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(B) To find the ratio of $\sigma$ to $\pi$ bonds,we analyze the structures:$1$. For $HCO_3^{-}$: The structure is $[HO-C(=O)-O]^{-}$. It has $5 \sigma

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$CH_2(CN)_2$

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(B) To find the ratio of $\sigma$ to $\pi$ bonds,we analyze the structures:$1$. For $HCO_3^{-}$: The structure is $[HO-C(=O)-O]^{-}$. It has $5 \sigma$ and $1 \pi$ bond. Ratio = $5:1$.$2$. For $CH_2(CN)_2$: The structure is $NC-CH_2-CN$. It has $9 \sigma$ and $4 \pi$ bonds. Ratio = $9:4$.$3$. For $HClO_4$: The structure is $HO-ClO_3$. It has $7 \sigma$ and $3 \pi$ bonds. Ratio = $7:3$.$4$. For $XeO_3$: The structure is a trigonal pyramidal molecule with $3 \sigma$ and $3 \pi$ bonds. Ratio = $1:1$.Wait,let us re-evaluate $CH_2(CN)_2$: $H-C(H)(C \equiv N)_2$. Total $\sigma$ bonds: $2$ $(C-H)$ + $1$ $(C-C)$ + $2$ $(C-C)$ + $2$ $(C-N)$ = $7$. Total $\pi$ bonds: $2 	imes 2 = 4$. Ratio $7:4$.Let us check $HCO_3^{-}$ again: $H-O-C(=O)-O^{-}$. $\sigma$ bonds: $1$ $(O-H)$ + $1$ $(C-O)$ + $1$ $(C=O)$ + $1$ $(C-O^-)$ = $4$. $\pi$ bonds: $1$. Ratio $4:1$.Let us check $HClO_4$: $H-O-Cl(=O)_3$. $\sigma$ bonds: $1$ $(O-H)$ + $1$ $(Cl-O)$ + $3$ $(Cl=O)$ = $5$. $\pi$ bonds: $3$. Ratio $5:3$.Re-evaluating the question: If we consider $CH_2(CN)_2$ as $CH_2(CN)_2$,the ratio is $9:4$. If we consider $C_3O_2$ (not in options),it is $3:2$. Given the options,let us re-examine $CH_2(CN)_2$ structure: $N \equiv C-CH_2-C \equiv N$. $\sigma$ bonds: $N-C$ $(2)$,$C-C$ $(2)$,$C-H$ $(2)$ = $6$. $\pi$ bonds: $N \equiv C$ $(4)$. Ratio $6:4 = 3:2$. Thus,the correct option is $B$.

List-$I$ (Species)	List-$II$ (Number of lone pairs)
$A. \ CH_3COCH_3$	$I. \ 2$
$B. \ CH_3CO^+$	$II. \ 0$
$C. \ CH_3CH_2^+$	$III. \ 1$
	$IV. \ 3$

Identify the molecule / ion in which the ratio of $\sigma$ to $\pi$ bonds is $3: 2$.

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