JEE Mains Chemistry · Coordination Compounds

High and Low Spin Configurations and CFSE

Electrons fill t₂g before eg, and pair up in t₂g only when the splitting beats the pairing energy; the resulting configuration fixes the unpaired electrons and the crystal field stabilisation energy, −0.4Δₒ for each t₂g electron and +0.6Δₒ for each eg electron.

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Why this matters

Thirty-four PYQs, twenty-five of them multiple choice, and eight from 2026. Fifteen write the t₂g/eg configuration of a complex and count its unpaired electrons or its electrons in one set; five do the same for a tetrahedral complex; fourteen compute or compare crystal field stabilisation energies.

Concept 1 of 3: High-spin and low-spin octahedral configurations

The first three electrons go one each into the three t₂g orbitals. The fourth has a choice: pair up in t₂g, paying the pairing energy P, or climb to eg, paying

\Delta_o

. It takes the cheaper route. A strong-field ligand makes

\Delta_o > P

, so electrons pair in t₂g (low spin); a weak-field ligand makes

\Delta_o < P

, so they spread out (high spin).

Definition

$\Delta_o > P$ : low spin. $\Delta_o < P$ : high spin.
d¹, d², d³, d⁸, d⁹, d¹⁰ have only one octahedral configuration. The choice exists only for d⁴ to d⁷.
High spin: d⁴ $t_{2g}^3e_g^1$ (4 unpaired), d⁵ $t_{2g}^3e_g^2$ (5), d⁶ $t_{2g}^4e_g^2$ (4), d⁷ $t_{2g}^5e_g^2$ (3).
Low spin: d⁴ $t_{2g}^4$ (2 unpaired), d⁵ $t_{2g}^5$ (1), d⁶ $t_{2g}^6$ (0), d⁷ $t_{2g}^6e_g^1$ (1).
Cobalt(II) with ammonia and air is oxidised to diamagnetic $\mathrm{[Co(NH_3)_6]^{3+}}$ , $t_{2g}^6$ .
Jahn–Teller: d⁹ Cu²⁺ has an unevenly filled eg set and distorts; the distortion is largest when the ligand set itself is uneven, as in trans- $\mathrm{[Cu(en)_2Cl_2]}$ .

Spin state criterion

\Delta_o > P \Rightarrow \text{low spin } (t_{2g} \text{ filled first}) \qquad \Delta_o < P \Rightarrow \text{high spin}

Worked example

Write the high-spin and the low-spin octahedral configurations of Co²⁺ and give the unpaired electrons in each.

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JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q26Moderate

Example 1 · Coordination Compounds · High and Low Spin Configurations and CFSE

Consider the transition metal ions

Mn^{3 +},Cr^{3 +},Fe^{3 +}

and

Co^{3 +}

and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is

t₂g³eg¹ is the weak-field configuration

Putting the fourth electron in eg means the pairing energy was larger than

\Delta_o

: a weak-field ligand and a high-spin complex. The strong-field d⁴ configuration is

t_{2g}^4

Pairs and unpaired electrons are different counts

Low-spin

\mathrm{[Fe(CN)_6]^{4-}}

has six t₂g electrons: 3 PAIRS and 0 unpaired. Read whether the question wants electrons, pairs or unpaired electrons.

Drill 14 more on high-spin and low-spin octahedral configurations

Concept 2 of 3: Tetrahedral configurations and their CFSE

In a tetrahedron the lower set is e (two orbitals) and the upper set is t₂ (three). The gap is only 4/9 of the octahedral one, too small to force pairing, so every tetrahedral complex is high spin. Fill e singly, then t₂ singly, then pair e, then pair t₂.

