What are the shapes of XeF_2 and XeF_4? | vedclass

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(A) For $XeF_2$: The central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms and has $3$ lone pairs. According to $VSEPR$ theor

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$XeF_2$: Linear,$XeF_4$: Square planar

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(A) For $XeF_2$: The central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms and has $3$ lone pairs. According to $VSEPR$ theory,the steric number is $2 + 3 = 5$,which corresponds to $sp^3d$ hybridization. The geometry is trigonal bipyramidal,but due to the presence of $3$ lone pairs in the equatorial positions,the shape is $Linear$.For $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridization. The geometry is octahedral,but due to the presence of $2$ lone pairs in the axial positions,the shape is $Square \ planar$.

List-$I$ Molecule/ion	List-$II$ Bond pair : lone pair (on the central atom)
$A. ICl_2^-$	$I. 4 : 2$
$B. H_2O$	$II. 4 : 1$
$C. SO_2$	$III. 2 : 3$
$D. XeF_4$	$IV. 2 : 2$

What are the shapes of $XeF_2$ and $XeF_4$?

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Match the List-$I$ with List-$II$.

List-$I$ Molecule/ion List-$II$ Bond pair : lone pair (on the central atom)

$A. ICl_2^-$ $I. 4 : 2$

$B. H_2O$ $II. 4 : 1$

$C. SO_2$ $III. 2 : 3$

$D. XeF_4$ $IV. 2 : 2$

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