Coordination Chemistry

Question 1 (12)

(a) (4) In a spherical field, all 5 d-orbitals are raised equally in energy (repulsion from ligand electron cloud) — no splitting (1). In octahedral geometry ligands approach along $\pm x, \pm y, \pm z$ axes (1). The $e_g$ orbitals ($d_{z^2}, d_{x^2-y^2}$) point directly at the ligands → greater repulsion → higher energy (1). The $t_{2g}$ orbitals ($d_{xy}, d_{yz}, d_{xz}$) point between axes → less repulsion → lower energy (1).

(b) (4) Let $e_g$ energy $= x$, $t_{2g}$ energy $= y$ relative to barycentre, with $x - y = \Delta_o$ (1). Barycentre conservation (total energy unchanged): $2x + 3y = 0$ (1) (2 orbitals in $e_g$, 3 in $t_{2g}$). Solve: from $x = y + \Delta_o$, $2(y+\Delta_o) + 3y = 0 \Rightarrow 5y = -2\Delta_o \Rightarrow y = -0.4\Delta_o$ (1), then $x = +0.6\Delta_o$ (1).

(c) (4) In tetrahedral geometry there are only 4 ligands (vs 6) and none point directly at the d-orbitals → smaller splitting (1). Geometric/statistical factor gives $\Delta_t = \tfrac{4}{9}\Delta_o \approx 0.44\Delta_o$ (1). Because $\Delta_t$ is small, it is almost always less than the pairing energy $P$ (1), so electrons prefer to occupy higher orbitals singly rather than pair → high-spin dominates (1).

Question 2 (10)

(a) (3) $[\text{Fe(CN)}_6]^{4-}$: Fe(II) = $d^6$, low spin $t_{2g}^6 e_g^0$ (1). CFSE $= 6(-0.4) + 0(0.6) = -2.4\Delta_o$; extra pairs formed vs free ion = 2 → $+2P$ (1). CFSE $= -2.4\Delta_o + 2P$ (1).

(b) (3) $[\text{Fe(H}_2\text{O)}_6]^{2+}$: $d^6$ high spin $t_{2g}^4 e_g^2$ (1). CFSE $= 4(-0.4) + 2(0.6) = -1.6 + 1.2 = -0.4\Delta_o$ (1); no extra pairing beyond free ion → $+0P$. CFSE $= -0.4\Delta_o$ (1).

(c) (2) $[\text{Cr(H}_2\text{O)}_6]^{3+}$: Cr(III) = $d^3$, $t_{2g}^3$ (1). CFSE $= 3(-0.4) = -1.2\Delta_o$ (1).

(d) (2) $\text{CN}^-$ is a strong-field ligand → large $\Delta_o > P$, so pairing in $t_{2g}$ is energetically favoured → low spin (1). $\text{H}_2\text{O}$ is weak-field → $\Delta_o < P$, so electrons stay unpaired (high spin) (1).

Question 3 (10)

(a) (5) Co(III) = $d^6$.

• $[\text{CoF}_6]^{3-}$: F⁻ weak field → electrons stay unpaired, uses outer $4d$ orbitals → $sp^3d^2$ (outer orbital), 4 unpaired e⁻ (1). $\mu = \sqrt{4\cdot6} = \sqrt{24} = 4.90$ BM (1.5).
• $[\text{Co(NH}_3)_6]^{3+}$: NH₃ stronger → pairing, uses inner $3d$ → $d^2sp^3$ (inner orbital), 0 unpaired e⁻ (1). $\mu = 0$ BM (diamagnetic) (1.5).

(b) (3) CFT: for F⁻, $\Delta_o < P$ → high spin $t_{2g}^4 e_g^2$, 4 unpaired (1). For NH₃, $\Delta_o > P$ → low spin $t_{2g}^6$, 0 unpaired (1). Order follows spectrochemical series $\text{F}^- < \text{NH}_3$ (1).

(c) (2) VBT cannot explain the colour / spectra of complexes (d-d transitions require the $\Delta_o$ energy gap concept) nor the spectrochemical series ordering; CFT accounts for both (2) (any valid: colour, or quantitative $\Delta_o$).

