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(A) The shapes of the given species are determined using $VSEPR$ theory:$A$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ ato

Vedclass · Accepted Answer

$A(3), B(4), C(2), D(1)$

Vedclass · Answer

(A) The shapes of the given species are determined using $VSEPR$ theory:$A$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The geometry is trigonal bipyramidal with one equatorial position occupied by a lone pair,resulting in a See-Saw shape $(A-3)$.$B$. $BrF_3$: Bromine has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The geometry is trigonal bipyramidal with two equatorial positions occupied by lone pairs,resulting in a Bent $T$-shaped structure $(B-4)$.$C$. $BrO_3^-$: Bromine has $7$ valence electrons. It forms $3$ double bonds with $O$ atoms and has $1$ lone pair. The geometry is tetrahedral with one position occupied by a lone pair,resulting in a Pyramidal shape $(C-2)$.$D$. $NH_4^+$: Nitrogen has $5$ valence electrons. It forms $4$ bonds with $H$ atoms and has no lone pairs. The geometry is Tetrahedral $(D-1)$.Thus,the correct matching is $A-3, B-4, C-2, D-1$.

Column-$I$	Column-$II$ (Shape)
$A$. $SF_4$	$1$. Tetrahedral
$B$. $BrF_3$	$2$. Pyramidal
$C$. $BrO_3^-$	$3$. See-Saw shaped
$D$. $NH_4^+$	$4$. Bent $T$-shaped

Match the species given in Column-$I$ with the shape given in Column-$II$ and mark the correct option:-

Column-$I$ Column-$II$ (Shape)

$A$. $SF_4$ $1$. Tetrahedral

$B$. $BrF_3$ $2$. Pyramidal

$C$. $BrO_3^-$ $3$. See-Saw shaped

$D$. $NH_4^+$ $4$. Bent $T$-shaped

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Match the species given in Column-$I$ with the shape given in Column-$II$ and mark the correct option:- Column-$I$ Column-$II$ (Shape) $A$. $SF_4$ $1$. Tetrahedral $B$. $BrF_3$ $2$. Pyramidal $C$. $BrO_3^-$ $3$. See-Saw shaped $D$. $NH_4^+$ $4$. Bent $T$-shaped