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-$I$	List-$II$
$A. XeF_4$	$I. See-saw$
$B. SF_4$	$II. Square planar$
$C. NH_4^{+}$	$III. Bent T-shaped$
$D. BrF_3$	$IV. Tetrahedral$

The group having triangular planar structures is

Similar Questions

Match List-$I$ with List-$II$:
List-$I$ List-$II$
$A. XeF_4$ $I. See-saw$
$B. SF_4$ $II. Square planar$
$C. NH_4^{+}$ $III. Bent T-shaped$
$D. BrF_3$ $IV. Tetrahedral$

Choose the correct answer from the options given below:

If lone pairs of electrons occupy axial positions,then which of the following statements about $ClF_3$ is incorrect?

What is the correct order of bond angles in $H_2O, NH_3, CH_4, CO_2$?

The incorrect geometry is represented by

In which of the following molecules/ions are all the bonds not equal?

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