Chemical Bonding

Question 1

(a) Born–Haber cycle (6 marks)

By Hess's law around the cycle, forming NaCl(s) either directly or via gaseous ions: $\Delta_f H = \Delta_{sub}H + IE + \tfrac12 D(\text{Cl}_2) + EA + U$ where $U$ = lattice enthalpy (formation of solid from gaseous ions, negative). (2 marks for correct cycle/equation.)

Solve for $U$: $U = \Delta_f H - \Delta_{sub}H - IE - \tfrac12 D - EA$ $U = -411 - 108 - 496 - 122 - (-349)$ $U = -411 - 108 - 496 - 122 + 349 = -788\ \text{kJ mol}^{-1}$ (3 marks arithmetic; 1 mark for stating convention: $U$ negative = exothermic formation of lattice from ions.)

(b) Kapustinskii (6 marks)

$r_+ + r_- = (102+181)\,\text{pm} = 283\,\text{pm} = 2.83\times10^{-10}\,\text{m}$. (1)

$U = \frac{1.2025\times10^{-4}\times 2\times1\times1}{2.83\times10^{-10}}\left(1 - \frac{3.45\times10^{-11}}{2.83\times10^{-10}}\right)$

Prefactor: $\dfrac{2.405\times10^{-4}}{2.83\times10^{-10}} = 8.498\times10^{5}\,\text{J mol}^{-1}$. (2)

Correction: $1 - \dfrac{3.45\times10^{-11}}{2.83\times10^{-10}} = 1 - 0.1219 = 0.8781$. (1)

$U = 8.498\times10^{5}\times0.8781 = 7.462\times10^{5}\,\text{J mol}^{-1} = 746\ \text{kJ mol}^{-1}$ (1 mark)

By convention Kapustinskii gives magnitude; so $U_{\text{Kap}} \approx -746\,\text{kJ mol}^{-1}$ vs Born–Haber $-788\,\text{kJ mol}^{-1}$; difference $\approx 42\,\text{kJ mol}^{-1}$ (~5%). Discrepancy: Kapustinskii uses averaged/empirical constants and assumes purely ionic point charges, ignoring partial covalent character and specific Madelung structure; Born–Haber is experimental. (1 mark)

(c) Fajan's rules (6 marks)

Covalent character increases with (i) higher cation charge, (ii) smaller cation size, (iii) larger/more polarizable anion. (2)

Across Na⁺ → Mg²⁺ → Al³⁺: charge rises (+1→+2→+3) and ionic radius falls (102→72→53 pm), so polarizing power (charge/radius) increases sharply. The common anion Cl⁻ is increasingly distorted. Therefore covalent character increases: NaCl < MgCl₂ < AlCl₃. (2)

AlCl₃ is essentially covalent (exists as Al₂Cl₆ dimers, low m.p., sublimes) because Al³⁺ is small and highly charged → extreme polarization of Cl⁻ electron cloud. Hence its purely ionic lattice-energy model overestimates the true (partly covalent) bonding, giving large deviation, whereas Na⁺ (low charge, large) barely polarizes Cl⁻, so NaCl is well-described as ionic. (2)

Question 2

(a) MO occupancies (8 marks — 2 each)

• N₂ (14 e⁻, s-p mixing order): $\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p}^4\sigma_{2p}^2$. $n_b=8,\ n_a=2$ → BO $=\tfrac{8-2}{2}=3$. Diamagnetic.
• O₂ (16 e⁻): $\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*2}$. $n_b=10,\ n_a=6$ → BO $=\tfrac{10-6}{2}=2$. Paramagnetic (2 unpaired).
• NO (15 e⁻): $\ldots\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*1}$. $n_b=10,\ n_a=5$ → BO $=\tfrac{10-5}{2}=2.5$. Paramagnetic (1 unpaired).
• CO (14 e⁻, isoelectronic with N₂): $\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p}^4\sigma_{2p}^2$. $n_b=8,\ n_a=2$ → BO $=3$. Diamagnetic.

