Based on VSEPR theory,the number of 90^{circ} | vedclass

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(B) The molecule $BrF_5$ has a square pyramidal geometry with one lone pair on the central $Br$ atom.Due to the strong repulsion between the lone pair

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(B) The molecule $BrF_5$ has a square pyramidal geometry with one lone pair on the central $Br$ atom.Due to the strong repulsion between the lone pair and the bonding pairs,the equatorial $F$ atoms are pushed downwards.Consequently,the bond angles between the axial $F$ and the equatorial $F$ atoms are reduced from $90^{\circ}$ to approximately $84.8^{\circ}$.Therefore,there are no $90^{\circ}$ $F-Br-F$ bond angles in the molecule.

$A$. $SnCl_2$	$I$. Trigonal planar
$B$. $NH_3$	$II$. Linear
$C$. $I_3^{-}$	$III$. Angular
$D$. $SO_3$	$IV$. Trigonal pyramidal

List-$I$	List-$II$
$A. ICl$	$i. \text{Linear}$

$B. ICl_3$	$ii. \text{T-Shape}$
$C. ClF_5$	$iii. \text{Square pyramidal}$
$D. IF_7$	$iv. \text{Pentagonal bipyramidal}$

Based on $VSEPR$ theory,the number of $90^{\circ}$ $F-Br-F$ bond angles in $BrF_5$ is

Similar Questions

What is the geometry of $ICl_2^-$?

Match the following:
$A$. $SnCl_2$ $I$. Trigonal planar
$B$. $NH_3$ $II$. Linear
$C$. $I_3^{-}$ $III$. Angular
$D$. $SO_3$ $IV$. Trigonal pyramidal

The correct answer is:

In the $BrF_3$ molecule,the equatorial lone pairs minimize which of the following?

Which of the following molecules has a trigonal pyramidal shape?

Match List-$I$ with List-$II$.
List-$I$ List-$II$
$A. ICl$ $i. \text{Linear}$
$B. ICl_3$ $ii. \text{T-Shape}$
$C. ClF_5$ $iii. \text{Square pyramidal}$
$D. IF_7$ $iv. \text{Pentagonal bipyramidal}$

Choose the correct answer from the options given below:

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