p-Block

Q1. (3 marks) Anomalous behaviour of boron due to (any two, 1.5 each):

• Small atomic size and high charge/size ratio → high polarising power (1.5)
• Absence of d-orbitals in valence shell → restricted to maximum covalency of 4, cannot expand octet (1.5)
• Very high ionisation enthalpy / non-metallic character vs metallic rest of group. Why: these arise from B being in period 2 with only 2s, 2p available.

Q2. (3 marks) Order: $BF_3 < BCl_3 < BBr_3 < BI_3$ (2) Reason: Lewis acidity increases as back-bonding (p$\pi$–p$\pi$) from halogen lone pair into empty 2p of boron decreases. $F$ (2p) overlaps best with B(2p), so $BF_3$ has strongest back-bonding → least electron deficient → weakest acid; overlap weakens F→Cl→Br→I. (1)

Q3. (4 marks)

• $B_2H_6$ has 4 terminal B–H bonds (normal 2c–2e) and 2 bridging H atoms. (1)
• The two bridging B–H–B bonds are three-centre two-electron (3c–2e) "banana" bonds (1)
• Each 3c–2e bond: two boron sp³ orbitals + one H 1s orbital share only 2 electrons over 3 atomic centres. (1)
• Diborane is electron-deficient (12 valence e⁻, needs 14 for normal bonding), which is why bridging bonds form. (1)

Q4. (5 marks) (a) Borazine = $B_3N_3H_6$, a six-membered planar ring of alternating B and N atoms (1); isoelectronic and isostructural with benzene $C_6H_6$ → "inorganic benzene" (1). (b) (i) diamond: sp³ (0.5); (ii) graphite: sp² (0.5). Graphite conducts (1) because each C uses only 3 of 4 valence electrons in σ-bonds; the 4th delocalised π-electron per atom is free to move along layers (1).

Q5. (4 marks) Ostwald process (1 + 2 + 1): $4NH_3 + 5O_2 \xrightarrow{Pt/Rh,\ 500\,K} 4NO + 6H_2O$ $2NO + O_2 \rightarrow 2NO_2$ $3NO_2 + H_2O \rightarrow 2HNO_3 + NO$ (1 mark per balanced equation; 1 for correct overall/conditions)

Q6. (5 marks) (a) $H_3PO_3$ is dibasic (2 ionisable P–OH), $H_3PO_4$ is tribasic (3 P–OH) (1.5). Reason: in $H_3PO_3$ one H is bonded directly to P (P–H, non-ionisable); in $H_3PO_4$ all three H are on –OH groups (1.5). (b) $N_2$ inert because of very strong N≡N triple bond with high bond dissociation enthalpy ($\approx 941$ kJ/mol), requiring large energy to break. (2)

Q7. (6 marks) (a) Contact process (1.5 each): $2SO_2 + O_2 \underset{V_2O_5}{\rightleftharpoons} 2SO_3$ $SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \ (\text{oleum});\quad H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4$ (b) Ozone absorbs harmful UV radiation, shielding life (1). Catalytic destruction (2): $Cl + O_3 \rightarrow ClO + O_2$ $ClO + O \rightarrow Cl + O_2$

Q8. (3 marks) Order: $HClO < HClO_2 < HClO_3 < HClO_4$ (1.5) Reason: as number of terminal O atoms increases, the higher oxidation state of Cl and greater delocalisation/stabilisation of the conjugate base (anion) increases → stronger acid (1.5).

Q9. (3 marks) (1 each) (a) $XeF_2$: linear, sp³d. (b) $XeF_4$: square planar, sp³d². (c) $XeO_3$: pyramidal (trigonal pyramidal), sp³.

Q10. (4 marks) (a) HF weaker despite high electronegativity because of its very high H–F bond dissociation enthasty (strong bond) and strong hydrogen bonding, so it ionises less; bond strength decreases HF→HI, making HI strongest. (2) (b) Interhalogens = compounds formed between two different halogens, e.g. $ClF_3$ (or $BrF_3$/$ICl_3$) (1); $AB_3$ type has T-shaped geometry (1).

[
  {"claim":"Ostwald step 1 balanced: 4NH3+5O2 -> 4NO+6H2O (N,H,O balance)","code":"N_l=4*1;N_r=4*1;H_l=4*3;H_r=6*2;O_l=5*2;O_r=4*1+6*1;result=(N_l==N_r and H_l==H_r and O_l==O_r)"},
  {"claim":"Contact step 2SO2+O2->2SO3 balanced","code":"S_l=2;S_r=2;O_l=2*2+2;O_r=2*3;result=(S_l==S_r and O_l==O_r)"},
  {"claim":"Diborane valence electrons = 12 (2B*3 + 6H*1)","code":"e=2*3+6*1;result=(e==12)"},
  {"claim":"Cl catalytic cycle net: O3 + O -> 2 O2 (O atoms balance)","code":"lhs=3+1;rhs=2*2;result=(lhs==rhs)"}
]