Question

Predict the compounds given in column I with the shape given in column II and mark the correct option.

Column I	Column II
(a) $Xe{F}_6$	(i) Distorted octahedral
(b) $Xe{O}_3$	(ii) Square planar
(c) $XeO{F}_4$	(iii) Pyramidal
(d) $Xe{F}_4$	(iv) Square pyramidal

• A. $\left(a\right)-\left(i\right),\left(b\right)-\left(iii\right),\left(c\right)-\left(iv\right),\left(d\right)-\left(ii\right)$
• B. $\left(a\right)-\left(i\right),\left(b\right)-\left(ii\right),\left(c\right)-\left(iv\right),\left(d\right)-\left(iii\right)$
• C. $\left(a\right)-\left(iv\right),\left(b\right)-\left(iii\right),\left(c\right)-\left(i\right),\left(d\right)-\left(ii\right)$
• D. $\left(a\right)-\left(iv\right),\left(b\right)-\left(i\right),\left(c\right)-\left(ii\right),\left(d\right)-\left(iii\right)$

Answer 1

The correct option is A $\left(a\right)-\left(i\right),\left(b\right)-\left(iii\right),\left(c\right)-\left(iv\right),\left(d\right)-\left(ii\right)$

In $Xe{F}_6$, the central Xe atom undergoes $s{p}^3{d}^3$ hybridisation and the molecular geometry is distorted octahedral with 1 lone pair and 6 bond pairs of electrons.

In $Xe{O}_3$, the central Xe atom undergoes $s{p}^3$ hybridisation and the molecular geometry is pyramidal with 1 lone pair and 3 bonding domains of electrons.

In $XeO{F}_4$, the central Xe atom undergoes $s{p}^3{d}^2$ hybridization and the molecular geometry is square pyramidal with 1 lone pair and 5 $\sigma$ bond pairs of electrons.

In $Xe{F}_4$, the central Xe atom undergoes $s{p}^3{d}^2$ hybridization and the molecular geometry is square planar with 2 lone pairs and 4 bond pairs of electrons.

Hence, option (A) is the correct answer.