The d-orbital electronic configuration of the complex among $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+},\left[\mathrm{CoF}_6\right]^{3-}$, $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ that has the highest CFSE is :

<p>Identify the oxidation state and d-electron count for each complex:</p> <p><p>For $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$:</p> <p><p>Ethylenediamine (en) is neutral.</p></p> <p><p>Oxidation state of cobalt is +3.</p></p> <p><p>Cobalt (atomic number 27) in the +3 state gives a $d^6$ configuration.</p></p> <p><p>With strong/moderate field ligands like en, the complex is low spin, leading to a configuration of $t_{2g}^6 e_g^0$.</p></p></p> <p><p>For $\left[\mathrm{CoF}_6\right]^{3-}$:</p> <p><p>Each fluoride ion is $\mathrm{F}^-$ so, with six of them, Co must be in the +3 state again.</p></p> <p><p>However, since fluoride is a weak field ligand, this complex adopts a high spin configuration for $d^6$, resulting in $t_{2g}^4 e_g^2$.</p></p></p> <p><p>For $\left[\mathrm{Mn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$:</p> <p><p>Water is neutral.</p></p> <p><p>Manganese in the +2 state gives a $d^5$ configuration.</p></p> <p><p>With weak field water, the $d^5$ configuration is high spin: $t_{2g}^3 e_g^2$.</p></p></p> <p><p>For $\left[\mathrm{Zn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$:</p> <p><p>Zinc in the +2 state has a $d^{10}$ configuration.</p></p> <p><p>In an octahedral field, all 10 electrons fill the orbitals as $t_{2g}^6 e_g^4$.</p></p></p> <p><p>Calculate the crystal field stabilization energy (CFSE) for each case:</p> <p>The CFSE in an octahedral field can be estimated as:</p> <p>$\mathrm{CFSE} = (n_{t_{2g}}\times -0.4\Delta_o) + (n_{e_g}\times 0.6\Delta_o)$</p></p> <p><p>For $t_{2g}^6 e_g^0$ (low-spin $d^6$ in $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$):</p> <p>$\mathrm{CFSE} = 6(-0.4\Delta_o) + 0(0.6\Delta_o) = -2.4\Delta_o$</p></p> <p><p>For $t_{2g}^4 e_g^2$ (high-spin $d^6$ in $\left[\mathrm{CoF}_6\right]^{3-}$):</p> <p>$$\mathrm{CFSE} = 4(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.6\Delta_o + 1.2\Delta_o = -0.4\Delta_o$$</p></p> <p><p>For $t_{2g}^3 e_g^2$ (high-spin $d^5$ in $\left[\mathrm{Mn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$):</p> <p>$$\mathrm{CFSE} = 3(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.2\Delta_o + 1.2\Delta_o = 0$$</p></p> <p><p>For $t_{2g}^6 e_g^4$ (for $d^{10}$ in $\left[\mathrm{Zn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$):</p> <p>$$\mathrm{CFSE} = 6(-0.4\Delta_o) + 4(0.6\Delta_o) = -2.4\Delta_o + 2.4\Delta_o = 0$$</p></p> <p>Compare the CFSE values:</p> <p><p>$t_{2g}^6 e_g^0$ gives a CFSE of $-2.4\Delta_o$ (most negative, hence highest stabilization).</p></p> <p><p>The others result in less stabilization (or net zero).</p></p> <p><p>Conclusion:</p> <p>The complex with the highest CFSE is the one with the configuration $t_{2g}^6 e_g^0$, which corresponds to $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$.</p></p> <p>Thus, the answer is:</p> <p>Option A: $t_{2g}^6 e_g^0$.</p>