For the tetrahedral complex ${\left[{\mathrm{MnBr}}_4\right]}^{2-}$, the spin only magnetic moment value is: [At. no. of Mn= 25]

${\left[{\mathrm{MnBr}}_4\right]}^{-2}$

Ox.State of Mn

$\begin{array}{l}\mathrm{x}+4\left(-1\right)=-2\\ \mathrm{x}=+2\end{array}$

$$\begin{gathered} \mathrm{Mn}(\mathrm{Z} = 25) \dots \dots . \left[\mathrm{Ar}\right]4{\mathrm{s}}^{2}3{\mathrm{d}}^5 \\ {\mathrm{Mn}}^{+2} \dots \dots . \left[\mathrm{Ar}\right]3{\mathrm{d}}^5 \end{gathered}$$

Given that ${\left[{\mathrm{MnBr}}_4\right]}^{-2}$  is a tetrahedral complex. $\therefore$  No rearrangement occurs.

Number of unpaired electrons(n) = 5

$\begin{array}{l}{\mathrm{\mu }}_{\mathrm{S}}=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}\mathrm{B}.\mathrm{M}\\ {\mathrm{\mu }}_{\mathrm{S}}=\sqrt{5\left(5+2\right)}\mathrm{B}.\mathrm{M}\\ {\mathrm{\mu }}_{\mathrm{S}}=5.9\mathrm{B}.\mathrm{M}\end{array}$

Note:–  In the formation of complex only vacant orbitals involve in hybridization.