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$ (A) $ The shapes of the given molecules/ions are determined by $VSEPR$ theory:
$A. I_3^-$: The central iodine atom has $3$ lone pairs and $2$ bond

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$A-4, B-1, C-3, D-2$

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$ (A) $ The shapes of the given molecules/ions are determined by $VSEPR$ theory:
$A. I_3^-$: The central iodine atom has $3$ lone pairs and $2$ bond pairs, resulting in a linear geometry ($sp^3d$ hybridization). Thus, $A-4$.
$B. ClF_3$: The central chlorine atom has $2$ lone pairs and $3$ bond pairs, resulting in a $T$-shaped geometry ($sp^3d$ hybridization). Thus, $B-1$.
$C. H_2O$: The central oxygen atom has $2$ lone pairs and $2$ bond pairs, resulting in an angular or bent geometry ($sp^3$ hybridization). Thus, $C-3$.
$D. SF_4$: The central sulfur atom has $1$ lone pair and $4$ bond pairs, resulting in a see-saw geometry ($sp^3d$ hybridization). Thus, $D-2$.
Therefore, the correct matching is $A-4, B-1, C-3, D-2$.

Match the following:
$($Molecule$/$Ion$)$ $($Shape$)$
$A. I_3^- \rightarrow 4.$ Linear
$B. ClF_3 \rightarrow 1.$ $T-$shaped
$C. H_2O \rightarrow 3.$ Angular
$D. SF_4 \rightarrow 2.$ See$-$Saw

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Match the following: $($Molecule$/$Ion$)$ $($Shape$)$ $A. I_3^- \rightarrow 4.$ Linear $B. ClF_3 \rightarrow 1.$ $T-$shaped $C. H_2O \rightarrow 3.$ Angular $D. SF_4 \rightarrow 2.$ See$-$Saw