According to VSEPR theory,in which species do | vedclass

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(A) For $CH^{+}_3$ (methyl carbocation):The carbon atom is $sp^2$ hybridized with three bond pairs and no lone pair.Its geometry is trigonal planar,me

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(A) For $CH^{+}_3$ (methyl carbocation):The carbon atom is $sp^2$ hybridized with three bond pairs and no lone pair.Its geometry is trigonal planar,meaning all atoms lie in the same plane.For $CH^{-}_3$ (methyl carbanion):The carbon atom is $sp^3$ hybridized with three bond pairs and one lone pair.Its geometry is trigonal pyramidal,meaning the atoms do not lie in the same plane.Therefore,only $CH^{+}_3$ has all atoms in the same plane.

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

According to $VSEPR$ theory,in which species do all the atoms lie in the same plane?
$1.$ $CH^{+}_3$
$2.$ $CH^{-}_3$

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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 number of lone pairs of electrons present on $Xe$ in $XeF_4$ is

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According to $VSEPR$ theory,in which species do all the atoms lie in the same plane?$1.$ $CH^{+}_3$$2.$ $CH^{-}_3$