Chemistry - CHEMICAL BONDING Question with Solution | TestHub

(A) ${\mathrm{XeF}}_4$:

$\mathrm{Xe}$ has 4 bond pairs along with 2 lone pairs in the ${\mathrm{XeF}}_4$. Thus, the hybridisation of $\mathrm{Xe}$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$. Now since it contains two lone pairs it will show square planar geometry.

(B) ${\mathrm{SF}}_4$:

The lone pair is an equatorial position, and there are two lone-pair—bond pair repulsions. Hence, it is more stable. So, the shape is described as a distorted tetrahedron, a folded square or a see-saw.

(C) ${\mathrm{NH}}_4^{+}$:

In this case steric number = lone pair + sigma bond = 0 + 4 = 4. 

So, it is sp³ hybridisation. There is no lone pair present. Thus, the shape of this ion is tetrahedral.

(D) ${\mathrm{BrF}}_3$:

To decrease repulsion between the lone pairs, the molecule's structure is bent, making it T-shaped.