The spin only magnetic moment of `[MnBr_(4)]^(2–)` is 5.9 BM. Predict the geometry of the complex ion ?

Explanation

Oxidation state of is the complex `=+2` Since the coordination number of `Mn^(+)` ion the complex `=4` Therefore it will either tetrahedral `(sp^(3) "hybridisation")` or square planner `(dsp^(2) "hybridisation")` Bit the fact that the magntic moment of the complex is `5.9BM` (given) which corresponds to n `("number of unpaired electrons") =5` Due to the presence of `5` unpaired electrons in the `3d` orbitals of `Mn^(+2)` it should be tetrahedral in shape rather than square planar . if the geometry of the complex ion is tetrahedral `Mn(Z=25)implies 3b^(5)4s^(2)`
Since `Br^(Θ)` is a weak ligand so pairing of electrons in the five half-filled `3d` orbitals will not occur .

If the geometry of the complex ion is square planar If the pairing of electrons in the five half-filled `3d` orbitals occurs

The number of unpaired electron in the complex should be equal to one `Somu_(s)` should be equal to `sqrt3 =3.83 BM` which is not given Hence hybridisation in not `dsp^(2)` but `sp^(3)` .