What is the hybridization of Xe in XeF_2? | vedclass

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

Question

(C) The hybridization of $Xe$ in $XeF_2$ is calculated using the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electro

Vedclass · Accepted Answer

$sp^3d$

Vedclass · Answer

(C) The hybridization of $Xe$ in $XeF_2$ is calculated using the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $XeF_2$,$V = 8$ (valence electrons of $Xe$),$M = 2$ (two $F$ atoms).$H = \frac{1}{2} (8 + 2) = 5$.$A$ value of $5$ corresponds to $sp^3d$ hybridization.

What is the hybridization of $Xe$ in $XeF_2$?

Similar Questions

Which of the following is a linear molecule?

The hybrid orbitals used by central atoms in $BeCl_2$,$BCl_3$ and $CCl_4$ molecules are respectively

Which of the following pairs have the same type of hybridization but different shapes?

What is the hybridization state of the central atom in the conjugate base of $NH_4^{+}$ ion?

Which of the following hybridisation results in non-planar orbitals?

Vedclass Test Series

Exam Paper Generator

Online Exam Module