The value of CFSE in a tetrahedral complex ha | vedclass

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Question

(A) In a tetrahedral complex,the $d$-orbitals split into $e$ and $t_2$ sets,where the $e$ set is lower in energy and the $t_2$ set is higher in energy

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

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(A) In a tetrahedral complex,the $d$-orbitals split into $e$ and $t_2$ sets,where the $e$ set is lower in energy and the $t_2$ set is higher in energy.For a $d^4$ configuration,the electrons fill the orbitals according to the energy levels.Since tetrahedral complexes are generally high-spin (because $\Delta_t The $CFSE$ formula for tetrahedral complexes is $CFSE = [(-0.6 	imes n_e) + (0.4 	imes n_{t_2})] \Delta_t$.Substituting $n_e = 2$ and $n_{t_2} = 2$:$CFSE = [(-0.6 	imes 2) + (0.4 	imes 2)] \Delta_t = [-1.2 + 0.8] \Delta_t = -0.4 \Delta_t$.$-0.4 \Delta_t$ is equivalent to $-\frac{2}{5} \Delta_t$.

The value of $CFSE$ in a tetrahedral complex having a $3d^4$ configuration of the metal ion,surrounded by strong field ligands,will be:

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