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(B) The $CFSE$ is calculated using the formula: $CFSE = [(-0.4 	imes n_{t_{2g}}) + (0.6 	imes n_{e_g})] \Delta_0$.For a complex to have zero $CFSE$,

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$[Fe(H_2O)_6]^{3+}$

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(B) The $CFSE$ is calculated using the formula: $CFSE = [(-0.4 	imes n_{t_{2g}}) + (0.6 	imes n_{e_g})] \Delta_0$.For a complex to have zero $CFSE$,the configuration must be such that the stabilization energy of electrons in $t_{2g}$ orbitals is exactly cancelled by the destabilization energy of electrons in $e_g$ orbitals.Let's analyze the electronic configurations of the metal ions in the given complexes (all are high-spin complexes with $H_2O$ as a weak field ligand):$A) [Mn(H_2O)_6]^{3+}: Mn^{3+} = 3d^4$. Configuration: $t_{2g}^3 e_g^1$. $CFSE = [(-0.4 	imes 3) + (0.6 	imes 1)] \Delta_0 = [-1.2 + 0.6] \Delta_0 = -0.6 \Delta_0$.$B) [Fe(H_2O)_6]^{3+}: Fe^{3+} = 3d^5$. Configuration: $t_{2g}^3 e_g^2$. $CFSE = [(-0.4 	imes 3) + (0.6 	imes 2)] \Delta_0 = [-1.2 + 1.2] \Delta_0 = 0$.$C) [Co(H_2O)_6]^{2+}: Co^{2+} = 3d^7$. Configuration: $t_{2g}^5 e_g^2$. $CFSE = [(-0.4 	imes 5) + (0.6 	imes 2)] \Delta_0 = [-2.0 + 1.2] \Delta_0 = -0.8 \Delta_0$.$D) [Co(H_2O)_6]^{3+}: Co^{3+} = 3d^6$. Configuration: $t_{2g}^4 e_g^2$. $CFSE = [(-0.4 	imes 4) + (0.6 	imes 2)] \Delta_0 = [-1.6 + 1.2] \Delta_0 = -0.4 \Delta_0$.Thus,$[Fe(H_2O)_6]^{3+}$ has zero $CFSE$.

Among the following complexes,the one which shows zero Crystal Field Stabilization Energy $(CFSE)$ is

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