Question

The correct order of increasing $s-$character (in percentage) in the hybrid orbitals of following molecules/ions is:
(I) $C{O}_3^{2-}$
(II) $Xe{F}_4$
(III) $I_3^{-}$
(IV) $NC{l}_3$
(V) $BeC{l}_2$

• A. II < III < IV < I < V
• B. II < IV < III < V < I
• C. III < II < I < V < IV
• D. II < IV < III < I < V

Answer 1

The correct option is A II < III < IV < I < V

The correct order of increasing $s-$character (in percentage) in the hybrid orbitals of following molecules/ions is:

(I) $C{O}_3^{2-}$ $C$ undergoes $s{p}^2$

(II) $Xe{F}_4$ $Xe$ undergoes $s{p}^3{d}^2$

(III) $I_3^{-}$ $I$ undergoes $s{p}^{3}d$

(IV) $NC{l}_3$ $N$ undergoes $s{p}^3$

(V) $BeC{l}_2$ $Be$ undergoes $sp$.

S-character in hybrid orbitals increases as the number of p orbitals and d orbitals in the hybridization.

So the correct order of S-character in hybrid orbitals is $sp>s{p}^2>s{p}^3>s{p}^{3}d>s{p}^3{d}^2$.

So the correct option is $II<III<IV<I<V$.

Hence option $A$ is correct.