GATE Coordination Chemistry — CFSE Calculations

By Aniket Bhardwaj · 3 September 2026 · GATE Chemistry

Crystal field stabilisation energy is a small piece of arithmetic that carries a lot of explanatory weight: it predicts spin state, rationalises the double-humped plot of hydration enthalpies across the first transition series, explains why some spinels are inverse, and tells you which complexes will distort. GATE Chemistry examines it as a direct calculation and as the reasoning step inside a longer question. This guide covers the calculation itself — carefully, including the pairing-energy term that most students drop.

The splitting pattern and the barycentre

In an octahedral field the five d orbitals split into a lower triply degenerate t_2g set (d_xy, d_yz, d_xz) and an upper doubly degenerate e_g set (d_z², d_x²−y²). The barycentre rule fixes the individual energies: the weighted average must stay at zero, so 3(−0.4) + 2(+0.6) = 0.

Octahedral: t_2g at −0.4Δ_o (= −4 Dq), e_g at +0.6Δ_o (= +6 Dq)
CFSE = (−0.4 n_t2g + 0.6 n_eg) Δ_o + (extra pairs) × P

Tetrahedral: e at −0.6Δ_t, t₂ at +0.4Δ_t, with Δ_t ≈ (4/9) Δ_o

In tetrahedral geometry the labels invert — the e set is now lower — and there is no subscript g because T_d has no inversion centre. Because Δ_t is only about 4/9 of Δ_o for the same metal and ligands, it is essentially never larger than the pairing energy, which is why tetrahedral complexes are effectively always high spin. That single sentence answers a surprising number of questions.

The pairing-energy term — do not skip it

"CFSE" strictly means the stabilisation from the splitting alone. But when a question asks you to decide between high spin and low spin, you must also count the extra electron–electron pairings that the low-spin arrangement forces, each costing P. Compare total energies, not bare CFSE values.

Which spin state for a d⁶ octahedral complex with Δ_o = 20 000 cm⁻¹ and P = 19 000 cm⁻¹?

High spin, t_2g⁴e_g²: CFSE = 4(−0.4) + 2(+0.6) = −1.6 + 1.2 = −0.4Δ_o = −8 000 cm⁻¹. Pairs formed: 1 (in t_2g).

Low spin, t_2g⁶: CFSE = 6(−0.4) = −2.4Δ_o = −48 000 cm⁻¹. Pairs formed: 3.

The low-spin form has 2 extra pairs, so its total is −48 000 + 2(19 000) = −10 000 cm⁻¹ against −8 000 cm⁻¹ for high spin.

Low spin wins by 2 000 cm⁻¹ (≈ 23.9 kJ/mol). Consistent with the quick rule: Δ_o > P → low spin.

Worked example — CFSE in real energy units

[Ti(H₂O)₆]³⁺ is d¹ with Δ_o = 20 300 cm⁻¹. Find the CFSE.

Configuration t_2g¹, so CFSE = −0.4 × 20 300 = −8 120 cm⁻¹.

Converting with 1 cm⁻¹ = 11.9627 J/mol:
8 120 × 11.9627 = 97 137 J/mol → −97.1 kJ/mol.

The 20 300 cm⁻¹ figure is itself read off the single d–d absorption band of this ion near 493 nm, which is why the complex is violet — it absorbs in the green-yellow.

[NiCl₄]²⁻ is tetrahedral d⁸. Find its CFSE.

Filling e then t₂ high spin: e⁴t₂⁴.
CFSE = 4(−0.6) + 4(+0.4) = −2.4 + 1.6 = −0.8 Δ_t

Expressed against the octahedral scale, −0.8 × (4/9) Δ_o = −0.356 Δ_o — far less than the −1.2 Δ_o an octahedral d⁸ would gain. That gap is why Ni(II) so strongly prefers six-coordination with most ligands, and why the tetrahedral chloride only forms with a bulky weak-field ligand.

