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-$I$ (Compound / Species)	List-$II$ (Shape / Geometry)
$A$. $SF_4$	$I$. Tetrahedral
$B$. $BrF_3$	$II$. Pyramidal
$C$. $BrO_3^{-}$	$III$. See-saw
$D$. $NH_4^{+}$	$IV$. Bent $T$-shape

The group having triangular planar structures is

Similar Questions

The correct order of the bond angles of the given compounds is

Which of the following represents the correct increasing order of bond angles for the given species?

The shape of the $XeF_2$ molecule is ..........

Match List-$I$ with List-$II$ :

List-$I$ (Compound / Species) List-$II$ (Shape / Geometry)

$A$. $SF_4$ $I$. Tetrahedral

$B$. $BrF_3$ $II$. Pyramidal

$C$. $BrO_3^{-}$ $III$. See-saw

$D$. $NH_4^{+}$ $IV$. Bent $T$-shape

Choose the correct answer from the options given below :

Decreasing order of bond angle is

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