1. Find the metal ion and its d-electron count in each complex

For all complexes, first determine oxidation state and hence the $d^n$ configuration of the metal ion.

• $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$

  Ethylenediamine $(\mathrm{en})$ is a neutral ligand. So cobalt is in $+3$ oxidation state.

  $\mathrm{Co}: [\mathrm{Ar}]3d^74s^2$

  $\mathrm{Co}^{3+} \Rightarrow 3d^6$

• $\left[\mathrm{CoF}_6\right]^{3-}$

  $6\mathrm{F}^- = -6$, overall charge $=-3$

  So cobalt is again $+3$. Hence $\mathrm{Co}^{3+} = 3d^6$.

• $\left[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}$

  Water is neutral, so manganese is $+2$.

  $\mathrm{Mn}: [\mathrm{Ar}]3d^54s^2$

  $\mathrm{Mn}^{2+} \Rightarrow 3d^5$

• $\left[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}$

  Zinc is $+2$.

  $\mathrm{Zn}: [\mathrm{Ar}]3d^{10}4s^2$

  $\mathrm{Zn}^{2+} \Rightarrow 3d^{10}$

2. Determine whether each octahedral complex is high spin or low spin

• $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$

  $\mathrm{en}$ is a relatively strong-field ligand, and $\mathrm{Co}^{3+}$ also favors pairing.

  So this is low-spin octahedral: $t_{2g}^6e_g^0$

• $\left[\mathrm{CoF}_6\right]^{3-}$

  $\mathrm{F}^-$ is a weak-field ligand.

  So this is high-spin octahedral $d^6$: $t_{2g}^4e_g^2$

• $\left[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}$

  $\mathrm{H_2O}$ is weak/intermediate field, and $\mathrm{Mn}^{2+}$ usually forms high-spin complexes.

  High-spin $d^5$: $t_{2g}^3e_g^2$

• $\left[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}$

  $d^{10}$ octahedral: $t_{2g}^6e_g^4$

3. Calculate CFSE for each configuration

For octahedral complexes, $\text{CFSE} = (-0.4\Delta_o)(n_{t_{2g}}) + (+0.6\Delta_o)(n_{e_g})$

Now evaluate:

• Option A: $t_{2g}^6e_g^0$ $\text{CFSE} = 6(-0.4\Delta_o)= -2.4\Delta_o$ Magnitude $=2.4\Delta_o$

• Option B: $t_{2g}^3e_g^2$ $\text{CFSE} = 3(-0.4\Delta_o)+2(+0.6\Delta_o)=-1.2\Delta_o+1.2\Delta_o=0$

• Option C: $t_{2g}^4e_g^2$ $\text{CFSE} = 4(-0.4\Delta_o)+2(+0.6\Delta_o)=-1.6\Delta_o+1.2\Delta_o=-0.4\Delta_o$ Magnitude $=0.4\Delta_o$

• Option D: $t_{2g}^6e_g^4$ $\text{CFSE} = 6(-0.4\Delta_o)+4(+0.6\Delta_o)=-2.4\Delta_o+2.4\Delta_o=0$

4. Compare the CFSE values

The greatest stabilization corresponds to the most negative CFSE (largest magnitude of stabilization):

• A: $-2.4\Delta_o$
• B: $0$
• C: $-0.4\Delta_o$
• D: $0$

Hence, the highest CFSE is for $t_{2g}^6e_g^0$

5. Final answer

The complex with highest CFSE is $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$, whose d-orbital configuration is: $t_{2g}^6e_g^0$ So, Option A is correct.