Which of the following has zero Crystal Field | vedclass

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(A) The $CFSE$ for an octahedral complex is calculated as: $CFSE = (-0.4 	imes n_{t2g} + 0.6 	imes n_{eg}) \Delta_o$.For $Mn^{2+}$ $(d^5)$ with a we

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$Mn^{2+}$ in the presence of a weak field ligand such as $H_2O$

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(A) The $CFSE$ for an octahedral complex is calculated as: $CFSE = (-0.4 	imes n_{t2g} + 0.6 	imes n_{eg}) \Delta_o$.For $Mn^{2+}$ $(d^5)$ with a weak field ligand: The configuration is $t_{2g}^3 e_g^2$. $CFSE = (-0.4 	imes 3 + 0.6 	imes 2) \Delta_o = (-1.2 + 1.2) \Delta_o = 0$.For $Co^{3+}$ $(d^6)$ with a strong field ligand: The configuration is $t_{2g}^6 e_g^0$. $CFSE = (-0.4 	imes 6 + 0.6 	imes 0) \Delta_o = -2.4 \Delta_o$.For $Fe^{3+}$ $(d^5)$ with a strong field ligand: The configuration is $t_{2g}^5 e_g^0$. $CFSE = (-0.4 	imes 5 + 0.6 	imes 0) \Delta_o = -2.0 \Delta_o$.For $Ni^{2+}$ $(d^8)$ with a weak field ligand: The configuration is $t_{2g}^6 e_g^2$. $CFSE = (-0.4 	imes 6 + 0.6 	imes 2) \Delta_o = (-2.4 + 1.2) \Delta_o = -1.2 \Delta_o$.Thus,$Mn^{2+}$ has a $CFSE$ of $0$.

$I$. $NCS^{-}$	$II$. $S^{2-}$
$III$. $en$	$IV$. $SCN^{-}$

Which of the following has zero Crystal Field Stabilization Energy $(CFSE)$ value in an octahedral field?

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