Statement - I : The spin only magnetic moment of [MnBr₄]^2– is 5.9 BM
Statement - II : [MnBr₄]^2– is tetrahedral with sp³ hybridisation of Mn²⁺ having 3d⁵ configuration

Magnetic moment=  ​$\sqrt{\mathrm{n}\left(\left(\mathrm{n}+2\right)\right)}$= 5.9 (given)

∴35 = n(n+2) or n=5

Magnetic moment is 5.9BM means unpaired electrons are 5

Bromine is a type of weak field ligand, so back pairing will not occur.

So, the geometry of the compound is tetrahedral and hybridization is sp³