The complete CFSE table

dⁿ	Oct. high spin	CFSE (Δ_o)	Oct. low spin	CFSE (Δ_o)	Tet. CFSE (Δ_t)
d¹	t_2g¹	−0.4	same	−0.4	−0.6
d²	t_2g²	−0.8	same	−0.8	−1.2
d³	t_2g³	−1.2	same	−1.2	−0.8
d⁴	t_2g³e_g¹	−0.6	t_2g⁴	−1.6	−0.4
d⁵	t_2g³e_g²	0	t_2g⁵	−2.0	0
d⁶	t_2g⁴e_g²	−0.4	t_2g⁶	−2.4	−0.6
d⁷	t_2g⁵e_g²	−0.8	t_2g⁶e_g¹	−1.8	−1.2
d⁸	t_2g⁶e_g²	−1.2	same	−1.2	−0.8
d⁹	t_2g⁶e_g³	−0.6	same	−0.6	−0.4
d¹⁰	t_2g⁶e_g⁴	0	same	0	0

Three configurations give zero CFSE in both geometries: d⁰, high-spin d⁵ and d¹⁰. Ions such as Ca²⁺, Mn²⁺, high-spin Fe³⁺ and Zn²⁺ therefore show no crystal-field preference at all, which is the key to the next section.

Where CFSE actually gets used

Hydration enthalpies. Plotted across Ca²⁺ to Zn²⁺, hydration enthalpy does not fall smoothly with ionic radius — it shows two humps with minima at d⁰, d⁵ and d¹⁰. Subtract the CFSE from each point and the curve straightens into the smooth line the simple electrostatic argument predicts. That correction is the evidence for crystal field theory.

Octahedral site preference energy (OSPE) = CFSE(oct) − CFSE(tet), and it decides spinel structures. In Fe₃O₄, Fe³⁺ is high-spin d⁵ with zero CFSE in either site, so it is indifferent; Fe²⁺ is d⁶ with CFSE(oct) = −0.4Δ_o and CFSE(tet) = −0.6Δ_t = −0.267Δ_o, giving OSPE = −0.133Δ_o. Fe²⁺ takes the octahedral hole and half the Fe³⁺ is pushed into tetrahedral sites — an inverse spinel. d³ and d⁸ ions have the largest OSPE and are the strongest octahedral seekers.

Jahn–Teller distortion. A degenerate ground state distorts to remove the degeneracy. Unequal occupation of the e_g set gives a strong distortion — high-spin d⁴ (t_2g³e_g¹), d⁹ (t_2g⁶e_g³) and low-spin d⁷. Unequal t_2g occupation gives only a weak distortion, because those orbitals point between the ligands. This is why Cu(II) complexes are so persistently tetragonally elongated.

Spectrochemical series — which end of Δ you are on

I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < NH₃ < en < bipy < phen < NO₂⁻ < CN⁻ < CO

Δ also rises down a group (3d < 4d < 5d, roughly 30–50% per row) and with oxidation state. So second- and third-row complexes are almost always low spin regardless of ligand.

Common mistakes

Dropping the pairing term when comparing spin states. Comparing −2.4Δ_o with −0.4Δ_o and concluding "low spin always" is wrong; the extra pairs cost real energy.
Counting total pairs instead of extra pairs. Only the pairings that the low-spin arrangement forces additionally enter the comparison.
Using +0.4/−0.6 for octahedral. The lower t_2g set is −0.4Δ_o; the inverted values belong to the tetrahedral case.
Applying low-spin filling to a tetrahedral complex. Δ_t is too small; treat T_d as high spin.
Forgetting the g subscript rules. t_2g/e_g in O_h, but e/t₂ in T_d — dropping this in a written answer is an easy mark to lose.
Mixing Dq and Δ_o. Δ_o = 10 Dq. A CFSE of −4 Dq and −0.4 Δ_o are the same thing; writing −4 Δ_o is a factor of ten out.

Convert your answer properly. CFSE questions routinely mix cm⁻¹, kJ/mol and eV, and the conversion factors (1 cm⁻¹ = 11.9627 J/mol; 1 eV = 96.485 kJ/mol) are where correct chemistry turns into a wrong final number. Do the conversion on the calculator, not in your head.

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