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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.

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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.

Another thing to consider is the field split. Its size is influenced by the metal’s charge. Higher charge typically means greater field split. Square planar complexes typically require a large field split to stabilise their low-spin state.

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.

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