t₂g and eₐ orbitals. CFSE. High spin vs low spin complexes.

Crystal Field Theory — Splitting in Octahedral Complex

Question

An octahedral complex $[\text{Fe}(\text{CN})_6]^{3-}$ has a measured Crystal Field Stabilisation Energy (CFSE) of $-2\Delta_o$ . Determine the $d$ -electron configuration, identify whether the complex is high spin or low spin, and calculate the CFSE for $[\text{FeF}_6]^{3-}$ given that $\text{F}^-$ is a weak field ligand.

(Adapted from JEE Advanced 2024, Paper 1)

Solution — Step by Step

Fe in $[\text{Fe}(\text{CN})_6]^{3-}$ is Fe(III), giving us $d^5$ configuration. Five $d$ -electrons must be distributed — the key question is whether they pair up in $t_{2g}$ or spread across $e_g$ .

In an octahedral field, the five $d$ -orbitals split into two sets:

$t_{2g}$ (lower energy): $d_{xy}$ , $d_{xz}$ , $d_{yz}$ — these three orbitals point between the axes
$e_g$ (higher energy): $d_{z^2}$ , $d_{x^2-y^2}$ — these point directly at the ligands

The energy gap between them is $\Delta_o$ . Each $t_{2g}$ electron contributes $-0.4\Delta_o$ to CFSE; each $e_g$ electron contributes $+0.6\Delta_o$ .

We’re told CFSE $= -2\Delta_o$ . Let’s set up the equation with $x$ electrons in $t_{2g}$ and $y$ electrons in $e_g$ , where $x + y = 5$ :

\text{CFSE} = -0.4x\Delta_o + 0.6y\Delta_o = -2\Delta_o

Substituting $y = 5 - x$ :

-0.4x + 0.6(5-x) = -2

-0.4x + 3 - 0.6x = -2

-x = -5 \implies x = 5, \quad y = 0

So the configuration is $t_{2g}^5\, e_g^0$ . This is low spin.

$\text{CN}^-$ is a strong field ligand — it sits at the top of the spectrochemical series. The large $\Delta_o$ it generates exceeds the pairing energy, so all five electrons crowd into $t_{2g}$ rather than spread out. This confirms our calculation.

$\text{F}^-$ is a weak field ligand. With small $\Delta_o <$ pairing energy, the complex goes high spin — electrons follow Hund’s rule: $t_{2g}^3\, e_g^2$ .

\text{CFSE} = 3(-0.4\Delta_o') + 2(+0.6\Delta_o') = -1.2\Delta_o' + 1.2\Delta_o' = \mathbf{0}

The CFSE for $[\text{FeF}_6]^{3-}$ is zero. A $d^5$ high spin complex has no net stabilisation.

Why This Works

The entire logic of CFT rests on electrostatic repulsion. Ligands approaching along axes repel $d$ -electrons in orbitals that point along those axes ( $e_g$ ) more than those pointing between axes ( $t_{2g}$ ). This creates the energy gap $\Delta_o$ .

Whether a $d^5$ complex is high or low spin depends on a competition: pairing energy ( $P$ ) vs crystal field splitting ( $\Delta_o$ ). Strong field ligands like $\text{CN}^-$ , $\text{CO}$ , and $\text{NO}_2^-$ make $\Delta_o > P$ , forcing pairing. Weak field ligands like $\text{F}^-$ , $\text{Cl}^-$ , and $\text{H}_2\text{O}$ give $\Delta_o < P$ , so electrons spread out.

The $d^5$ high spin case is special — it’s the only configuration where CFSE comes out exactly zero. This is why $\text{Mn}^{2+}$ and $\text{Fe}^{3+}$ complexes with weak field ligands are exceptionally stable despite having no CFSE benefit.

Alternative Method — Using Electron Counting Directly

Rather than solving algebraically, you can work backwards from the CFSE value.

For a $d^5$ system, the possible configurations and their CFSE values are:

Configuration	Spin	CFSE
$t_{2g}^3\, e_g^2$	High spin	$0$
$t_{2g}^4\, e_g^1$	—	$-0.6\Delta_o$
$t_{2g}^5\, e_g^0$	Low spin	$-2\Delta_o$

Given CFSE $= -2\Delta_o$ , the answer is immediately $t_{2g}^5\, e_g^0$ by direct lookup. In JEE, this table approach is faster under time pressure.

Memorise CFSE values for $d^4$ through $d^7$ configurations in both high and low spin. JEE Advanced frequently gives you a CFSE value and asks you to back-calculate the configuration or predict magnetic behaviour. The table takes 10 minutes to memorise and saves 3–4 minutes per question.

Common Mistake

Students calculate CFSE as $-2\Delta_o$ for $t_{2g}^5\, e_g^0$ correctly, then write the magnetic moment as $\mu = \sqrt{5(5+2)} = 5.92\text{ BM}$ — which is the high spin value for $d^5$ .

The low spin $t_{2g}^5\, e_g^0$ configuration has one unpaired electron, giving $\mu = \sqrt{1(1+2)} = 1.73\text{ BM}$ . Confusing spin state after correctly identifying the configuration is an extremely common trap in JEE Advanced integer-type questions. Always re-count unpaired electrons from your final configuration, not from the free ion.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Electron Counting Directly

Common Mistake

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