Noesis
Level 4 — Applicationp-Block

p-Block

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Level: 4 (Application — novel/unseen problems) Time Limit: 60 minutes Total Marks: 50

Instructions: Attempt all questions. Reasoning must be shown; unsupported answers earn no credit.

Q1. (10 marks) An unknown boron hydride X has the empirical composition BH₃ but exists as a discrete dimeric molecule with molar mass ≈ 27.7 g mol⁻¹ (B = 10.8, H = 1.0).

(a) Determine the molecular formula of X from the molar mass. (2) (b) The molecule contains two types of B–H bonds. Classify them and explain the bonding of the bridging bonds in terms of electron counting. (3) (c) Count the total number of valence electrons available and show that a conventional Lewis structure with only 2c–2e bonds is impossible. (3) (d) When X reacts with excess NH₃ under controlled heating, an aromatic six-membered ring Y ("inorganic benzene") forms. Give the formula of Y and one way its reactivity differs from benzene. (2)

Q2. (10 marks) Consider the four chlorine oxoacids: HClO, HClO₂, HClO₃, HClO₄.

(a) Arrange them in order of increasing acid strength and justify using oxidation state / number of terminal O atoms. (3) (b) The successive pKa\mathrm{p}K_a values (approx.) are 7.5, 2.0, −1.0, −10. Using Pauling's rule pKa≈8−5m\mathrm{p}K_a \approx 8 - 5m (where m = number of terminal =O atoms), predict pKa\mathrm{p}K_a for each and comment on agreement. (4) (c) A solution is 0.10 M in HClO (pKa=7.5\mathrm{p}K_a = 7.5). Calculate the pH. (3)

Q3. (12 marks) In the industrial manufacture of nitric acid (Ostwald process):

(a) Write the three balanced equations for: catalytic oxidation of NH₃, oxidation of NO, and absorption of NO₂ in water. (3) (b) In the final step 3NO2+H2O→2HNO3+NO3NO_2 + H_2O \rightarrow 2HNO_3 + NO, the liberated NO is recycled. If 1000 mol of NH₃ enters and every N atom is ultimately converted to HNO₃ with the NO fully recycled, determine the theoretical moles of HNO₃ produced and the moles of O₂ consumed overall. (5) (c) Assign the oxidation state of nitrogen in NH₃, NO, NO₂ and HNO₃, and state the net number of electrons lost per N atom going from NH₃ to HNO₃. (4)

Q4. (10 marks) Xenon forms several fluorides and an oxide.

(a) Using VSEPR, predict the shape and describe hybridization for XeF₂, XeF₄ and XeO₃. State the number of lone pairs on Xe in each. (6) (b) XeF₆ reacts with water in two stages. Write balanced equations for partial hydrolysis giving XeOF₄ and complete hydrolysis giving XeO₃. (2) (c) Explain, using electronegativity/oxidising-power arguments, why xenon forms stable fluorides but no stable chlorides. (2)

Q5. (8 marks) Two phosphorus oxoacids, H₃PO₃ and H₃PO₄, are compared.

(a) Draw the structure of each showing P–H and P–OH bonds, and state the basicity (number of replaceable H) of each. (4) (b) Explain why H₃PO₃ is dibasic despite having three H atoms. (2) (c) 0.050 mol of H₃PO₃ is exactly neutralised by NaOH. Calculate the volume of 0.20 M NaOH required (use correct basicity). (2)

Answer keyMark scheme & solutions

Q1 (10)

(a) Molar mass of BH₃ unit = 10.8 + 3 = 13.8 g mol⁻¹. Ratio 27.7 / 13.8 ≈ 2 → X = B₂H₆ (diborane). (2)

(b) Two types: 4 terminal B–H bonds (normal 2c–2e σ bonds) and 2 bridging B–H–B bonds, each a three-centre two-electron (3c–2e) bond. In each bridge, one H shares a single electron pair over two B atoms; the electron pair spreads across the B–H–B triangle. (3) (1 for identifying two types, 1 for 3c–2e term, 1 for explanation)

(c) Valence electrons = 2(B: 3) + 6(H: 1) = 6 + 6 = 12 electrons = 6 pairs. A classical structure with 8 B–H bonds (4 terminal + 4 in two bridges) would require 8 pairs = 16 electrons. Only 12 available → electron-deficient → cannot be all 2c–2e. Four bonds must be the two 3c–2e bridges (2 pairs cover 4 B–H links). (3)

(d) Y = borazine, B₃N₃H₆. Unlike benzene, borazine is more reactive toward addition (e.g. adds HCl across B–N because of the polar B(δ+)–N(δ−) bonds); its π system is less uniformly delocalised. (2)

Q2 (10)

(a) Increasing acid strength: HClO < HClO₂ < HClO₃ < HClO₄. As oxidation state of Cl rises (+1→+7) and number of terminal =O atoms increases, the O atoms withdraw electron density, stabilising the conjugate base (charge delocalisation) → stronger acid. (3)

(b) Pauling: pKa=8−5m\mathrm{p}K_a = 8 - 5m.

