The F-Br-F bond angles in BrF_5 and the Cl-P- | vedclass

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(D) $BrF_5$ has a square pyramidal geometry with one lone pair on the $Br$ atom. Due to the lone pair-bond pair repulsion,the equatorial $F$ atoms are

Vedclass · Accepted Answer

non-identical in $BrF_5$ and non-identical in $PCl_5$

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(D) $BrF_5$ has a square pyramidal geometry with one lone pair on the $Br$ atom. Due to the lone pair-bond pair repulsion,the equatorial $F$ atoms are pushed slightly away from the lone pair,making the $F_{axial}-Br-F_{equatorial}$ bond angles less than $90^{\circ}$ (approximately $84.8^{\circ}$),while the $F_{equatorial}-Br-F_{equatorial}$ angles in the square plane are $90^{\circ}$. Thus,the $F-Br-F$ bond angles are non-identical.$PCl_5$ has a trigonal bipyramidal geometry. It has two types of $Cl-P-Cl$ bond angles: equatorial-equatorial $(120^{\circ})$ and axial-equatorial $(90^{\circ})$. Thus,the $Cl-P-Cl$ bond angles are non-identical.

List-$I$	List-$II$
$I$. $XeF_4$	$A$. See-saw
$II$. $I_3^-$	$B$. Tetrahedral
$III$. $XeO_2F_2$	$C$. Bond angle $90^o$
$IV$. $SO_4^{2-}$	$D$. Linear

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

The $F-Br-F$ bond angles in $BrF_5$ and the $Cl-P-Cl$ bond angles in $PCl_5$,respectively,are

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The correct order of bond angles of the molecules $SiCl_4$,$SO_3$,$NH_3$,and $HgCl_2$ is:

Which of the following pairs of compounds are isostructural?

Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists.

List-$I$ List-$II$

$I$. $XeF_4$ $A$. See-saw

$II$. $I_3^-$ $B$. Tetrahedral

$III$. $XeO_2F_2$ $C$. Bond angle $90^o$

$IV$. $SO_4^{2-}$ $D$. Linear

$A$ pair of molecules with see-saw shape and linear shape,respectively,is

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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