Question

The correct order of increasing s-character

(in percentage) in the hybrid orbitals of following

molecules/ions is :-

(I) 2 CO3

 (II) XeF4

(III) 3 I

 (IV) NCl3

(V) BeCl2

(1) II < III < IV < I < V

(2) II < IV < III < V < I

(3) III < II < I < V < IV

(4) II < IV < III < I < V​

Answer 1

answer for this question is (1)

Answer 2

Answer: Option (1) is the correct answer.

Explanation:

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.  

$CO_{3}$ has hybridization as $sp^{2}$.

$XeF_{4}$ has hybridization as $sp^{3}d^{2}$.

$I^{3-}$ has hybridization as $sp^{3}d$

$NCl_{3}$ has hybridization as $sp^{3}$.

$BeCl_{2}$ has hybridization as sp as it has linear geometry.

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

Hence, correct order of increasing s-character is as follows.

        $XeF_{4}$ < $I^{3-}$ < $NCl_{3}$ < $CO_{3}$ <  $BeCl_{2}$

Thus, we can conclude that the correct order of increasing s-character is II < III < IV < I < V.