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

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(B) $1$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ and has $1$ lone pair. The hybridization is $sp^3d$,and the shape is Se

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

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(B) $1$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ and has $1$ lone pair. The hybridization is $sp^3d$,and the shape is See-saw $(III)$.$2$. $BrF_3$: Bromine has $7$ valence electrons. It forms $3$ bonds with $F$ and has $2$ lone pairs. The hybridization is $sp^3d$,and the shape is Bent $T$-shape $(IV)$.$3$. $BrO_3^{-}$: Bromine has $7$ valence electrons. It forms $3$ bonds with $O$ and has $1$ lone pair. The hybridization is $sp^3$,and the shape is Pyramidal $(II)$.$4$. $NH_4^{+}$: Nitrogen has $5$ valence electrons. It forms $4$ bonds with $H$ and has $0$ lone pairs. The hybridization is $sp^3$,and the shape is Tetrahedral $(I)$.Thus,the correct matching is $A-III, B-IV, C-II, D-I$.

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

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 :

Similar Questions

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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 :