1. Find oxidation state and d-electron count of Fe in each complex

For octahedral complexes, CFSE is zero for configurations like high-spin $d^5$, $d^0$, or $d^{10}$.

We examine each option.

2. Option A: $\mathrm{K}_3[\mathrm{Fe}(\mathrm{SCN})_6]$

The complex ion is $[\mathrm{Fe}(\mathrm{SCN})_6]^{3-}$.

Let oxidation state of Fe be $x$.

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

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

Atomic number of Fe = 26

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

$\mathrm{SCN}^-$ is a weak field ligand here, so the complex is high-spin octahedral.

For high-spin $d^5$ octahedral:

$t_{2g}^3 e_g^2$

CFSE:

$3(-0.4\Delta_o) + 2(+0.6\Delta_o) = -1.2\Delta_o + 1.2\Delta_o = 0$

So CFSE = 0.

3. Option B: $[\mathrm{Fe}(\mathrm{en})_3]\mathrm{Cl}_3$

The complex cation is $[\mathrm{Fe}(\mathrm{en})_3]^{3+}$.

Since en is neutral,

$x = +3$

So Fe is again $\mathrm{Fe}^{3+} = d^5$.

en is a stronger ligand than $\mathrm{SCN}^-$ and generally gives a low-spin complex with $\mathrm{Fe}^{3+}$.

Low-spin $d^5$ octahedral:

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

CFSE:

$5(-0.4\Delta_o) = -2.0\Delta_o$

So CFSE is not zero.

4. Option C: $[\mathrm{Fe}(\mathrm{NH}_3)_6]\mathrm{Br}_2$

The complex cation is $[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}$.

Since $\mathrm{NH}_3$ is neutral,

$x = +2$

So Fe is $\mathrm{Fe}^{2+} = d^6$.

For octahedral $d^6$, whether high spin or low spin, CFSE is not zero:

• high spin: $t_{2g}^4 e_g^2$
• low spin: $t_{2g}^6 e_g^0$

Thus CFSE is not zero.

5. Option D: $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$

The complex ion is $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$.

Let oxidation state of Fe be $x$:

$x + 6(-1) = -4$ $x = +2$

So Fe is $\mathrm{Fe}^{2+} = d^6$.

$\mathrm{CN}^-$ is a strong field ligand, so this is low-spin octahedral:

$t_{2g}^6 e_g^0$

CFSE:

$6(-0.4\Delta_o) = -2.4\Delta_o$

So CFSE is not zero.

6. Conclusion

Only Option A has octahedral CFSE equal to zero.

$\boxed{\mathrm{K}_3[\mathrm{Fe}(\mathrm{SCN})_6]}$