# Geometry of tetrabromidomanganate(II)

What is geometry and hybridization of $\ce{[MnBr4]^2-}$ for $\ce{Mn}$ ($Z = 25$)?

I am not sure if it is tetrahedral or square planar.
I assumed $\ce{Br}$ is a weak ligand and got its geometry to be tetrahedral.
Am I right?

I am a 12th grader and I only know the basics of coordination chemistry.

A square planar geometry is somewhat constrained, if you wish. There need to be compelling electronic reasons to move four coordinating anions (especially large ones like bromide) closer together than tetrahedral — which is the case in square planar, cf the smaller bond angles. Usually, this geometry is associated with $\mathrm{d^8}$ ions but it can happen for other compounds, too.
Manganese(II) is a $\mathrm{d^5}$ ion. Since it only has a low charge, it is very likely to be in the high-spin state. The high-spin state of a $\mathrm{d^5}$ ion does not stabilise any geometry whatsoever. Thus, no reasons exist to go square planar and the tetrabromidocomplex must be tetrahedral.