Question 4 (10)

(a) (2) $\mu_s = \sqrt{n(n+2)}$ BM (1); assumes only spin angular momentum contributes, orbital contribution quenched (1).

(b) (4) $2.83 = \sqrt{n(n+2)} \Rightarrow n(n+2) = 8 \Rightarrow n = 2$ (2). Ni²⁺ is $d^8$; 2 unpaired e⁻ → paramagnetic → tetrahedral ($e^4 t_2^4$ gives 2 unpaired) (1). Square-planar $d^8$ would be diamagnetic ($\mu=0$), so geometry is tetrahedral (1).

(c) (4) $[\text{Mn(H}_2\text{O)}_6]^{2+}$ is $d^5$ high spin, all 5 electrons unpaired, one per orbital (1). Any d-d transition requires a spin flip → forbidden by the spin selection rule ($\Delta S = 0$) (1). Also Laporte (orbital) forbidden in octahedral field (1). Both rules violated → extremely weak absorption → very pale colour (1).

Question 5 (10)

(a) (3) Chelate effect: multidentate (chelating) ligands form more stable complexes than comparable monodentate ligands (1). Driven principally by entropy (1): replacing several monodentate ligands with fewer polydentate ones increases the number of free molecules → $\Delta S > 0$ (1).

(b) (3) Left side: 1 complex + 3 en = 4 particles; right side: 1 complex + 6 H₂O = 7 particles (1). $\Delta n = +3$ (net increase in free particles) (1) → more disorder → $\Delta S > 0$ (1).

(c) (4) $\Delta G^\circ = -RT\ln\beta = -2.303RT\log\beta$ (1). $= -2.303 \times 8.314 \times 298 \times 18.3$ (1) $= -2.303 \times 8.314 \times 298 = 5706$ J; $\times 18.3 = 104{,}400$ J (1) $\Delta G^\circ \approx -104.4\ \text{kJ/mol}$ (1).

Question 6 (8)

(a) (4) $[\text{Co(en)}_2\text{Cl}_2]^+$ has cis and trans geometrical isomers (1). Trans has a plane of symmetry → optically inactive (1). Cis lacks a plane/centre of symmetry → chiral, exists as a pair of enantiomers (d and l) → optically active (1). Total: 3 stereoisomers (trans, cis-d, cis-l) (1).

(b) (4) $[\text{Cu(H}_2\text{O)}_6]^{2+}$ is $d^9$: configuration $t_{2g}^6 e_g^3$ → the $e_g$ set is unevenly (asymmetrically) occupied (1). The $e_g$ orbitals point at the ligands, so unequal occupation of $d_{z^2}$ vs $d_{x^2-y^2}$ makes the geometry unstable → Jahn–Teller theorem forces distortion (usually axial elongation) to lower energy (1). $[\text{Zn(H}_2\text{O)}_6]^{2+}$ is $d^{10}$: $t_{2g}^6 e_g^4$ → $e_g$ symmetrically filled → no orbital degeneracy to lift → no distortion (1). J–T applies only to degenerate ground states (asymmetric $e_g$/$t_{2g}$ occupation) (1).

[
  {"claim":"Q2b CFSE d6 high spin = -0.4 Δo", "code":"cfse = 4*(-0.4) + 2*(0.6); result = abs(cfse - (-0.4)) < 1e-9"},
  {"claim":"Q2a CFSE d6 low spin = -2.4 Δo (before pairing term)", "code":"cfse = 6*(-0.4) + 0*(0.6); result = abs(cfse - (-2.4)) < 1e-9"},
  {"claim":"Q4b n=2 gives mu approx 2.83 BM", "code":"n=2; mu = sqrt(n*(n+2)); result = abs(float(mu) - 2.828) < 0.01"},
  {"claim":"Q3a CoF6 mu with 4 unpaired = sqrt24 = 4.90 BM", "code":"n=4; mu=sqrt(n*(n+2)); result = abs(float(mu)-4.899)<0.01"},
  {"claim":"Q5c Delta G = -104.4 kJ/mol", "code":"R=8.314; T=298; logb=18.3; dG = -2.303*R*T*logb/1000; result = abs(dG - (-104.4)) < 0.5"}
]