The spin only magnetic moment of [MnBr₄]^2– is 5.9 BM. The geometry of the complex ion is

$$\begin{gathered} \sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}=5.9 \\ \mathrm{n}\left(\mathrm{n}+2\right)=35, \mathrm{n}=5 \end{gathered}$$

n=5 means Mg has unpaired electrons are 5. Bromine is a weak field ligand, so back pairing will not occur.

So, the geometry of the compound is tetrahedral.