Definition

Order of filling: e¹, e², then t₂¹ to t₂³, then e³, e⁴, then t₂⁴ to t₂⁶.
CFSE $= (-0.6\,n_e + 0.4\,n_{t_2})\Delta_t$ .
Examples: d⁰ $\mathrm{TiCl_4}$ , $\mathrm{[MnO_4]^-}$ ; d¹ $\mathrm{[MnO_4]^{2-}}$ ; d² $\mathrm{[FeO_4]^{2-}}$ ; d⁵ $\mathrm{[FeCl_4]^-}$ , $\mathrm{[MnBr_4]^{2-}}$ ; d⁶ $\mathrm{[FeCl_4]^{2-}}$ ; d⁷ $\mathrm{[CoCl_4]^{2-}}$ ; d⁸ $\mathrm{[NiCl_4]^{2-}}$ , $\mathrm{[Ni(PPh_3)_2Cl_2]}$ ; d¹⁰ $\mathrm{Ni(CO)_4}$ .
A tetrahedral CFSE is in $\Delta_t$ , not $\Delta_o$ : d⁸ gives $-0.8\Delta_t$ .

d count	Configuration	Unpaired electrons	CFSE
d⁰	$e^0t_2^0$	0	0
d¹	$e^1t_2^0$	1	$-0.6\Delta_t$
d²	$e^2t_2^0$	2	$-1.2\Delta_t$
d³	$e^2t_2^1$	3	$-0.8\Delta_t$
d⁴	$e^2t_2^2$	4	$-0.4\Delta_t$
d⁵	$e^2t_2^3$	5	0
d⁶	$e^3t_2^3$	4	$-0.6\Delta_t$
d⁷	$e^4t_2^3$	3	$-1.2\Delta_t$
d⁸	$e^4t_2^4$	2	$-0.8\Delta_t$
d¹⁰	$e^4t_2^6$	0	0

Tetrahedral complexes are always high spin, so each d count has exactly one row.

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JEE Mains · 2023 · 24 Jan 2023 · Q59Moderate

Example 2 · Coordination Compounds · High and Low Spin Configurations and CFSE

The d-electronic configuration of

\left\lbrack CoCl_{4} \right\rbrack^{2 -}

in tetrahedral crystal field is

e^{m}t_{2}^{n}

. Sum of "m" and "number of unpaired electrons" is

In a tetrahedron, e is filled first

The lower set is e, not t₂. Writing a tetrahedral d⁷ ion as

t_2^4e^3

copies the octahedral order and gets the configuration and the CFSE wrong. It is

e^4t_2^3

, CFSE

-1.2\Delta_t

A tetrahedral CFSE is measured in Δt

Paramagnetic

\mathrm{[Ni(PPh_3)_2Cl_2]}

is tetrahedral, so its CFSE is

-0.8\Delta_t

, not

-0.8\Delta_o

. A statement giving it in

\Delta_o

is incorrect.

Drill 4 more on tetrahedral configurations and their cfse

Concept 3 of 3: Crystal field stabilisation energy of octahedral complexes

Each electron in a lower t₂g orbital saves

0.4\Delta_o

; each in an upper eg orbital costs

0.6\Delta_o

. Add them up and you have the crystal field stabilisation energy. It is largest for low-spin d⁶, where six electrons all sit low, and zero for high-spin d⁵ and for d¹⁰, where the gains and costs cancel.

Definition

$\text{CFSE} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\Delta_o$ , plus $mP$ for each extra pair forced by low spin if the question asks for it.
High spin: d¹ −0.4, d² −0.8, d³ −1.2, d⁴ −0.6, d⁵ 0, d⁶ −0.4, d⁷ −0.8, d⁸ −1.2, d⁹ −0.6, d¹⁰ 0 (in $\Delta_o$ ).
Low spin: d⁴ −1.6, d⁵ −2.0, d⁶ −2.4, d⁷ −1.8.
To compare real complexes, the CFSE also scales with $\Delta_o$ : stronger ligand, higher metal charge and chelation all raise it. $\mathrm{[Co(en)_3]^{3+} > [Co(NH_3)_6]^{3+}}$ .
Working backwards: CFSE $-0.8\Delta_o$ with 3 unpaired electrons is high-spin d⁷ ( $t_{2g}^5e_g^2$ ), as in Co²⁺.

