Question

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

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

Answer 1

The correct option is A

$H_{2}C{O}_3$

In this problem, first let's get the structure of each compound by using VSEPR theory. Let's take $H_{2}C{O}_3$ (C is central atom) electron pairs (ep) = $(\frac{V+L+A+C}{2}$) = $\frac{4+2}{2}$ so, trigonal planar geometry $\Rightarrow$ ${sp}^2$ hybridization

Now number of hybrid orbitals=Number of bonded atoms + number of lone pairs

$\Rightarrow$ Number of hybrid orbitals = 3 $\Rightarrow$ ${sp}^2$

(b) $Si{F}_4$ (Si is central atom)

ep = $\frac{4+4}{z}$ = 4 $\Rightarrow$ Tetradral $\Rightarrow$ ${sp}^3$

$\Rightarrow$ Number of hybrid orbitals = 4 $\Rightarrow$ ${sp}^3$

(c) $B{F}_3$ (B is central atom)

ep = $\frac{z+z}{z}$ = 3 $\Rightarrow$ Trigonal Planar $\Rightarrow$ ${sp}^2$

$\Rightarrow$ Number of hybrid orbitals = 3 $\Rightarrow$ ${sp}^2$ hybridization

(d)$HCl{O}_2$ (Cl is central atom)

ep = $\frac{7+2}{z}$ = 4 $\Rightarrow$ Tetrahedral $\Rightarrow$ ${sp}^3$

Number of hybrid orbitals = 2 + 2 = 4 $\Rightarrow$ ${sp}^3$

Number of hybrid oribtals = 2 + 2 = 4 $\Rightarrow$ ${sp}^3$

$\to$Now, we know that in $H_{2}C{O}_3$and $B{F}_3$, central atom (C and B) are ${sp}^2$ hybridised.

Now, to look for which one is polar, lets take a look at the structure of both.

Now, check for symmetry in both the molecules.

In $H_{2}C{O}_3$, the dipole moments will not cancel out each other.

While in $B{F}_3$, the dipole moments will cancel out each other and the net dipole moment will be 0.