On the basic `VBT` answer the following questions for the 4-coordinated complex compounds
(a) `[CoBr_(4)]^(2-)`, (b) `[Zn(CN)_(4)]^(2-)`, (c ) `[MnCI_(4)]^(2-)`
(i) What is the oxidation state of the central metal atom//ion?
(ii) What type of hybridisation is involved?
(iii) What is the geometry and magnetic behaviour of the complex ion/compound
(iv) Calculate the value of `mu_(spin)` only .

Explanation

Let the oxidation state of `Co` in `[CoBr_(4)]^(2-)` be x
Then `x + (-1) xx 4 = - 2`
`x = - 2 + 4 = + 2`
Therefore oxidation state of Co in `[CoBr_(4)]^(2-)` is `2+`

`Br^(Θ)` is a weak ligand so no pairing of `3d` electrons occurs
(iii) Since `Co^(2+)` ion in `[CoBr_(4)]^(2-)` shown `sp^(3)` hybridisation geometry of complex ion is tetrahedral further as evident three unpaired electrons are present so the complex ion is paramagnetic
(iv) `mu_(sp i n) = sqrt(n (n +2)) BM`
`mu_(sp i n) = sqrt(n (n +2)) = sqrt(15 )\:' (n = 3)`
`= 3.87 BM`
(b) `[Zn(CN)_(4)]^(2-)` ion
(i) Let oxidation state of Zn in `[Zn(CN)_(4)]^(2-)` be x
`x + (-1) xx 4 = 2`
`x = + 2`

(iii) As evident `Zn^(2+)` ion in `(Zn(CN)_(4)]^(2-)` is `sp^(3)` hybridised with no unpaired electrons therefore geometry of `[Zn(CN)_(4)]^(2-)` is tetrahedral and it is dimagnetic
(iv) `mu_("spin")` only `= sqrt(n (n + 2)) BM`
`:' n = 0, = mu_("spin")` only = 0
(c ) `[Mn(CI_(4)]^2-)`
Let oxidation state of Mn in `[MnCI_(4)]^(2-)` be x
`x + (-1) xx 4 =2`
`x = + 2`
Threrefore, oxidation state of Mn in `[MnCI_(4)]^(2-)` is +2
(ii) Electronic configuration of

`CI^(Θ)` is weak ligand so no pairing of `3d` electrons occurs
(iii) Since `Mn^(2+)` ion is `p^(3)` hybridised in `[MnCI_(4)]^(2-)`
geometry of `[MnCI_(4)]^(2-)` is tetrahedral Magnetic behaviour of `[MnCI_(4)]^(2-)` is paramagnetic `(n =5)`
`(iv) mu_("spin") only = sqrt(5(5 +2)) BM = sqrt(35 )BM = 5.91 BM` .