Coordination Chemistry

Question 1

(a) (6 marks) High-spin filling follows Hund: fill all five d-orbitals singly (3 in $t_{2g}$, 2 in $e_g$) before pairing. Let $t =$ electrons in $t_{2g}$, $e =$ electrons in $e_g$; $t+e=n$. CFSE $= (-0.4t + 0.6e)\,\Delta_o$. (2)

For high spin:

• $n=1..3$: $t=n$, $e=0$
• $n=4..5$: $t=3$, $e=n-3$
• $n=6..8$: $t=n-2$... explicitly $t=\min(n,3)$ up to 5, then after 5 the 6th–10th electrons pair in $t_{2g}$ then $e_g$.

General: for $n\le5$, $t=\min(n,3),\ e=\max(n-3,0)$; for $n>5$, $t=3+\min(n-5,3),\ e=2+\max(n-8,0)$. (2)

Extra pairing term: high-spin has extra pairs only for $n>5$. Number of paired electrons beyond the free-ion count = $\max(n-5,0)$, so $m = \max(n-5,0)$ and CFSE(total) $=$ (field term) $+ mP$. (2)

(b) (8 marks) $\mathrm{Fe^{2+}}$ is $d^6$.

High spin $[\mathrm{Fe(H_2O)_6}]^{2+}$: $t_{2g}^4 e_g^2$. CFSE $= (-0.4\times4 + 0.6\times2)\Delta_o = (-1.6+1.2)\Delta_o = -0.4\,\Delta_o$; extra pairs vs free ion $= 1 \Rightarrow +1P$. Total $= -0.4\,\Delta_o + P$. Unpaired $n=4 \Rightarrow \mu=\sqrt{4\cdot6}=\sqrt{24}=4.90\,\mu_B$. (4)

Low spin $[\mathrm{Fe(CN)_6}]^{4-}$: $t_{2g}^6 e_g^0$. CFSE $=(-0.4\times6+0)\Delta_o=-2.4\,\Delta_o$; three pairs vs free-ion's zero extra... free ion $d^6$ has 1 pair; low-spin has 3 pairs $\Rightarrow$ extra $=2P$. Total $=-2.4\Delta_o+2P$. Unpaired $n=0 \Rightarrow \mu=0$ (diamagnetic). (2)

Explanation: $\mathrm{CN^-}$ is a strong-field ligand high in the spectrochemical series ($\mathrm{H_2O}$ is weak-field). For $\mathrm{CN^-}$, $\Delta_o > P$ so pairing in $t_{2g}$ is favoured (low spin); for $\mathrm{H_2O}$, $\Delta_o < P$ so electrons occupy $e_g$ (high spin). (2)

(c) (6 marks) Filling rule encoded: high spin fills singly to 5 then pairs; low spin fills $t_{2g}$ completely (up to 6) before any $e_g$. Award 2 marks for correct rule statement.

def cfse_octahedral(n, spin):
    if spin == "high":
        t = min(n, 3) + max(min(n-5, 3), 0)   # t2g electrons
        e = n - t                              # eg electrons
        # unpaired: singly filled orbitals
        unpaired = 5 - abs(5 - n) if n <= 5 else 10 - n
    else:  # low spin
        t = min(n, 6)
        e = max(n - 6, 0)
        unpaired = min(t, 6-t) + ... # see below
        # t2g unpaired = t if t<=3 else 6-t ; eg unpaired = e if e<=2 else 4-e
        up_t = t if t <= 3 else 6 - t
        up_e = e if e <= 2 else 4 - e
        unpaired = up_t + up_e
    cfse = round(-0.4*t + 0.6*e, 2)
    return cfse, unpaired

Full marks for correct t,e split, correct CFSE formula, and correct unpaired count logic. (4)

(d) (4 marks) No high/low-spin distinction exists when there is only one way to distribute electrons: $d^1, d^2, d^3$ (all in $t_{2g}$, no pairing choice) and $d^8, d^9, d^{10}$ ($t_{2g}$ full, remaining forced into $e_g$). For these, unpaired count is fixed. High/low spin differs only for $d^4$–$d^7$. (4) (2 for identifying $d^{1,2,3,8,9,10}$; 2 for justification.)

Question 2

(a) (8 marks) $[\mathrm{Co(en)_2Cl_2}]^+$ octahedral, two bidentate en + two Cl. Geometrical isomers: cis (Cl adjacent, 90°) and trans (Cl axial, 180°). (2)

• trans: has a mirror plane (and $S_2/\sigma_h$); superimposable on its mirror image ⇒ achiral. (2)
• cis: point group $C_2$; no $\sigma$ plane, no improper axis ($S_n$); non-superimposable mirror images ⇒ optically active, exists as $\Delta/\Lambda$ enantiomer pair. (2)

So three stereoisomers total: trans (1) + cis pair (2). Symmetry argument: chirality requires absence of any $S_n$ (including $\sigma=S_1$, $i=S_2$); trans possesses these, cis does not. (2)

(b) (6 marks) Isomers:

