Question

Which of the following compounds is polar and has a central atom with $s{p}^3$ hybridisation?

• A. $H_{2}C{O}_3$
• B. $Si{F}_4$
• C. $B{F}_3$
• D. $HCl{O}_2$

Answer 1

The correct option is D $HCl{O}_2$

Hybridization of the central atom in a molecule can be calculated by calculating its steric number. So, the steric number for carbon atom in $H_{2}C{O}_3$ = number of atoms bonded + lone pairs = 3 + 0 = 3.
$\therefore$ hybridisation of the central atom $\left(C\right)$ = $s{p}^2$, and $H_{2}C{O}_3$ is a polar molecule.

Similarly, the steric number for silicon atom in $Si{F}_4$ = 4 + 0 = 4.
$\therefore$ hybridisation of the central atom $\left(Si\right)$ = $s{p}^3$, and $Si{F}_4$ has a tetrahedral geometry. So, due to symmetry, it is a non-polar molecule.

The steric number for boron atom in $B{F}_3$ = 3 + 0 = 3.
$\therefore$ hybridisation of the central atom $\left(B\right)$ = $s{p}^2$, and $B{F}_3$ has a trigonal planar geometry. So, due to symmetry, it is a non-polar molecule.

The steric number for chlorine atom in $HCl{O}_2$ = 2 + 2 = 4.
$\therefore$ hybridisation of the central atom $\left(Cl\right)$ = $s{p}^3$, and $HCl{O}_2$ has a bent shape. So, it is a polar molecule.