One the basic of `CFT` predict the geometry of the compund `K_(3)[Mn(CN)_(6)]` Also calculate the value of `mu_(spin)` only of the compund .

Explanation

In`k_(3)[Mn(CN)_(6)],[Mn(CN)_(6)]^(3-)` is the complex ion In `[Mn(CN)_(6)]^(3-)` the oxidation state of central metal ion `Mn is +3(Mn = 3d^(5) 4s^(2) , Mn^(3+) = 3d^(4))`
`CN` of Mn in `[Mn(CN)_(6)]^(3-)` the oxidation state of central metal ion
`Mn` is `+3 (Mn =3d^(5) 4s^(2), Mn^(3+) = 3d^(4))`
`CN` of Mn in `[Mn(CN)_(6)]^(3-)` is 6 hence it is an octahedral complex `CN^(Θ)` ion is a strong field ligand therefore `[Mn(CN)_(6)]^(3-)` is low spin octahedral complex and hence `Mn^(3+)` ion `(Mn(CN)_(6)]^(3-))` is
Splitting of d-orbitals in octahedral geometry with `n =2`
`mu_(spin) = sqrt(2 (2 +2)) = 2.83BM` .