p-Block

Question 1

(a) Diborane (6 marks)

• Valence electrons: each B contributes 3, each H contributes 1: $2(3) + 6(1) = 12$ electrons. (1)
• A classical ethane-like structure ($H_3B–BH_3$) needs 7 bonds (6 B–H + 1 B–B) = 14 electrons in 7 σ-bonds, but only 12 are available → 2 short. Hence classical structure impossible. (2)
• Actual structure: 4 terminal B–H bonds (2c–2e) = 8 electrons; 2 bridging B–H–B 3c–2e bonds = 4 electrons. Total $8+4=12$ ✓. (2)
• Each 3c–2e bridge bond holds 2 electrons shared over 3 atoms. (1)

(b) Borazine (4 marks)

• Benzene $C_6H_6$: $6(4)+6(1)=30$ valence electrons. (1)
• Borazine $B_3N_3H_6$: $3(3)+3(5)+6(1)=9+15+6=30$ valence electrons → isoelectronic ✓. (2)
• Borazine is more reactive to addition because the B–N π system is polarized (N more electronegative → uneven charge distribution, δ⁻ on N, δ⁺ on B); the ring is not as aromatically stabilized/uniform, so it undergoes addition (e.g. with HCl) more readily. (1)

(c) Xe compounds VSEPR (8 marks) — use $\text{lone pairs} = \tfrac{1}{2}(V - B)$ where $V$ = Xe valence electrons (8), $B$ = number of bonded atoms (each single bond uses 1 Xe electron; for XeO₃ treat each O as taking part in bonding using 2 Xe electrons as double bonds — accept either standard treatment). Marks: 2 per species.

(Lone pairs on Xe: XeF₂ → $\tfrac12(8-2)=3$; XeF₄ → $\tfrac12(8-4)=2$; XeF₆ → $\tfrac12(8-6)=1$; XeO₃ → 3 σ + 1 lp with π using remaining pairs.) Award full marks for correct shape + polarity.

(d) BF₃ vs BCl₃ Lewis acidity (6 marks)

• Both are electron-deficient (B has empty 2p). (1)
• In $BF_3$, F 2p orbitals are similar in size to B 2p → effective $p\pi$–$p\pi$ back-donation into the empty B orbital, partially satisfying B's electron deficiency. (2)
• In $BCl_3$, Cl 3p is larger/more diffuse → poorer overlap with B 2p, weaker back-bonding, so B is more electron-deficient and more eager to accept a lone pair. (2)
• Therefore $BCl_3 > BF_3$ in Lewis acidity, despite F's higher electronegativity. (1)

Question 2

(a) Ostwald (8 marks)

• $4NH_3 + 5O_2 \to 4NO + 6H_2O$ (1)
• $2NO + O_2 \to 2NO_2$ (1)
• $3NO_2 + H_2O \to 2HNO_3 + NO$ (1) (accept $4NO_2+O_2+2H_2O\to4HNO_3$)
• Overall (100% conversion): $NH_3 \to HNO_3$, mole ratio 1:1. (1)
• Moles $NH_3 = 1000/17.0 = 58.82$ mol → moles $HNO_3 = 58.82$. (2)
• Mass $HNO_3 = 58.82 \times 63.0 = 3705.9$ g $\approx 3.71$ kg. (2)

(b) Contact process (6 marks)

• (i) Raising T: shifts equilibrium backward (exothermic forward) → decreases SO₃ yield. (1)
• (ii) Raising P: 3 mol gas → 2 mol gas, so higher P shifts forward → increases yield. (1)
• (iii) Removing SO₃: shifts forward → increases conversion. (1)
• Compromise at ~450 °C: low T favours yield but rate too slow; high T fast but poor yield. 450 °C (with V₂O₅ catalyst) gives acceptable yield at practical rate. (3)

(c) Kinetics + coding (6 marks)

• Integrated first-order law: $\ln\frac{[A]_0}{[A]} = kt$, or $[A]=[A]_0 e^{-kt}$. (1)
• $12.5\% = 1/8 = (1/2)^3$ → 3 half-lives. (1)
• $t_{1/2} = \ln2/k = 0.693/3.0\times10^{-3} = 231.0$ s → $t = 3\times231.0 = 693.1$ s. (2) (Equivalently $t=\ln(8)/k = 2.079/0.003 = 693.1$ s.)
• Euler pseudocode: (2)

A = A0; t = 0; dt = 0.1
while t < t_end:
    A = A - k*A*dt      # update: A_{n+1}=A_n - k A_n dt
    t = t + dt

Question 3

(a) Oxoacid acidity (5 marks)

• Oxidation states of Cl: HClO +1, HClO₂ +3, HClO₃ +5, HClO₄ +7. (2)
• More O atoms → greater delocalisation/stabilisation of the negative charge on the conjugate base (ClO₄⁻ most stabilised) + higher oxidation state increases inductive electron withdrawal from O–H → easier proton loss → stronger acid. (3)

(b) H₃PO₃ vs H₃PO₄ (4 marks)

• $H_3PO_3$ (phosphorous acid): structure has one P–H bond, two P–OH, one P=O → only 2 ionisable H (the P–H is not acidic) → basicity 2. (2)
• $H_3PO_4$: three P–OH and one P=O → all 3 H ionisable → basicity 3. (2)

(c) B–Si diagonal (4 marks) — any 3 with justification:

• Both form acidic oxides (B₂O₃, SiO₂).
• Both form covalent, hydrolysable halides (BF₃/BCl₃, SiCl₄).
• Both form covalent hydrides that are electron-deficient/volatile.
• Both have similar charge/radius ratio and electronegativity (~2.0) → similar polarising power. (1 each, max 4, +justification)

(d) Ozone (3 marks)

• Bond order = 1.5 (average of one single + one double over two positions). (1)
• Two resonance structures interchange the double bond → both O–O bonds equivalent. (1)
• Observed length ~128 pm, intermediate between O–O single (~148) and O=O double (~121). (1)

[
  {"claim":"Diborane has 12 valence electrons","code":"result = (2*3 + 6*1) == 12"},
  {"claim":"Benzene and borazine both have 30 valence electrons (isoelectronic)","code":"benzene = 6*4 + 6*1; borazine = 3*3 + 3*5 + 6*1; result = (benzene == 30) and (borazine == 30) and (benzene == borazine)"},
  {"claim":"1.00 kg NH3 gives ~3.71 kg HNO3","code":"mass = (1000/17.0)*63.0; result = abs(mass - 3705.88) < 1.0"},
  {"claim":"Time for N2O5 to reach 12.5% is ~693 s at k=3e-3","code":"import sympy as sp; k=sp.Rational(3,1000); t = sp.log(8)/k; result = abs(float(t) - 693.1) < 1.0"}
]