Based upon VSEPR theory,match the shape (geom | vedclass

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(B) To determine the shape of the molecules,we use the $VSEPR$ theory:$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bond

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

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(B) To determine the shape of the molecules,we use the $VSEPR$ theory:$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $3 + 2 = 5$ ($sp^3d$ hybridization). Due to $2$ lone pairs in equatorial positions,the shape is $T$-shaped $(A-III)$.$2$. $BF_3$: The central atom $B$ has $3$ valence electrons. It forms $3$ bonds with $F$ atoms and has $0$ lone pairs. The steric number is $3$ ($sp^2$ hybridization). The shape is trigonal planar $(B-IV)$.$3$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$ ($sp^3d^2$ hybridization). Due to $2$ lone pairs in axial positions,the shape is square planar $(C-I)$.$4$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The steric number is $4 + 1 = 5$ ($sp^3d$ hybridization). Due to $1$ lone pair in an equatorial position,the shape is see-saw $(D-II)$.Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

List-$I$ (Shape)	List-$II$ (Molecules)
$A$. $T$-shaped	$I$. $XeF_4$
$B$. Trigonal planar	$II$. $SF_4$
$C$. Square planar	$III$. $ClF_3$
$D$. See-saw	$IV$. $BF_3$

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