p-Block

Q1 (10 marks)

(a) Each B has 3 valence e⁻, each H has 1 → total = 2(3) + 6(1) = 12 valence electrons = 6 bonds worth of pairs. An ethane-like structure needs a B–B bond + 6 B–H bonds = 7 bonds = 14 electrons. Only 12 are available — electron deficient by 2. So 2 of the 8 B–H "links" must share electrons: two bridging H each held by a 3c-2e bond (B–H–B sharing one electron pair over three atoms). (2 for electron count, 2 for deficiency argument)

(b) 4 terminal H (2 per B, in the plane) and 2 bridging H (above/below the B–B axis). Terminal H–B–H ≈ 120°(≈117–119°); bridging B–H–B ≈ 97°; terminal region ~sp³ B. (3: correct 4 terminal + 2 bridge = 2, angles = 1)

(c) Borazine B₃N₃H₆: 6 ring atoms, 3 B (3e each) + 3 N (5e each) contribute; it is isoelectronic/isostructural with benzene C₆H₆ (planar 6-membered ring, delocalised π-type electrons). Difference: borazine is more reactive / undergoes addition reactions (e.g., adds HCl) because the B–N π delocalisation is unequal (N more electronegative → charge separation, δ⁺ on B, δ⁻ on N), making it less aromatic/more polar than benzene. (isoelectronic reasoning 1.5, reactivity difference + reason 1.5)

Q2 (8 marks)

Increasing Lewis acidity: BF₃ < BCl₃ < BBr₃ < BI₃. (2 for order)

Reasoning: Boron is electron deficient (empty 2p). The halogen lone pair can donate into this empty p-orbital forming pπ–pπ back-bonding, which reduces boron's electron deficiency and lowers Lewis acidity. (2)
Back-bonding is strongest when orbital sizes match: F(2p) matches B(2p) best → strongest back-donation → BF₃ least acidic. Down the group (Cl 3p, Br 4p, I 5p) overlap with B 2p worsens → back-bonding weakens → boron stays more electron deficient → stronger Lewis acid. (3)
This overrides the electronegativity argument (which would wrongly predict BF₃ strongest). (1)

Q3 (10 marks)

(a) H₃PO₃ (phosphorous acid): structure has P with one P–H bond, one P=O, and two P–OH. Only the two O–H protons are ionisable → diprotic. (2.5)
H₃PO₄: P bonded to one P=O and three P–OH (no P–H) → all three O–H ionisable → triprotic. (1.5)
Rule: basicity = number of O–H (hydroxyl) groups; H bonded directly to P is not acidic (P–H does not ionise). (1)

(b) Acidity order: HOCl < HClO₂ < HClO₃ < HClO₄ (increasing). (2)
Reasoning from scratch: as more terminal (non-OH) O atoms are added, Cl oxidation state rises (+1, +3, +5, +7). More electronegative terminal O atoms withdraw electron density from the O–H bond (inductive) and, crucially, delocalise/stabilise the negative charge of the conjugate base (ClO₄⁻ spreads charge over 4 equivalent O). Greater conjugate-base stability → stronger acid. (3)

Q4 (8 marks)

Ostwald process: (3 equations × 1.5 = 4.5)

1. $4NH_3 + 5O_2 \xrightarrow{Pt/Rh,\ \sim 500\,^\circ C} 4NO + 6H_2O$
2. $2NO + O_2 \rightarrow 2NO_2$
3. $3NO_2 + H_2O \rightarrow 2HNO_3 + NO$

Catalyst role: Pt/Rh gauze catalyses selective oxidation of NH₃ to NO (fast, avoids over-oxidation to N₂); the reaction is exothermic. (1)

Mass calculation (2.5):
Overall: 1 mol NH₃ → 1 mol HNO₃ (N conserved).
68 kg NH₃ = 68/17 = 4 kmol NH₃ → 4 kmol HNO₃.
Mass HNO₃ = 4 × 63 = 252 kg.

Q5 (8 marks)

(a) (2 each = 6)

• XeF₂: Xe has 8 valence e⁻; 2 bond pairs + 3 lone pairs = 5 pairs → sp³d → trigonal bipyramidal e⁻ geometry; lone pairs equatorial → linear shape.
• XeF₄: 4 bond pairs + 2 lone pairs = 6 pairs → sp³d² → octahedral e⁻ geometry; lone pairs trans (axial) → square planar.
• XeF₆: 6 bond pairs + 1 lone pair = 7 pairs → sp³d³ → distorted octahedral (capped octahedron).

(b) XeF₆ has a stereochemically active lone pair (7th pair) which occupies space and distorts the geometry away from a regular octahedron (fluxional/distorted structure). (2)

Q6 (6 marks)

(a) $2SO_2 + O_2 \xrightarrow{V_2O_5,\ \sim 450\,^\circ C} 2SO_3$. (1.5)
SO₃ + water directly is highly exothermic and forms a dense acid mist/fog that is hard to condense; instead SO₃ is absorbed in conc. H₂SO₄ giving oleum (H₂S₂O₇), then diluted → efficient, no mist. (1.5)

(b) Formation: $O_2 \xrightarrow{UV} 2O$; $O + O_2 \rightarrow O_3$. Destruction: $O_3 \xrightarrow{UV} O_2 + O$ (natural balance). (1.5)
Cl acts catalytically: $Cl + O_3 \rightarrow ClO + O_2$; $ClO + O \rightarrow Cl + O_2$ — Cl is regenerated, so one Cl destroys thousands of O₃ (chain). (1.5)

[
  {"claim":"68 kg NH3 (M=17) = 4 kmol","code":"result = (68/17)==4"},
  {"claim":"HNO3 molar mass = 63","code":"M=1+14+3*16; result = M==63"},
  {"claim":"Theoretical HNO3 mass = 252 kg (1:1 N)","code":"n=68/17; result = (n*63)==252"},
  {"claim":"Diborane valence electrons = 12","code":"result = (2*3+6*1)==12"},
  {"claim":"XeF4 total pairs around Xe = 6 (4 bp + 2 lp)","code":"bp=4; lp=(8-4)//2; result = (bp+lp)==6"}
]