The d-electronic configuration of CoCl$$_4^{2-}$$ in tetrahedral crystal field is $$e^m t_2^n$$. The sum of m and number of unpaired electrons is

$$CoCl_4^{2-}$$: Co²⁺ has d⁷ configuration. In tetrahedral field (always high spin):

$$e^4 t_2^3$$ (4 electrons in e, 3 in t₂). $$m = 4$$.

Unpaired electrons: e has 4 electrons (2 pairs, 0 unpaired), t₂ has 3 (all unpaired). Total unpaired = 3.

Sum = $$m$$ + unpaired = 4 + 3 = 7.