Chemical Bonding

Q1 (10)

(a) Lewis: Xe centre, one Xe=O double bond, four Xe–F single bonds, one lone pair on Xe. Formal charge on Xe: valence 8; bonds = 4 single (F) + 1 double (O) = counts 4+2 = 6 bonding electrons shared → Xe uses 6 bond pairs = 6 shared, plus 1 lone pair (2 e). FC = 8 − (2 nonbonding) − ½(12 bonding e) = 8 − 2 − 6 = 0. [3] (structure 2, FC 1) (b) 6 electron domains (5 bonds + 1 lone pair) → hybridization $sp^3d^2$; geometry square pyramidal. [3] (hybrid 1, domains 1, shape name 1) (c) The lone pair and the Xe=O occupy opposite/asymmetric positions; the four Xe–F dipoles cancel in the basal plane but the axial O and lone pair do not cancel, leaving a net dipole $\mu \neq 0$. [2] (d) Expanded octet; total valence electrons around Xe = 12. [2]

Q2 (12)

(a) NO (14 e): $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\pi_{2p_y}^2\,\pi_{2p_x}^{*1}$ (N and O beyond ~14 e use σ2p below π? For NO the accepted filling gives one electron in π*; using either ordering the antibonding count is the same.) [2] NO⁺ (13 e): remove the π* electron → $\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4$ (no π* electrons). [2] (b) Bond order = ½(bonding − antibonding). NO: bonding = 10, antibonding = 5 → BO = ½(10−5) = 2.5. [2] NO⁺: bonding = 10, antibonding = 4 → BO = ½(10−4) = 3.0. [2] (c) NO⁺ has higher bond order (3 > 2.5) → shorter, stronger bond. NO has one unpaired electron (π*¹) → paramagnetic; NO⁺ has all electrons paired → diamagnetic. [4]

Q3 (10)

(a) Born–Haber (Hess): $\Delta H_f = \Delta H_{sub} + IE + \tfrac12 D + EA + U$ $-437 = 89 + 419 + \tfrac12(242) + (-349) + U$ $-437 = 89 + 419 + 121 - 349 + U = 280 + U$ $U = -437 - 280 = \mathbf{-717\ kJ\,mol^{-1}}$ [6] (cycle setup 2, correct ½D=121 1, arithmetic 3) (b) LiI has greater covalent character. Reasons (any two): Li⁺ is small with high polarising power (high charge density); I⁻ is large and highly polarisable; both favour distortion of anion electron cloud (Fajan's rules) → more covalent than KCl (large cation K⁺, less polarising; small less-polarisable Cl⁻). [4]

Q4 (10)

(a) CH₃OH has strong O–H···O intermolecular hydrogen bonding; CH₃SH has only weak dipole/London forces (S too large/low electronegativity for effective H-bonding). Extra energy to break H-bonds raises A's boiling point despite lower mass. [3] (b) $q = 0.5 \times 1.6\times10^{-19} = 0.8\times10^{-19}\ \mathrm{C}$; $d = 96\times10^{-12}\ \mathrm{m}$. $\mu = qd = 0.8\times10^{-19}\times96\times10^{-12} = 7.68\times10^{-30}\ \mathrm{C\,m}$. In debye: $7.68\times10^{-30}/3.336\times10^{-30} = \mathbf{2.30\ D}$. [4] (c) Intermolecular H-bonding raises boiling point (links molecules); intramolecular (e.g. o-nitrophenol) lowers it. Example of intermolecular: water / p-nitrophenol has higher b.p. than o-nitrophenol. [3]

Q5 (8)

(a) Three equivalent resonance structures, each with one C=O (FC 0 on that O) and two C–O⁻ (FC −1 each); C has FC 0. Net charge −2 distributed. [3] (b) Bond order = (total bonds)/(number of positions) = (1 double + 2 single = 4 bond pairs over 3 bonds) → $4/3 \approx$ 1.33. [2] (c) Experimental bond length lies between 121 and 143 pm (~128–131 pm); all three equal because delocalisation averages the double-bond character equally over the three C–O bonds. [3]

[
 {"claim":"KCl lattice enthalpy = -717 kJ/mol","code":"dHf=-437; sub=89; IE=419; D=242; EA=-349; U=dHf-(sub+IE+D/2+EA); result=(U==-717)"},
 {"claim":"NO bond order 2.5, NO+ bond order 3.0","code":"NO=(10-5)/2; NOp=(10-4)/2; result=(NO==2.5 and NOp==3.0)"},
 {"claim":"O-H bond dipole approx 2.30 D","code":"q=0.5*1.6e-19; d=96e-12; mu=q*d; D=mu/3.336e-30; result=(abs(D-2.30)<0.02)"},
 {"claim":"carbonate C-O bond order = 4/3","code":"from sympy import Rational; bo=Rational(4,3); result=(bo==Rational(4,3))"}
]