The type of hybrid orbitals used by the chlor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The hybridization of the central atom is calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electro

Vedclass · Accepted Answer

$sp^3$

Vedclass · Answer

(A) The hybridization of the central atom is calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $ClO_2^-$,$V = 7$ (for $Cl$),$M = 0$ (oxygen is divalent),$C = 0$,and $A = 1$.$H = \frac{1}{2}(7 + 0 - 0 + 1) = \frac{8}{2} = 4$.$A$ value of $4$ corresponds to $sp^3$ hybridization.Therefore,the chlorine atom in $ClO_2^-$ is $sp^3$ hybridized,resulting in an angular (bent) shape due to the presence of two lone pairs.

The type of hybrid orbitals used by the chlorine atom in $ClO_2^-$ is

Similar Questions

$BF_3$ is a planar and electron-deficient compound. The hybridization and the number of electrons around the central atom,respectively,are:

The hybridisation of the central atom in $NF_3$ is:

What is the hybridization of the nitrogen atomic orbitals in $NO_2^+$,$NO_3^-$,and $NH_4^+$ respectively?

The hybridisation of the central atom $I$ in $IF_7$ is:

If pure unhybridised orbitals are used in forming $CH_4$ then which statement is incorrect :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module