In the structure of SF_{4},the lone pair of e | vedclass

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(A) The central atom $S$ in $SF_{4}$ has $5$ electron pairs ($4$ bond pairs and $1$ lone pair),resulting in $sp^{3}d$ hybridization.According to $VSEP

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equatorial position and there are two lone pair-bond pair repulsions at $90^{\circ}$

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(A) The central atom $S$ in $SF_{4}$ has $5$ electron pairs ($4$ bond pairs and $1$ lone pair),resulting in $sp^{3}d$ hybridization.According to $VSEPR$ theory,the lone pair occupies the equatorial position in a trigonal bipyramidal geometry to minimize repulsion.In the equatorial position,the lone pair experiences repulsion from two axial bond pairs at $90^{\circ}$ and two equatorial bond pairs at $120^{\circ}$.Therefore,there are two lone pair-bond pair repulsions at $90^{\circ}$.

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $XeF_4$	$I$. Square pyramidal
$B$. $ClF_3$	$II$. Pentagonal bipyramidal
$C$. $BrF_5$	$III$. Square planar
$D$. $IF_7$	$IV$. Bent $T$-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

In the structure of $SF_{4}$,the lone pair of electrons on $S$ is in ....

Similar Questions

Match the following:

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $XeF_4$ $I$. Square pyramidal

$B$. $ClF_3$ $II$. Pentagonal bipyramidal

$C$. $BrF_5$ $III$. Square planar

$D$. $IF_7$ $IV$. Bent $T$-shape

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

Which of the following is the correct decreasing order of bond angles?

$SbF_5$ reacts with $XeF_4$ and $XeF_6$ to form ionic compounds $[XeF_3^+][SbF_6^-]$ and $[XeF_5^+][SbF_6^-]$. Determine the molecular shape of $[XeF_3^+]$ ion and $[XeF_5^+]$ ion respectively.

Which of the following is the correct order of bond angles?

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