1. Find oxidation state and d-electron count

For $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}:$

Let oxidation state of Fe be $x$.

$x + 6(-1) = -3$ $x - 6 = -3$ $x = +3$

So the metal ion is $\mathrm{Fe}^{3+}$.

Atomic number of Fe = 26.

Neutral Fe: $[\mathrm{Ar}]\,3d^6 4s^2$

$\mathrm{Fe}^{3+}$ loses 2 electrons from 4s and 1 from 3d:

$\mathrm{Fe}^{3+} : [\mathrm{Ar}]\,3d^5$

So it is a $d^5$ system.

2. Decide high spin or low spin

Ligand $\mathrm{CN}^-$ is a strong field ligand, so the octahedral complex is low spin (inner orbital complex).

Thus the electron arrangement in octahedral field is:

$t_{2g}^5 e_g^0$

3. Calculate CFSE

In an octahedral field:

• each electron in $t_{2g}$ contributes $-0.4\Delta_0$
• each electron in $e_g$ contributes $+0.6\Delta_0$

Therefore,

$\text{CFSE} = 5(-0.4\Delta_0) + 0(+0.6\Delta_0)$

$\text{CFSE} = -2.0\Delta_0$

Ignoring pairing energy, the value is

$(-)2\Delta_0$

So the required integer is:

$\boxed{2}$

4. Comparison with stored answer

Stored correct answer = 2.

Our derived answer = 2.

They agree.