Discuss the molecular structure of $BrF_3$ based on the $VSEPR$ theory.
(N/A) The central atom $Br$ has seven valence electrons. Three of these electrons form bond pairs with three fluorine atoms,leaving four electrons behind.
Thus,it contains three bond pairs and two lone pairs.
According to the $VSEPR$ theory,these are arranged at the corners of a trigonal bipyramid.
The two lone pairs occupy equatorial positions to minimize lone pair-lone pair and bond pair-lone pair repulsions,which are greater than bond pair-bond pair repulsions.
Additionally,the axial fluorine atoms bend towards the equatorial fluorine atom to minimize lone pair-lone pair repulsions.
Therefore,its shape is slightly bent,resembling a $'T'$ shape.
Thus,it contains three bond pairs and two lone pairs.
According to the $VSEPR$ theory,these are arranged at the corners of a trigonal bipyramid.
The two lone pairs occupy equatorial positions to minimize lone pair-lone pair and bond pair-lone pair repulsions,which are greater than bond pair-bond pair repulsions.
Additionally,the axial fluorine atoms bend towards the equatorial fluorine atom to minimize lone pair-lone pair repulsions.
Therefore,its shape is slightly bent,resembling a $'T'$ shape.
Explore More
Similar Questions
What is the shape of the $BrF_5$ molecule?
Which of the following sets is not correct?
Medium
View SolutionThe incorrect order of bond angle is:
Medium
View SolutionThe molecular shapes of $SF_4, CF_4$,and $XeF_4$ are:
Medium
View SolutionThe geometries of $XeF_4$ and $XeOF_4$ respectively are
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo