Discuss the molecular structure of BrF_3 base | vedclass

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

Question

(N/A) The central atom $Br$ has seven valence electrons. Three of these electrons form bond pairs with three fluorine atoms,leaving four electrons beh

Vedclass · Accepted Answer

Vedclass · Answer

(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.

List-$I$ (Molecule)	List-$II$ (Shape)
$A. SF_4$	$I. T\text{-shaped}$
$B. ClF_3$	$II. \text{Square planar}$
$C. BrF_5$	$III. \text{See-saw}$
$D. XeF_4$	$IV. \text{Square pyramidal}$

Discuss the molecular structure of $BrF_3$ based on the $VSEPR$ theory.

Similar Questions

Provide the different spatial arrangements of $ClF_3$ including its lone pairs.

$SF_6$ is a kinetically inert substance because

Which molecule is not linear?

Which of the following molecules is linear and possesses the maximum number of lone pairs on the central atom?

Match the molecules in List-$I$ with their respective shapes in List-$II$.
List-$I$ (Molecule) List-$II$ (Shape)
$A. SF_4$ $I. T\text{-shaped}$
$B. ClF_3$ $II. \text{Square planar}$
$C. BrF_5$ $III. \text{See-saw}$
$D. XeF_4$ $IV. \text{Square pyramidal}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module