The group having triangular planar structures | vedclass

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(B) To determine the shape of a molecule,we calculate the hybridization of the central atom using the formula: $H = \frac{1}{2}(V + M - C + A)$,where

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$CO_3^{2-}, NO_3^-, SO_3$

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(B) To determine the shape of a molecule,we calculate the hybridization of the central atom using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge. $1$. For $CO_3^{2-}$: $H = \frac{1}{2}(4 + 0 - 0 + 2) = 3$ ($sp^2$ hybridization,triangular planar). $2$. For $NO_3^-$: $H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$ ($sp^2$ hybridization,triangular planar). $3$. For $SO_3$: $H = \frac{1}{2}(6 + 0 - 0 + 0) = 3$ ($sp^2$ hybridization,triangular planar). All species in option $B$ have $sp^2$ hybridization with no lone pairs on the central atom,resulting in a triangular planar geometry.

List-$I$ (molecules/ions)	List-$II$ (No. of lone pairs of $e^-$ on central atom)
$A$. $IF_7$	$I$. Three
$B$. $ICl_4^-$	$II$. One
$C$. $XeF_6$	$III$. Two
$D$. $XeF_2$	$IV$. Zero

The group having triangular planar structures is

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List-$I$ (molecules/ions) List-$II$ (No. of lone pairs of $e^-$ on central atom)

$A$. $IF_7$ $I$. Three

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$C$. $XeF_6$ $III$. Two

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