Question

The correct order of increasing s character (in perencetage) in the hybrid orbitals in below molecules/ions is (assume all hybrid orbitals are exactly equivalent):

$\underset{I}{C{O}_3^{2-}};\underset{II}{Xe{F}_4};\underset{III}{I_3^{-}};\underset{IV}{NC{l}_3};\underset{V}{BeC{l}_2\left(g\right)}$

• 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

More is the percentage of s-character in a compound more will be the electronegativity of central metal atom. The hybridization of given compound will be as follows.
$$\begin{array}{cc}\text{Species}& \text{Hybridisation}\\ C{O}_3^{2-}& s{p}^2\\ Xe{F}_4& s{p}^3{d}^2\\ I_3^{-}& s{p}^{3}d\\ NC{l}_3& s{p}^3\\ BeC{l}_2& sp\end{array}$$

Lesser is the number of p-orbitals more will be the percentage of s-character.

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.