Magnetic moment value of `[Mn(CN)_(6)]^(3-)` ion in `2.8BM` Predict the type of hybridisation and geometry of the ion .

Explanation

Short method to calculate 'n' `("number of unpaired electrons if" mu_(s) "is given")`
Take `2.8BM~~3B.M` then take its square which is `9` and then take its underroot sqrt9` which is nearly equal to `sqrt8BM` when `mu_(s) = sqrt8Bm` then `n = 2`
Thus Mn in `[Mn(CN)_(6)]^(3-)` has two unpaired electrons `(n=2)` Oxidation state of Mn in `[Mn(CN)_(6)]^(3-)` is `+3 Mn (Z=25)`
`implies 3d^(5)4s^(2)` and `Mn^(+3) =3d^(4)`

`CN^(Θ)` being a strong ligand forces only one pairing of `4`half-filled 3dorbitals
Remember (i) In `d^(3)` type electronic configuration no pairing occurs whether the ligand is strong or weak
(ii) In `d^(4)` type electronic configuration only one pairing occurs if the ligand is strong
(iii) Thus the hybridisation is `d^(2)sp^(3)` with octahedral geometry
(iv) Inner orbital complex .