Question:

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 :

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CFSE is largest for the strong-field low-spin d6 complex.
Updated On: Oct 26, 2026
  • $t_{2 g}{ }^6 e_g^0$
  • $t_{2 g}{ }^3 e_g^2$
  • $\mathrm{t}_{2 \mathrm{~g}}{ }^4 \mathrm{e}_{\mathrm{g}}{ }^2$
  • $t_{2 g}{ }^6 e_g^4$
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The Correct Option is A

Solution and Explanation

$[Co(en)_3]^{3+}$: $Co^{3+}$ is $d^6$ and en is a strong-field ligand, so $t_{2g}^6 e_g^0$ and $CFSE = -2.4\Delta_o$ (largest).
$[CoF_6]^{3-}$: weak-field $F^-$, $t_{2g}^4 e_g^2$, $CFSE = -0.4\Delta_o$.
$[Mn(H_2O)_6]^{2+}$: $d^5$ high spin $t_{2g}^3 e_g^2$, $CFSE = 0$.
$[Zn(H_2O)_6]^{2+}$: $d^{10}$, $CFSE = 0$.
The highest CFSE belongs to the configuration $t_{2g}^6 e_g^0$: option (1).
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