Chemical Bonding

Question 1 (12 marks)

(a) The cycle (6 marks — 1 per correctly named+signed term):

Forming $\text{NaCl}(s)$ from elements: $\Delta H_f = \Delta H_{sub}(\text{Na}) + IE(\text{Na}) + \tfrac{1}{2}D(\text{Cl}_2) + EA(\text{Cl}) + U$

Terms:

• $\Delta H_{sub}$: Na(s)→Na(g), + (endothermic)
• $IE$: Na(g)→Na⁺(g)+e⁻, +
• $\tfrac12 D$: ½Cl₂(g)→Cl(g), +
• $EA$: Cl(g)+e⁻→Cl⁻(g), −
• $U$: Na⁺(g)+Cl⁻(g)→NaCl(s), − (lattice formation)
• $\Delta H_f$: net, −

Why: Hess's law — the enthalpy of formation equals the sum of all step enthalpies since enthalpy is a state function.

(b) Solve for $U$ (4 marks): $U = \Delta H_f - \left[\Delta H_{sub} + IE + \tfrac12 D + EA\right]$ $U = -411 - [108 + 496 + 121 + (-349)]$ Note $\tfrac12 D = \tfrac12(242) = 121$. $U = -411 - [376] = -787 \text{ kJ/mol}$ (2 marks setup, 1 for the ½ bond term, 1 for correct answer $-787$.)

(c) (2 marks): $U$ is negative because forming a lattice from gaseous ions releases energy (electrostatic attraction stabilizes). If "lattice energy" is defined as the breaking process (lattice → gaseous ions), it is reported as $+787$ kJ/mol.

Question 2 (14 marks)

(a) (2): LCAO: atomic orbitals of comparable energy and correct symmetry combine to form the same number of molecular orbitals (bonding + antibonding). Bond order $= \tfrac12(N_b - N_a)$.

(b) (6): O₂/F₂ ordering (no dominant mixing): $\sigma_{1s},\sigma^*_{1s},\sigma_{2s},\sigma^*_{2s},\sigma_{2p_z},(\pi_{2p_x}=\pi_{2p_y}),(\pi^*_{2p_x}=\pi^*_{2p_y}),\sigma^*_{2p_z}$ O₂ has 16 electrons. Filling: valence config (2s and 2p, 12 valence e⁻): \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4 \pi^*_{2p}^2

• $N_b = 8$, $N_a = 4$ ⇒ Bond order $= \tfrac12(8-4)=2$.
• The two $\pi^*$ electrons occupy degenerate orbitals singly (Hund) ⇒ 2 unpaired electrons, paramagnetic.

(2 for correct ordering, 2 for filling, 1 bond order, 1 magnetism.)

(c) (3): MOT places the last two electrons in two degenerate $\pi^*$ antibonding orbitals; by Hund's rule they occupy separately with parallel spins, giving 2 unpaired electrons → paramagnetic. Lewis structure (O=O with all lone pairs paired) wrongly predicts all electrons paired (diamagnetic); MOT succeeds because it accounts for the degenerate antibonding level.

(d) (3): NO: 15 valence electrons → config gives $N_b=10, N_a=5$, bond order $=2.5$. Removing the single $\pi^*$ electron to form NO⁺ raises bond order to 3, so the bond gets shorter and stronger (removed an antibonding electron).

Question 3 (10 marks)

(a) (6): SCN⁻, 16 valence electrons. Formal charge $= V - L - B/2$.

Working the three canonical forms:

• (I) S=C=N (each end double bond): S: $6-4-2=0$; C: $4-0-4=0$; N: $6-4-2 = 0$? recompute N with double bond + 2 lp: $5-4-2=-1$. → S 0, C 0, N −1.
• (II) S–C≡N (S single, 3 lp; N triple, 1 lp): S: $6-6-1=-1$; C: $4-0-4=0$; N: $5-2-3=0$. → S −1, C 0, N 0.
• (III) S≡C–N (S triple 1 lp; N single 3 lp): S: $6-2-3=+1$; C: $4-0-4=0$; N: $5-6-1=-2$. → S +1, C 0, N −2.

(1 mark per correct FC set × structures; 2 for structures drawn.)

(b) (2): Structure (II) S–C≡N dominates: negative formal charge sits on the more electronegative N (0 on N, −1 on less electronegative S is the alternative)... best is minimal charge magnitude and negative charge on most electronegative atom. Structure II places −1 on S; Structure I places −1 on N (most electronegative). Structure I (S=C=N with −1 on N) is favored by electronegativity, II by low separation. Accept: dominant = the one with −1 on N and small FCs (Structure I), since negative charge on most electronegative atom is preferred. (Full marks for correct FC-based reasoning either way.)

(c) (2): Resonance energy = the extra stabilization (lower energy) of the real delocalized structure relative to the most stable single contributing Lewis structure. It increases stability.

Question 4 (12 marks)

(a) (6) & (b) (3):

(c) (3):

• XeF₄: non-polar — the 4 Xe–F dipoles in a square-planar arrangement cancel (lone pairs axial, symmetric).
• SF₄: polar — seesaw shape is asymmetric; the lone pair in the equatorial plane leaves an uncancelled net dipole.
• PCl₅: non-polar — trigonal bipyramidal symmetry; axial and equatorial P–Cl dipoles vector-sum to zero.

Question 5 (6 marks)

(a) (3): Covalent character increases with: (1) high charge on cation and/or anion; (2) small cation size; (3) large anion size — plus cations with pseudo-noble-gas ($d^{10}$) configuration polarize more (bonus).

(b) (3):

• AlCl₃ > AlF₃ in covalent character: Cl⁻ is larger/more polarizable than F⁻ → more covalent.
• LiCl > KCl in covalent character: Li⁺ is smaller → higher polarizing power → more covalent.

Question 6 (6 marks)

(a) (3): Boiling points: $\text{HF} > \text{HI} > \text{HBr} > \text{HCl}$. Down the group (HCl→HI) BP rises with molar mass/London dispersion. HF anomaly: strong intermolecular hydrogen bonding (F highly electronegative, small) raises its BP above all the others despite lowest mass.

(b) (3): In ice each water forms 4 hydrogen bonds in a fixed tetrahedral arrangement, producing an open hexagonal lattice with large empty spaces → lower density. On melting, this lattice partly collapses, molecules pack closer, so liquid water is denser (max density at 4 °C).

[
  {"claim":"Born-Haber: U = -411 - (108+496+121-349) = -787 kJ/mol","code":"dHf=-411; sub=108; IE=496; halfD=242/2; EA=-349; U=dHf-(sub+IE+halfD+EA); result=(U==-787)"},
  {"claim":"O2 bond order = 1/2(Nb-Na) = 1/2(8-4) = 2","code":"Nb=8; Na=4; bo=Rational(1,2)*(Nb-Na); result=(bo==2)"},
  {"claim":"NO bond order = 1/2(10-5) = 2.5","code":"Nb=10; Na=5; bo=Rational(1,2)*(Nb-Na); result=(bo==Rational(5,2))"},
  {"claim":"SCN- structure I formal charges: S=0, C=0, N=-1","code":"def fc(V,L,B): return V-L-B/2\nS=fc(6,4,4); C=fc(4,0,8); N=fc(5,4,4); result=(S==0 and C==0 and N==-1)"}
]