In BrF_3 molecule,the lone pairs occupy equat | vedclass

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Question

(A) In $BrF_3$,the central atom $Br$ has $7$ valence electrons. It forms $3$ $Br-F$ bonds and has $2$ lone pairs,resulting in a total of $5$ electron

Vedclass · Accepted Answer

Lone pair-lone pair repulsion and lone pair-bond pair repulsion

Vedclass · Answer

(A) In $BrF_3$,the central atom $Br$ has $7$ valence electrons. It forms $3$ $Br-F$ bonds and has $2$ lone pairs,resulting in a total of $5$ electron pairs ($sp^3d$ hybridization).According to $VSEPR$ theory,lone pairs prefer equatorial positions in a trigonal bipyramidal geometry to minimize lone pair-lone pair repulsion and lone pair-bond pair repulsion.By placing both lone pairs in equatorial positions,the $90^{\circ}$ lone pair-lone pair interactions are avoided,which significantly reduces the overall repulsion in the molecule.

List-$I$	List-$II$
$A$. $XeF_4$	$1$. Pyramidal
$B$. $XeF_6$	$2$. $T$-shape
$C$. $XeO_3$	$3$. Distorted octahedral
$D$. $XeOF_2$	$4$. Square planar

In $BrF_3$ molecule,the lone pairs occupy equatorial positions to minimize

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Select the correct matching:

List-$I$ List-$II$

$A$. $XeF_4$ $1$. Pyramidal

$B$. $XeF_6$ $2$. $T$-shape

$C$. $XeO_3$ $3$. Distorted octahedral

$D$. $XeOF_2$ $4$. Square planar

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