Octahedral CFSE

\text{CFSE} = \left(-0.4\,n_{t_{2g}} + 0.6\,n_{e_g}\right)\Delta_o

Worked example

Find the CFSE, in units of

\Delta_o

, of

\mathrm{[V(H_2O)_6]^{3+}}

and of low-spin

\mathrm{[Mn(CN)_6]^{3-}}

, ignoring pairing energy.

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The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q102Moderate

Example 3 · Coordination Compounds · High and Low Spin Configurations and CFSE

The d-orbital electronic configuration of the complex among

\left\lbrack Co(en)_{3} \right\rbrack^{3 +},\ \left\lbrack CoF_{6} \right\rbrack^{3 -}

\left\lbrack Mn\left( H_{2}O \right)_{6} \right\rbrack^{2 +}

and

\left\lbrack Zn\left( H_{2}O \right)_{6} \right\rbrack^{2 +}

that has the highest CFSE is :

CFSE is not the splitting energy

For

\mathrm{[Ti(H_2O)_6]^{3+}}

(d¹) the CFSE is

-0.4\Delta_o

, so

\Delta_o

is 2.5 times the CFSE magnitude. The light absorbed matches

\Delta_o

, not the CFSE.

Zero CFSE means high-spin d⁵ or d¹⁰

A complex with 'CFSE = 0' and a moment near 5.9 BM has five unpaired electrons: high-spin d⁵, such as Mn²⁺ or Fe³⁺ with a weak ligand like SCN⁻. The splitting itself is not zero.

Rules of thumb can clash with the numbers

A 2021 key ordered CFSE as

\mathrm{[Co(H_2O)_6]^{2+} < [CoF_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(en)_3]^{3+}}

, putting the +2 ion lowest by charge. In

\Delta_o

units the Co²⁺ aqua ion is

-0.8\Delta_o

and

\mathrm{[CoF_6]^{3-}}

only

-0.4\Delta_o

. Use the charge rule when comparing the SAME configuration.

Drill 13 more on crystal field stabilisation energy of octahedral complexes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

High-spin and low-spin octahedral configurations
Spin state criterion
$\Delta_o > P \Rightarrow \text{low spin } (t_{2g} \text{ filled first}) \qquad \Delta_o < P \Rightarrow \text{high spin}$
Crystal field stabilisation energy of octahedral complexes
Octahedral CFSE
$\text{CFSE} = \left(-0.4\,n_{t_{2g}} + 0.6\,n_{e_g}\right)\Delta_o$

Reference tables (1)

Tetrahedral configurations and their CFSE10 rows

d count	Configuration	Unpaired electrons	CFSE
d⁰	$e^0t_2^0$	0	0
d¹	$e^1t_2^0$	1	$-0.6\Delta_t$
d²	$e^2t_2^0$	2	$-1.2\Delta_t$
d³	$e^2t_2^1$	3	$-0.8\Delta_t$
d⁴	$e^2t_2^2$	4	$-0.4\Delta_t$
d⁵	$e^2t_2^3$	5	0
d⁶	$e^3t_2^3$	4	$-0.6\Delta_t$
d⁷	$e^4t_2^3$	3	$-1.2\Delta_t$
d⁸	$e^4t_2^4$	2	$-0.8\Delta_t$
d¹⁰	$e^4t_2^6$	0	0

Tetrahedral complexes are always high spin, so each d count has exactly one row.

Watch out for (7)

t₂g³eg¹ is the weak-field configuration
→ High-spin and low-spin octahedral configurations
Pairs and unpaired electrons are different counts
→ High-spin and low-spin octahedral configurations
In a tetrahedron, e is filled first
→ Tetrahedral configurations and their CFSE
A tetrahedral CFSE is measured in Δt
→ Tetrahedral configurations and their CFSE
CFSE is not the splitting energy
→ Crystal field stabilisation energy of octahedral complexes
Zero CFSE means high-spin d⁵ or d¹⁰
→ Crystal field stabilisation energy of octahedral complexes
Rules of thumb can clash with the numbers
→ Crystal field stabilisation energy of octahedral complexes

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