Question

Among the following compounds, the one that is polar and has the central atom with ${sp}^2$ hybridization is:

• A. $H_2{CO}_3$
• B. ${SiF}_4$
• C. ${BF}_3$
• D. ${HClO}_2$

Answer 1

The correct option is A $H_2{CO}_3$

$\left(a\right)$$H_{2}C{O}_3\to$ Polar compound.

No.of Bond pair$\frac{1\times 2+4+6\times 3}{8}=\frac{2+4+18}{8}=\frac{24}{8}=3\Rightarrow S{P}^2$ Hybridisation.

$\left(b\right)$ $Si{F}_4\to$ Non Polar

No.of Bond pair$\frac{4+7\times 4}{8}=\frac{4+28}{8}=\frac{32}{8}=4\Rightarrow S{P}^3$ Hybridisation.

$\left(c\right)$ $B{F}_3\to$ Non Polar

No.of Bond pair$\frac{3+7\times 3}{8}=\frac{3+21}{8}=\frac{24}{8}=3\Rightarrow S{P}^2$ Hybridisation.

$\left(d\right)$ $HCl{O}_2\to$ Non Polar

No.of Bond pair$\frac{1+7+6\times 2}{8}=\frac{20}{8}=2,\frac{4}{2}=2\Rightarrow S{P}^3$ Hybridisation.

Our requirement of $S{P}^2$ hybridisation and center atom begin polar.