Reduce the hybridisation, geometry and shape of the following
(i) `CH_(2)^(2+)` (ii) `Br_(3)^(Θ)` (iii) `CIO_(3)^(o+)`
(iv) `F_(2)SeO` (v) `IO_(2)F_(2)^(Θ)`
Either of the hybridisation (i) `sp^(2)` and (ii) `sp^(3)d^(2)` of a central atom can lead to a squar planar molecule Give one example of each .

Explanation

`CH_(2)^(2+)`
` H= (1)/(2) (V + M` - no. of + ve charge)
` = (1)/(2) (4 + 2 - 2) = 2 =sp`
The molecule is linear. There will be no multiple bonding since H-atoms have no orbitals available and C-atom has lost the extra electrons
`H = (1)/(2) (V + M+` on . of -ve charge)
` = (1)/(2) (7 + 2 + 1)`
` = 5 = sp^(3) d = Tbp`
`Br_(3)^(Θ)` has Tbp geometry but due to the presence of three `lp's` it is linear in shape
(iii) `C1O_(3)^(o+): :underset(..)O=overset(o+)underset( :underset(..)O)underset(||)(C1)=ddotO:`
`H = (11)/(2) (V +M` - no of +ve charge)
` (1)/(2) (7 +0 -1) = 3 = sp^(2) =` planar
`CIO_(3)^(o+)` is unstable ion, `(CI =O)` bonding is expected
`H= (1)/(2) (V +M) = (1)/(2) ( 6 + 2) = 4 = sp^(3) =TH`
`FeSeO` has terahedral geometry but due to the presence of one lp it has pyramidal in shape This structure is analogous
(v) `IO_(2)F_(2)^(Θ)`
`H=(1)/(2) (V +M +` no. of -ve charge)
`(1)/(2)(7 + 2 +1) =5 = sp^(3) d`
It has `sp^(3)d` hybridisation with Tbp geometry but due to the presence of one lp of `e^(-)` it has see-saw shape
(b) `[Ni(CN)_(4)]^(2-)]`
(ii) `:overset(..)X eF_(4)`