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(A) $1$. For $A$ (Tetracyanomethane,$C(CN)_4$): Structure is $C(CN)_4$. It has $4$ $C-C$ $\sigma$ bonds,$4$ $C\equiv N$ $\sigma$ bonds,and $8$ $\pi$ b

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$A = B < C < D$

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(A) $1$. For $A$ (Tetracyanomethane,$C(CN)_4$): Structure is $C(CN)_4$. It has $4$ $C-C$ $\sigma$ bonds,$4$ $C\equiv N$ $\sigma$ bonds,and $8$ $\pi$ bonds. Total $\sigma = 8$,$\pi = 8$. Ratio $\sigma/\pi = 8/8 = 1$.$2$. For $B$ (Carbon dioxide,$CO_2$): Structure is $O=C=O$. It has $2$ $\sigma$ bonds and $2$ $\pi$ bonds. Ratio $\sigma/\pi = 2/2 = 1$.$3$. For $C$ (Benzene,$C_6H_6$): It has $12$ $\sigma$ bonds and $3$ $\pi$ bonds. Ratio $\sigma/\pi = 12/3 = 4$.$4$. For $D$ ($1, 3$-Butadiene,$CH_2=CH-CH=CH_2$): It has $9$ $\sigma$ bonds and $2$ $\pi$ bonds. Ratio $\sigma/\pi = 9/2 = 4.5$.Comparing the ratios: $A(1) = B(1) Thus,the correct order is $A = B < C < D$.

Calculate the ratio of $\sigma$ and $\pi$ bonds for the following compounds and determine the correct order:
$A$: Tetracyanomethane $(C(CN)_4)$
$B$: Carbon dioxide $(CO_2)$
$C$: Benzene $(C_6H_6)$
$D$: $1, 3$-Butadiene $(CH_2=CH-CH=CH_2)$

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Calculate the ratio of $\sigma$ and $\pi$ bonds for the following compounds and determine the correct order:$A$: Tetracyanomethane $(C(CN)_4)$$B$: Carbon dioxide $(CO_2)$$C$: Benzene $(C_6H_6)$$D$: $1, 3$-Butadiene $(CH_2=CH-CH=CH_2)$