The number of lone pairs of electrons and the | vedclass

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

Question

(D) To determine the hybridization and number of lone pairs for $XeOF_2$:$1$. The central atom is Xenon $(Xe)$, which has $8$ valence electrons.$2$. I

Vedclass · Accepted Answer

$2, sp^3d$

Vedclass · Answer

(D) To determine the hybridization and number of lone pairs for $XeOF_2$:$1$. The central atom is Xenon $(Xe)$, which has $8$ valence electrons.$2$. In $XeOF_2$, $Xe$ forms two single bonds with two Fluorine $(F)$ atoms and one double bond with one Oxygen $(O)$ atom.$3$. The number of bonding electron pairs = $2$ (from $F$) + $2$ (from $O$) = $4$ electron pairs involved in bonding.$4$. The number of lone pairs on $Xe$ = (Total valence electrons - electrons used in bonding) / $2$ = $(8 - 6) / 2 = 1$ lone pair.$5$. The steric number = (Number of sigma bonds) + (Number of lone pairs) = $3$ (sigma bonds) + $1$ (lone pair) = $4$. Wait, let us re-evaluate: $Xe$ is bonded to $2$ $F$ atoms and $1$ $O$ atom. Total sigma bonds = $3$. Total lone pairs = $2$. Steric number = $3 + 2 = 5$.$6$. $A$ steric number of $5$ corresponds to $sp^3d$ hybridization.$7$. Thus, $XeOF_2$ has $2$ lone pairs and $sp^3d$ hybridization.

The number of lone pairs of electrons and the hybridization of Xenon $(Xe)$ in $XeOF_2$ are

Similar Questions

The hybridization of atomic orbitals of nitrogen in $NO_2^+$,$NO_3^-$,and $NH_4^+$ are:

The mode of hybridisation of carbon in $CO_2$ is

What is the correct mode of hybridization of the central atom in the following compounds: $NO_2^+$,$SF_4$,$PF_6^-$?

In which of the following combinations does the hybridisation of the central atom $(^*)$ $NOT$ change?

What are the shapes of ethyne and methane?

Vedclass Test Series

Exam Paper Generator

Online Exam Module