The blue complex [CoCl₄]²⁻ is tetrahedral…

Step 1. The electrons. $\text{Co}^{2+}$ has $27 - 2 - 18 = 7$ electrons in 3d. The lower $e$ set takes four of them, two in each orbital. The three $t_2$ orbitals take the other three, one each. The arrangement is $e^4 t_2^3$.

Step 2. Unpaired electrons. The three $t_2$ electrons are unpaired. So $n = 3$ and $\mu = \sqrt{3 \times 5} = 3.87\ \text{BM}$.

Step 3. Low spin? No. The gap is $\Delta_t = \tfrac{4}{9}\Delta_o$, less than half the octahedral gap, and it is smaller than the pairing energy. The electrons spread over all five orbitals instead of pairing.