Match List-I with List-II:List-IList-IIA. XeF | vedclass

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(D) To determine the geometry of the molecules,we use the $VSEPR$ theory:$1$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ b

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

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(D) To determine the geometry of the molecules,we use the $VSEPR$ theory:$1$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $6$ ($sp^3d^2$ hybridization),resulting in a square planar geometry $(A-II)$.$2$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The steric number is $5$ ($sp^3d$ hybridization),resulting in a see-saw geometry $(B-I)$.$3$. $NH_4^{+}$: The central atom $N$ has $5$ valence electrons. It forms $4$ bonds with $H$ atoms and has no lone pairs. The steric number is $4$ ($sp^3$ hybridization),resulting in a tetrahedral geometry $(C-IV)$.$4$. $BrF_3$: The central atom $Br$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $5$ ($sp^3d$ hybridization),resulting in a bent $T$-shaped geometry $(D-III)$.Thus,the correct matching is $A-II, B-I, C-IV, D-III$.

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

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

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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: