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(B) $1$. $SO_2Cl_2$: The central $S$ atom has $4$ bonding pairs (two $S=O$ and two $S-Cl$ bonds) and $0$ lone pairs. Hybridization is $sp^3$,resulting

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

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(B) $1$. $SO_2Cl_2$: The central $S$ atom has $4$ bonding pairs (two $S=O$ and two $S-Cl$ bonds) and $0$ lone pairs. Hybridization is $sp^3$,resulting in a Tetrahedral shape $(III)$.$2$. $NO$: Total valence electrons = $5 + 6 = 11$. It has an odd number of electrons,making it Paramagnetic $(I)$.$3$. $NO_2^-$: The central $N$ atom has $2$ bonding pairs and $1$ lone pair. It is Diamagnetic $(II)$.$4$. $I_3^-$: The central $I$ atom has $2$ bonding pairs and $3$ lone pairs. Hybridization is $sp^3d$,resulting in a Linear shape $(IV)$.Matching: $A-III, B-I, C-II, D-IV$.

List-$I$ (Molecule Species)	List-$II$ (Property/Shape)
$A$. $SO_2Cl_2$	$I$. Paramagnetic
$B$. $NO$	$II$. Diamagnetic
$C$. $NO_2^-$	$III$. Tetrahedral
$D$. $I_3^-$	$IV$. Linear

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

List-$I$ (Molecule Species) List-$II$ (Property/Shape)

$A$. $SO_2Cl_2$ $I$. Paramagnetic

$B$. $NO$ $II$. Diamagnetic

$C$. $NO_2^-$ $III$. Tetrahedral

$D$. $I_3^-$ $IV$. Linear

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Match List-$I$ with List-$II$ : List-$I$ (Molecule Species) List-$II$ (Property/Shape) $A$. $SO_2Cl_2$ $I$. Paramagnetic $B$. $NO$ $II$. Diamagnetic $C$. $NO_2^-$ $III$. Tetrahedral $D$. $I_3^-$ $IV$. Linear Choose the correct answer from the options given below :