(b) Function (6 marks)

$B(n) = \tfrac12(n_b - n_a)$. (1)

def bond_order(config):
    # config: {'orbital': (occupancy, is_bonding)}
    nb = sum(occ for occ,(bond) in
             [(v[0], v[1]) for v in config.values()] if bond)
    na = sum(v[0] for v in config.values() if not v[1])
    bo = (nb - na) / 2
    unpaired = sum(1 for v in config.values()
                   if v[0] == 1)          # simplified: singly-filled MOs
    magnetism = "paramagnetic" if unpaired > 0 else "diamagnetic"
    return bo, magnetism
 
O2 = {
  's2s':(2,True), 's2s*':(2,False),
  's2p':(2,True), 'pi2p_a':(2,True), 'pi2p_b':(2,True),
  'pi2p*_a':(1,False), 'pi2p*_b':(1,False)
}
# bond_order(O2) -> (2.0, 'paramagnetic')

(3 marks logic; 2 marks correct O₂ demonstration: $n_b=10,n_a=6$, BO=2, paramagnetic.)

(c) Proof (4 marks)

O₂ has 16 electrons. Filling to the antibonding $\pi^*$ level places the last 2 electrons into the two degenerate $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ orbitals. By Hund's rule these occupy separate orbitals with parallel spins → 2 unpaired electrons → paramagnetic. (2) The Lewis structure O=O shows all electrons paired (double bond + two lone pairs on each O), wrongly predicting diamagnetism because it cannot represent the degenerate antibonding pair. MOT correctly places the unpaired spins in the $\pi^*_{2p}$ set. (2)

Question 3

(a) (6 marks — 2 each)

• XeF₄: 6 domains (4 bp + 2 lp), $sp^3d^2$, octahedral e-geometry, square planar shape, nonpolar (lp trans, dipoles cancel).
• SF₄: 5 domains (4 bp + 1 lp), $sp^3d$, trigonal bipyramidal e-geometry, see-saw shape, polar.
• ClF₃: 5 domains (3 bp + 2 lp), $sp^3d$, TBP e-geometry, T-shaped, polar.

(b) Dipole vector sum (5 marks)

Two equal vectors of magnitude $\mu_{\text{OH}}$ at angle $\theta$; resultant along bisector: $\mu_{\text{net}} = \sqrt{\mu^2 + \mu^2 + 2\mu^2\cos\theta} = 2\mu_{\text{OH}}\cos\!\left(\frac{\theta}{2}\right)$ (3 marks derivation.)

With $\mu_{\text{OH}}=1.51$ D, $\theta=104.5°$, $\theta/2 = 52.25°$, $\cos 52.25° = 0.6122$: $\mu_{\text{net}} = 2\times1.51\times0.6122 = 1.85\ \text{D}$ (2 marks; matches experimental 1.85 D.)

(c) (3 marks)

CO₂ is linear ($\theta = 180°$): the two C=O bond dipoles are equal in magnitude, opposite in direction (antiparallel), so $\mu_{\text{net}} = 2\mu\cos(180°/2)=2\mu\cos90° = 0$ — they cancel exactly. (2) In H₂O the bent geometry ($104.5°$) means the two O–H dipoles do not oppose; their vector sum is nonzero, so water is polar. Geometry, not bond polarity alone, decides molecular polarity. (1)

[
  {"claim":"Born-Haber lattice energy of NaCl = -788 kJ/mol",
   "code":"dfH=-411; sub=108; IE=496; halfD=122; EA=-349; U=dfH-sub-IE-halfD-EA; result = (U == -788)"},
  {"claim":"Kapustinskii lattice energy of NaCl approx 746 kJ/mol magnitude",
   "code":"r=2.83e-10; pref=1.2025e-4*2*1*1/r; corr=1-3.45e-11/r; U=pref*corr/1000; result = (abs(U-746) < 3)"},
  {"claim":"O2 bond order = 2 with 10 bonding and 6 antibonding electrons",
   "code":"nb=10; na=6; BO=(nb-na)/2; result = (BO == 2)"},
  {"claim":"Water net dipole = 2*1.51*cos(104.5/2) approx 1.85 D",
   "code":"import sympy as sp; mu=sp.Float(1.51); th=sp.rad(104.5); net=2*mu*sp.cos(th/2); result = (abs(float(net)-1.85) < 0.02)"}
]