The d-electronic configuration of CoCl 4 2 - in tetrahedral crystal field is e m t 2 n . The sum of m and…

The oxidation state of Co in complex ${\left[{\mathrm{CoCl}}_4\right]}^{2-}$ is

$x-4=-2$

$x=+2$

Electronic configuration of $Co$ is  ${}_{27}Co=\left[Ar\right]4s^{2}3d^7$

So, $C{o}^{+2}=\left[\mathrm{Ar}\right]3d^7$

Since ${\mathrm{Cl}}^{-}$ is a weak field ligand, so the given complex will have a tetrahedral geometry with following Crystal-field splitting

So, value of $\left(\mathrm{m}\right)+$(unpaired electrons) is $\left(4\right)+\left(3\right)=7$.