Question

Among the following given compounds, the one that is polar and has a central atom with $s{p^3}$ hybridization is:
A. ${H_2}C{O_3}$
B. $Si{F_4}$
C. $B{F_3}$
D. $HCl{O_2}$

Answer 1

In $s{p^3}$ hybridization mixing of one 2s-orbital and three 2p-orbital takes place to form four hybrid orbitals. In order for a molecule to possess $s{p^3}$ hybridization, it should contain one s-orbital and three p orbital. The shape of the compound of $s{p^3}$ hybridization is tetrahedral.

Complete step by step answer:
The polarity of the compound is dependent on the ability of an atom present in a compound to attract another atom. Due to this attraction, there is a difference in electronegativity when one atom pulls the other atom more strongly and then the compound is said to be polar in nature.
${H_2}C{O_3}$
The structure of ${H_2}C{O_3}$ is shown below.

The structure shows that it contains a 3 sigma bond and 1 pi bond.
The hybridization is $s{p^2}$ and the structure is trigonal planar.
The compound is polar as there is an electronegativity difference between the carbon atom and oxygen atom.
$Si{F_4}$
The structure of $Si{F_4}$ is shown below.

It contains four sigma bonds and the geometry is tetrahedral. The hybridization is $s{p^3}$. Fluorine is a very electronegative element, the dipole moment of the four Si-F bonds will cancel out and the resulting dipole moment will be zero. Thus it is not a polar compound.
$B{F_3}$

It contains three sigma bonds and the geometry is trigonal planar. The hybridization is $s{p^2}$. The net dipole moment is zero. Thus it is not a polar compound.
$HCl{O_2}$

It contains three bond pairs and two lone pairs. The hybridization is $s{p^3}$. It is a polar compound.
Therefore, the correct option is D.

Note:
The lone pairs also take part in hybridization as it is localized at the central metal in pure orbital and the orbitals of the metal atom take part in hybridization.