Match List-I with List-II : List-I Li | vedclass

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(A) To determine the shape of the given species,we use the $VSEPR$ theory:$1$. $H_3O^+$: Oxygen has $6$ valence electrons. It forms $3$ bonds with $H$

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

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(A) To determine the shape of the given species,we use the $VSEPR$ theory:$1$. $H_3O^+$: Oxygen has $6$ valence electrons. It forms $3$ bonds with $H$ and has $1$ lone pair. Total electron pairs = $4$ ($sp^3$ hybridization). Due to $1$ lone pair,the shape is Pyramidal $(III)$.$2$. Acetylide anion $(C_2^{2-})$: The structure is $[C \equiv C]^{2-}$. The carbon atoms are $sp$ hybridized,resulting in a Linear $(II)$ shape.$3$. $NH_4^+$: Nitrogen has $5$ valence electrons. It forms $4$ bonds with $H$ and has $0$ lone pairs. Total electron pairs = $4$ ($sp^3$ hybridization). The shape is Tetrahedral $(I)$.$4$. $ClO_2^-$: Chlorine has $7$ valence electrons. It forms $2$ bonds with $O$ and has $2$ lone pairs. Total electron pairs = $4$ ($sp^3$ hybridization). Due to $2$ lone pairs,the shape is Bent $(IV)$.Matching the results: $A-III, B-II, C-I, D-IV$.

List-$I$	List-$II$
$A$. $H_3O^+$	$I$. Tetrahedral
$B$. Acetylide anion	$II$. Linear
$C$. $NH_4^+$	$III$. Pyramidal
$D$. $ClO_2^-$	$IV$. Bent

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

List-$I$ List-$II$

$A$. $H_3O^+$ $I$. Tetrahedral

$B$. Acetylide anion $II$. Linear

$C$. $NH_4^+$ $III$. Pyramidal

$D$. $ClO_2^-$ $IV$. Bent

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Match List-$I$ with List-$II$ : List-$I$ List-$II$ $A$. $H_3O^+$ $I$. Tetrahedral $B$. Acetylide anion $II$. Linear $C$. $NH_4^+$ $III$. Pyramidal $D$. $ClO_2^-$ $IV$. Bent Choose the correct answer from the options given below :