Match List-I with List-II: List-I (Molec | vedclass

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(C) The shapes of the molecules are determined by $VSEPR$ theory:$1$. $NH_3$: Central atom $N$ has $3$ bond pairs and $1$ lone pair,resulting in a tri

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$A-III, B-I, C-IV, D-II$

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(C) The shapes of the molecules are determined by $VSEPR$ theory:$1$. $NH_3$: Central atom $N$ has $3$ bond pairs and $1$ lone pair,resulting in a trigonal pyramidal shape $(A-III)$.$2$. $BrF_5$: Central atom $Br$ has $5$ bond pairs and $1$ lone pair,resulting in a square pyramidal shape $(B-I)$.$3$. $PCl_5$: Central atom $P$ has $5$ bond pairs and $0$ lone pairs,resulting in a trigonal bipyramidal shape $(C-IV)$.$4$. $CH_4$: Central atom $C$ has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral shape $(D-II)$.Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $NH_3$	$I$. Square pyramid
$B$. $BrF_5$	$II$. Tetrahedral
$C$. $PCl_5$	$III$. Trigonal pyramidal
$D$. $CH_4$	$IV$. Trigonal bipyramidal

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

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $NH_3$ $I$. Square pyramid

$B$. $BrF_5$ $II$. Tetrahedral

$C$. $PCl_5$ $III$. Trigonal pyramidal

$D$. $CH_4$ $IV$. Trigonal bipyramidal

Choose the correct answer from the options below:

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Match List-$I$ with List-$II$: List-$I$ (Molecule) List-$II$ (Shape) $A$. $NH_3$ $I$. Square pyramid $B$. $BrF_5$ $II$. Tetrahedral $C$. $PCl_5$ $III$. Trigonal pyramidal $D$. $CH_4$ $IV$. Trigonal bipyramidal Choose the correct answer from the options below: