1. Find the oxidation state and d-electron count of Fe

Given complex: $\left[\mathrm{Fe}(\mathrm{NH}_3)_x(\mathrm{CN})_y\right]^-$

Let oxidation state of Fe be $n$.

• $\mathrm{NH}_3$ is a neutral ligand.
• $\mathrm{CN}^-$ has charge $-1$.
• Overall charge on complex is $-1$.

So, $n + 0\cdot x - y = -1$ $n = y-1$

Since the metal is iron(III), we are told Fe is in the $+3$ oxidation state: $n=+3$ Thus, $y-1=3 \Rightarrow y=4$

Iron has atomic number $26$.

Fe$^{3+}$ electronic configuration: $\mathrm{Fe}: [\mathrm{Ar}]3d^64s^2$ $\mathrm{Fe}^{3+}: [\mathrm{Ar}]3d^5$

So it is a $d^5$ system.

2. Use the condition: no electron in $e_g$ orbital

In an octahedral crystal field, the $d$ orbitals split as:

• lower: $t_{2g}$
• higher: $e_g$

For a $d^5$ ion to have no electron in $e_g$, all 5 electrons must be in $t_{2g}$: $t_{2g}^5 e_g^0$

This is possible only for a low-spin octahedral complex.

Since $\mathrm{CN}^-$ is a strong-field ligand, it can produce pairing and low-spin configuration.

The very mention of $t_{2g}$ and $e_g$ splitting implies an octahedral coordination number 6. Thus, $x+y=6$

3. Check consistency

We already found: $y=4$ Therefore, $x=6-4=2$

Complex would be: $[\mathrm{Fe}(\mathrm{NH}_3)_2(\mathrm{CN})_4]^-$ which is octahedral and can be low-spin due to strong-field $\mathrm{CN}^-$ ligands.

4. Evaluate options

• A: 6 ✅
• B: 4 ❌
• C: 3 ❌
• D: 5 ❌

So the correct answer is: $\boxed{6}$