The magnetic moment of $[MnCl_{4}]^{2-}$ is $5.92 \ BM$. Explain the reason for this value.

(N/A) In $[MnCl_{4}]^{2-}$,the oxidation state of $Mn$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^{5}$.
Since $Cl^{-}$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ subshell.
Thus,there are $5$ unpaired electrons $(n=5)$.
The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
This confirms the high spin $sp^{3}$ hybridized tetrahedral complex.