The d-electronic configuration of [CoCl_4]^{2 | vedclass

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(C) The central metal ion is $Co^{2+}$,which has a $3d^7$ electronic configuration.$Cl^-$ is a weak field ligand $(WFL)$,so pairing does not occur.In

Vedclass · Accepted Answer

$7$

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(C) The central metal ion is $Co^{2+}$,which has a $3d^7$ electronic configuration.$Cl^-$ is a weak field ligand $(WFL)$,so pairing does not occur.In a tetrahedral crystal field,the $d$-orbitals split into $e$ (lower energy) and $t_2$ (higher energy) sets.According to Hund's rule,the $7$ electrons are filled as follows: $e^4 t_2^3$.Here,$m = 4$ and $n = 3$.The number of unpaired electrons is $3$.Sum of '$m$' and the number of unpaired electrons $= 4 + 3 = 7$.

The $d$-electronic configuration of $[CoCl_4]^{2-}$ in a tetrahedral crystal field is $e^m t_2^n$. The sum of '$m$' and the number of unpaired electrons is $.............$.

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