The correct option is
D (i - d), (ii - c), (iii - b), (iv - a)
Generally, we have two approaches to calculate the hybridisation for different structures:
Approach−1 : Using steric number approach which is ,
Steric Number
(x) =
12(s + v − q).
where, s = number of monovalent surrounding atoms
v = number of valence electrons on the central atom
q = charge (with sign)
Approach−2 : Counting the of total number of sigma bonds and lone pairs around a particular atom.
(i) In
XeF2, s = 2 ; v = 8 ; q = 0,
Hence, s = 5 i.e.,
sp3d hybridised. The molecule is linear shaped with three lone pairs occupying the equatorial positions on
Xe.
(ii) In
N−3, we determine the hybridisation by counting the number of sigma bonds and lone pairs around the central atom. Hence, the central nitrogen has 2
σ bonds and no lone pairs. So, the steric number of
N in
N−3 is 2. Therefore, the hybridisation of
N in
N−3 is
sp and it has a linear shape.
(iii) In
PCl−6, s = 6 ; v = 5 ; q = -1. Hence, x = 6 i.e.
sp3d2 hybridisation. If we arrange this compound in a geometry corresponding to its hybridisation, we get octahedral shape with no lone pairs on
P.
(iv) In
ICl+2, s = 2 ; v = 7 ; q = +1.
Hence, x = 4 i.e.
sp3 hybridisation. If we arrange this compound in a geometry corresponding to its hybridisation, we get bent shape with two lone pairs on
I.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/797221/original_Dia_24.28.png)
Hence, (i - d), (ii - c), (iii - b), (iv - a) is the right option.