• $[\mathrm{Co(NH_3)_5Br}]\mathrm{SO_4}$ — free $\mathrm{SO_4^{2-}}$; test with $\mathrm{BaCl_2}$ → white $\mathrm{BaSO_4}$ precipitate. Name: pentaamminebromidocobalt(III) sulfate. (3)
• $[\mathrm{Co(NH_3)_5SO_4}]\mathrm{Br}$ — free $\mathrm{Br^-}$; test with $\mathrm{AgNO_3}$ → pale-yellow $\mathrm{AgBr}$ precipitate. Name: pentaamminesulfatocobalt(III) bromide. (3)

(c) (4 marks) Tetrahedral $[\mathrm{MA_2B_2}]$: all four positions are equivalent (all at $\sim109.5°$, mutually adjacent), so there is no cis/trans distinction ⇒ no geometrical isomers. With two identical A and two identical B monodentate ligands the molecule retains a mirror plane ($C_{2v}$), so it is achiral ⇒ no optical isomerism. (2) Square-planar $[\mathrm{MA_2B_2}]$: the four positions lie in a plane with defined cis (90°) and trans (180°) relationships, so cis and trans geometrical isomers exist; the plane of the molecule is a $\sigma$ plane so each is achiral. (2)

Question 3

(a) (6 marks) $\mathrm{Ni^{2+}} = d^8$.

• $[\mathrm{Ni(CN)_4}]^{2-}$: strong-field CN⁻ pairs the two $e_g$-type electrons; one $3d$ orbital vacated → $dsp^2$ hybridization → square planar, all paired, $\mu = 0$ (diamagnetic). (3)
• $[\mathrm{NiCl_4}]^{2-}$: weak-field Cl⁻; $sp^3$ hybridization → tetrahedral, two unpaired electrons, $\mu=\sqrt{2\cdot4}=\sqrt8=2.83\,\mu_B$ (paramagnetic). (3) Geometry difference: strong-field ligand supplies enough energy to force pairing and adopt low-energy square-planar $dsp^2$; weak field cannot, so $sp^3$ tetrahedral is retained.

(b) (6 marks) EAN $= Z(\text{metal}) - (\text{oxidation state}) + 2\times(\text{coordination number})$.

• $[\mathrm{Co(NH_3)_6}]^{3+}$: Co $Z=27$, ox = +3, CN = 6. EAN $= 27 - 3 + 12 = 36$ (Kr). Obeys 18-e rule (36 = Kr closed shell). (3)
• $[\mathrm{Fe(CN)_6}]^{4-}$: Fe $Z=26$, ox = +2, CN = 6. EAN $= 26 - 2 + 12 = 36$ (Kr). Obeys EAN rule. (2) Implication: attainment of noble-gas configuration correlates with high kinetic (substitution) inertness—both are kinetically inert, low-spin complexes. (1)

(c) (6 marks) $\mathrm{Cu^{2+}} = d^9$: $t_{2g}^6 e_g^3$. The $e_g$ set ($d_{z^2}, d_{x^2-y^2}$) is degenerate but unequally occupied (one orbital doubly, one singly filled) → Jahn–Teller active. (2) Because $e_g$ orbitals point directly at ligands, distortion is large and observable. Elongation along $z$ lowers $d_{z^2}$ and raises $d_{x^2-y^2}$. Put the doubly-occupied orbital in the lowered ($d_{z^2}$) level and singly in the raised one. (2) Energy: with splitting $2\delta$ ($d_{z^2}$ down by $\delta$, $d_{x^2-y^2}$ up by $\delta$), occupancy $d_{z^2}^2 d_{x^2-y^2}^1$ gives net energy change $= 2(-\delta) + 1(+\delta) = -\delta < 0$. Hence elongation is stabilizing → axial elongation predicted. (2)

[
  {"claim":"HS Fe(II) d6 CFSE = -0.4 Delta_o (field term)","code":"t=4; e=2; result = (-0.4*t + 0.6*e) == -0.4"},
  {"claim":"LS Fe(II) d6 CFSE = -2.4 Delta_o","code":"t=6; e=0; result = (-0.4*t + 0.6*e) == -2.4"},
  {"claim":"spin-only mu for 4 unpaired = sqrt(24)","code":"n=4; result = sqrt(n*(n+2)) == sqrt(24)"},
  {"claim":"EAN of Co in [Co(NH3)6]3+ = 36","code":"Z=27; ox=3; CN=6; result = (Z - ox + 2*CN) == 36"},
  {"claim":"EAN of Fe in [Fe(CN)6]4- = 36","code":"Z=26; ox=2; CN=6; result = (Z - ox + 2*CN) == 36"},
  {"claim":"mu for tetrahedral NiCl4^2- (2 unpaired) = sqrt(8)","code":"n=2; result = sqrt(n*(n+2)) == sqrt(8)"},
  {"claim":"Jahn-Teller d9 elongation net energy = -delta","code":"delta=symbols('delta'); result = simplify(2*(-delta)+1*(delta) + delta) == 0"}
]