The shapes of SOF_4,XeO_2F_2 and ClF_3 are re | vedclass

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(A) $1$. For $SOF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ and $1$ double bond with $O$. The steric number is $

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Trigonal bipyramidal,see-saw and $T$-shape

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(A) $1$. For $SOF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ and $1$ double bond with $O$. The steric number is $5$ ($4$ sigma bonds + $1$ pi bond is not counted in $VSEPR$,but the double bond occupies one equatorial position). The geometry is trigonal bipyramidal with no lone pairs,so the shape is trigonal bipyramidal.$2$. For $XeO_2F_2$: The central atom $Xe$ has $8$ valence electrons. It forms $2$ single bonds with $F$ and $2$ double bonds with $O$. The steric number is $4$ (bonds) + $1$ lone pair = $5$. The lone pair occupies an equatorial position,resulting in a see-saw shape.$3$. For $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ and has $2$ lone pairs. The steric number is $3 + 2 = 5$. The lone pairs occupy equatorial positions,resulting in a $T$-shape.Thus,the shapes are trigonal bipyramidal,see-saw,and $T$-shape.

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 shapes of $SOF_4$,$XeO_2F_2$ and $ClF_3$ are respectively:

Similar Questions

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

The maximum number of lone pairs of electrons on the central atom from the following species is $........$. $ClO_3^-$,$XeF_4$,$SF_4$ and $I_3^-$

The number of molecules/ions having a trigonal bipyramidal shape is $........$.
$PF_5, BrF_5, PCl_5, [PtCl_4]^{2-}, BF_3, Fe(CO)_5$

Which set of species is arranged in order of increasing $O-N-O$ bond angle?

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