Value of CFSE,in a tetrahedral complex having | vedclass

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(A) In a tetrahedral complex,the $d$-orbitals split into two sets: $e$ (lower energy) and $t_2$ (higher energy).For a $d^4$ configuration,the electron

Vedclass · Accepted Answer

$- \frac{2}{5} \Delta_t$

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(A) In a tetrahedral complex,the $d$-orbitals split into two sets: $e$ (lower energy) and $t_2$ (higher energy).For a $d^4$ configuration,the electrons fill the orbitals according to Hund's rule.The energy of the $e$ set is $-0.6 \Delta_t$ and the $t_2$ set is $+0.4 \Delta_t$.For $d^4$,the configuration is $e^2 t_2^2$.$CFSE = (2 	imes -0.6 \Delta_t) + (2 	imes +0.4 \Delta_t) = -1.2 \Delta_t + 0.8 \Delta_t = -0.4 \Delta_t$.Since $-0.4 \Delta_t = -\frac{2}{5} \Delta_t$,the correct value is $-\frac{2}{5} \Delta_t$.

Value of $CFSE$,in a tetrahedral complex having $3d^4$ configuration of metal ion,surrounded by weak field ligands,will be

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