Question
The type of hybrid orbitals used by the chlorine atom in $$ClO_2^ - $$ is
A.
$$s{p^3}$$
B.
$$s{p^2}$$
C.
$$sp$$
D.
none of these
Answer :
$$s{p^3}$$
Solution :
$$H = \frac{1}{2}\left( {V + M - C + A} \right)$$
where $$H = $$ No. of orbitals involved in hybridisation $$\left( {viz.2\,,3\,,\,4,\,5,\,6} \right)$$ and hence nature of hybridisation $$\left( {viz.\,s{p^2},\,s{p^3},\,s{p^3}d,\,s{p^3}{d^2}} \right)$$ can be ascertained.
$$V = $$ No. of electrons in valence shell of the central atom, $$M = $$ No. of monovalent atoms,$$C = $$ Charge on cation, $$A = $$ Charge on anion,
For $$ClO_2^ - $$ we have, $$H = \frac{1}{2}\left( {7 + 0 - 0 + 1} \right)$$
$$ \Rightarrow H = \frac{1}{2}\left( {7 + 1} \right) = 4\,or\,s{p^3}$$ hybridisation as $$4\,$$ orbitals are involved