  • HClO, m = 0 → 8 (obs 7.5) ✓
  • HClO₂, m = 1 → 3 (obs 2.0) ✓ approx
  • HClO₃, m = 2 → −2 (obs −1) ✓ approx
  • HClO₄, m = 3 → −7 (obs −10) — rough Trend of decreasing pKₐ by ≈5 per O is well reproduced; agreement good for low m, poorer for very strong HClO₄. (4)

(c) HClO: Ka=10−7.5=3.16×10−8K_a = 10^{-7.5} = 3.16\times10^{-8}. [H+]=Ka⋅C=3.16×10−8×0.10=3.16×10−9=5.62×10−5[H^+] = \sqrt{K_a \cdot C} = \sqrt{3.16\times10^{-8}\times 0.10} = \sqrt{3.16\times10^{-9}} = 5.62\times10^{-5}. pH =−log⁡(5.62×10−5)=4.25= -\log(5.62\times10^{-5}) = 4.25. (3)

Q3 (12)

(a) (3) 4NH3+5O2→Pt/Rh4NO+6H2O4NH_3 + 5O_2 \xrightarrow{Pt/Rh} 4NO + 6H_2O 2NO+O2→2NO22NO + O_2 \rightarrow 2NO_2 3NO2+H2O→2HNO3+NO3NO_2 + H_2O \rightarrow 2HNO_3 + NO

(b) Since every N atom becomes HNO₃ (NO recycled), moles HNO₃ = moles N = 1000 mol. (2)

Overall reaction (recycling NO): combine so N conserved: NH3+2O2→HNO3+H2ONH_3 + 2O_2 \rightarrow HNO_3 + H_2O (Check: N balances; O: LHS 4, RHS 3+1=4 ✓; H: LHS 3, RHS 1+2=3 ✓; oxidation N −3→+5 loses 8e, 2O₂ gain 8e ✓.) So O₂ per N = 2 → O₂ = 2 × 1000 = 2000 mol. (3)

(c) Oxidation states: NH₃ = −3, NO = +2, NO₂ = +4, HNO₃ = +5. (2) Electrons lost per N: from −3 to +5 = 8 electrons. (2)

Q4 (10)

(a) (6, 1 each shape/hyb + LP)

  • XeF₂: 5 electron pairs (2 bond + 3 lone), sp³d, linear, 3 lone pairs.
  • XeF₄: 6 pairs (4 bond + 2 lone), sp³d², square planar, 2 lone pairs.
  • XeO₃: 4 electron domains (3 σ + 1 lone), sp³, trigonal pyramidal, 1 lone pair.

(b) (2) XeF6+H2O→XeOF4+2HFXeF_6 + H_2O \rightarrow XeOF_4 + 2HF XeF6+3H2O→XeO3+6HFXeF_6 + 3H_2O \rightarrow XeO_3 + 6HF

(c) Fluorine is the most electronegative element and a stronger oxidiser; the Xe–F bond is strong and stabilises high Xe oxidation states. Chlorine is less electronegative, Xe–Cl bond too weak, and Xe cannot oxidise/hold Cl effectively → no stable xenon chlorides. (2)

Q5 (8)

(a) (4)

  • H₃PO₃ (phosphorous acid): tetrahedral P with one P–H bond, one P=O, two P–OH. Basicity = 2.
  • H₃PO₄ (phosphoric acid): tetrahedral P with one P=O and three P–OH. Basicity = 3.

(b) Only the H atoms bonded to O (as –OH) are ionisable. In H₃PO₃ one H is bonded directly to P (P–H), which is non-acidic; hence only 2 of 3 H are replaceable → dibasic. (2)

(c) H₃PO₃ is dibasic: mol NaOH = 2 × 0.050 = 0.10 mol. Volume = 0.10 / 0.20 = 0.50 L = 500 mL. (2)

[
  {"claim":"B2H6 molar mass approx 27.7 and equals 2 BH3 units","code":"M=2*(10.8+3*1.0); result = abs(M-27.6)<0.2"},
  {"claim":"pH of 0.10 M HClO with pKa 7.5 is about 4.25","code":"Ka=Rational(1,1)*10**Rational(-15,2); C=Rational(1,10); H=sqrt(Ka*C); pH=-log(H,10); result = abs(float(pH)-4.25)<0.05"},
  {"claim":"Ostwald: 1000 mol NH3 gives 1000 mol HNO3 and consumes 2000 mol O2","code":"n=1000; HNO3=n; O2=2*n; result = (HNO3==1000) and (O2==2000)"},
  {"claim":"N oxidation change NH3 to HNO3 is 8 electrons","code":"result = (5-(-3))==8"},
  {"claim":"H3PO3 dibasic: 0.050 mol needs 500 mL of 0.20 M NaOH","code":"molNaOH=2*Rational(5,100); V=molNaOH/Rational(2,10); result = abs(float(V)-0.5)<1e-9"}
]