(10) = 5, sp³d; lone pairs = 5 − 2 = 3. (ii) SiF₆²⁻: V = 4, M = 6, A = 2; H = ½(12) = 6, sp³d²; lone pairs = 6 − 6 = 0.

• ICl₂⁻: H = 5, sp³d, 3 lone pairs (1)
• SiF₆²⁻: H = 6, sp³d², no lone pair (1)

C₂ (12 e): KK σ2s² σ*2s² (π2px² = π2py²); counting σ1s and σ*1s, N_b = 8, N_a = 4, b.o. = ½(8 − 4) = 2; all electrons paired, diamagnetic. C₂²⁻ (14 e, like N₂): the two extra electrons fill σ2pz, which lies above π2p in this order: N_b = 10, N_a = 4, b.o. = ½(10 − 4) = 3; all paired, diamagnetic. In C₂²⁻ the added electrons enter a bonding orbital (σ2pz), so N_b rises and the bond order rises; in O₂²⁻ they enter the antibonding π*2p orbitals, so N_a rises and the bond order falls.

• bond orders 2 (C₂) and 3 (C₂²⁻), worked (1)
• both diamagnetic (1)
• C₂²⁻: electrons enter bonding σ2pz; O₂²⁻: antibonding π* (1)

Carbon is sp hybridised; its two sp hybrids point in opposite directions, giving the linear shape. One sp hybrid overlaps axially with the 1s orbital of H (σ C–H); the other overlaps axially with an orbital of N (σ C–N). The two unhybridised 2p orbitals of carbon, at right angles to each other and to the axis, overlap sidewise with two p orbitals of N to form two π bonds, so C≡N is one σ and two π.

• carbon sp, linear (1)
• σ C–H from sp–1s and σ C–N by axial overlap (1)
• two π from sidewise overlap of unhybridised p orbitals (1)

Only orbitals of the valence shell are hybridised; the orbitals mixed must have almost equal energy; the number of hybrid orbitals equals the number of atomic orbitals mixed, and the hybrids are equivalent in energy and shape (promotion of an electron is not essential). Correction: filled orbitals of the valence shell can also take part, as the lone-pair orbitals of N in NH₃ do.

• any two correct conditions or features (1)
• a third correct condition or feature (1)
• correction: filled valence orbitals can also hybridise (1)