The d-electronic configuration of [ CoCl _ 4 ] ^ 2 - in tetrahedral...

The oxidation state of Co in complex ${\left[{CoCl}_{4}\right]}^{2 -}$ is

$x - 4 = - 2$

$x = + 2$

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

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

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

So, value of $( m ) +$(unpaired electrons) is $( 4 ) + ( 3 ) = 7$.