1. Find oxidation state of Co in $[\mathrm{CoCl_4}]^{2-}$

Let oxidation state of Co be $x$.

$x + 4(-1) = -2$ $x - 4 = -2$ $x = +2$

So, cobalt is $\mathrm{Co^{2+}}$.

2. Write the d-electron count of $\mathrm{Co^{2+}}$

Atomic number of Co is $27$.

Neutral Co: $\mathrm{Co}: [Ar] \, 3d^7 4s^2$

For $\mathrm{Co^{2+}}$, remove two electrons from $4s$ first: $\mathrm{Co^{2+}}: [Ar] \, 3d^7$

So the ion has 7 d-electrons.

3. Tetrahedral crystal field splitting

In a tetrahedral field, the lower energy set is $e$ and the higher energy set is $t_2$.

Thus, electrons fill as: $e^m t_2^n$

Since tetrahedral complexes are generally high spin (because $\Delta_t$ is small), for $d^7$ the arrangement is:

• First fill lower $e$ set: $4$ electrons maximum
• Then fill higher $t_2$ set with remaining $3$ electrons

So, $e^4 t_2^3$

Hence, $m = 4$

4. Find number of unpaired electrons

For tetrahedral $d^7$:

• In $e^4$, both orbitals are fully paired $\Rightarrow 0$ unpaired there
• In $t_2^3$, the three electrons occupy three orbitals singly $\Rightarrow 3$ unpaired electrons

Therefore, number of unpaired electrons $= 3$.

5. Required sum

$m + \text{number of unpaired electrons} = 4 + 3 = 7$

Final Answer

$\boxed{7}$

The derived answer matches the stored correct answer.