The crystal field stabilization energy (CFSE) | vedclass

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(A) $1$. For $[Fe(H_2O)_6]Cl_2$: The central metal ion is $Fe^{2+}$,which has a $d^6$ configuration. $H_2O$ is a weak field ligand,so the configuratio

Vedclass · Accepted Answer

$ -0.4 \, \Delta_o$ and $ -0.8 \, \Delta_t$

Vedclass · Answer

(A) $1$. For $[Fe(H_2O)_6]Cl_2$: The central metal ion is $Fe^{2+}$,which has a $d^6$ configuration. $H_2O$ is a weak field ligand,so the configuration is $t_{2g}^4 e_g^2$. The $CFSE$ is calculated as: $CFSE = [-0.4 	imes 4 + 0.6 	imes 2] \, \Delta_o = [-1.6 + 1.2] \, \Delta_o = -0.4 \, \Delta_o$.$2$. For $K_2[NiCl_4]$: The central metal ion is $Ni^{2+}$,which has a $d^8$ configuration. It forms a tetrahedral complex $[NiCl_4]^{2-}$. The configuration is $e^4 t_2^4$. The $CFSE$ for a tetrahedral complex is calculated as: $CFSE = [-0.6 	imes n_e + 0.4 	imes n_{t2}] \, \Delta_t = [-0.6 	imes 4 + 0.4 	imes 4] \, \Delta_t = [-2.4 + 1.6] \, \Delta_t = -0.8 \, \Delta_t$.$3$. Therefore,the values are $-0.4 \, \Delta_o$ and $-0.8 \, \Delta_t$.

The crystal field stabilization energy $(CFSE)$ of $[Fe(H_2O)_6]Cl_2$ and $K_2[NiCl_4]$,respectively,are

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