Which of the following hybrid orbitals is use | vedclass

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

Question

(A) To determine the hybridization of the central atom $Cl$ in $ClO_2^-$,we calculate the steric number $(X)$:$X = \frac{1}{2} [V + M - C + A]$Where $

Vedclass · Accepted Answer

$sp^3$

Vedclass · Answer

(A) To determine the hybridization of the central atom $Cl$ in $ClO_2^-$,we calculate the steric number $(X)$:$X = \frac{1}{2} [V + M - C + A]$Where $V$ is the number of valence electrons of the central atom $(Cl = 7)$,$M$ is the number of monovalent atoms attached ($O$ is divalent,so $M = 0$),$C$ is the cationic charge $(0)$,$A$ is the anionic charge $(1)$.$X = \frac{1}{2} [7 + 0 - 0 + 1] = \frac{8}{2} = 4$.$A$ steric number of $4$ corresponds to $sp^3$ hybridization.Thus,the chlorine atom in $ClO_2^-$ undergoes $sp^3$ hybridization.

Which of the following hybrid orbitals is used by the chlorine atom in the $ClO_2^-$ ion?

Similar Questions

$KI + MnO_2 + H_2SO_4 \to$ Vapours $(B) +$ other products
Vapours $(B) + K_2S_2O_8 \to (C) + (D)$
$(D) \xrightarrow{(B)}$ Brown $(E)$
Hybridisation of central atom in $(E)$ is

What is the $C - C$ bond length in molecules with $sp^2$ hybridization (in $\mathring{A}$)?

Identify the pair of molecules which have same hybridisation as the hybridisation in $Xenon$ $(II)$ fluoride.

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

Differentiate between atomic orbitals and hybrid orbitals.

Vedclass Test Series

Exam Paper Generator

Online Exam Module