BCl_3 is a planar molecule while NCl_3 is pyr | vedclass

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(A) According to $VSEPR$ theory,the shape of a molecule is determined by the number of bond pairs and lone pairs around the central atom.$BCl_3$: The

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$BCl_3$ has no lone pair of electrons but $NCl_3$ has a lone pair of electrons

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(A) According to $VSEPR$ theory,the shape of a molecule is determined by the number of bond pairs and lone pairs around the central atom.$BCl_3$: The central Boron atom has $3$ bond pairs and $0$ lone pairs,resulting in $sp^2$ hybridization and a trigonal planar geometry.$NCl_3$: The central Nitrogen atom has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization and a pyramidal geometry due to the repulsion between the lone pair and bond pairs.

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

$BCl_3$ is a planar molecule while $NCl_3$ is pyramidal,because

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The molecule which has a greater number of lone pairs of electrons than bond pairs of electrons on its central atom is

Apart from tetrahedral geometry,another possible geometry for $CH_4$ is square planar with the four $H$ atoms at the corners of the square and the $C$ atom at its centre. Explain why $CH_4$ is not square planar?

To which type does $ClF_3$ belong among $AB_4$,$AB_4E$,and $AB_3E_2$? Explain.

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 :

The structure of $PF_5$ molecule is

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