Chemistry - CHEMICAL BONDING Question with Solution | TestHub

To determine the ratio of $\sigma$ and $\pi$ bonds, we analyze each molecule:

A: Tetracyanomethane (C(CN)₄)

Structure: A central carbon atom bonded to four cyano groups. Each CN group has a C$\equiv$N triple bond.

$\sigma$ bonds: 4 (C-C) + 4 (C$\equiv$N, one $\sigma$ bond per triple bond) = 8

$\pi$ bonds: 4 (C$\equiv$N, two $\pi$ bonds per triple bond) $\times$ 2 = 8

Ratio $\sigma :\pi =8:8=1:1$

B: Carbon dioxide (CO₂)

Structure: O=C=O

$\sigma$ bonds: 2 (C=O, one $\sigma$ bond per double bond) = 2

$\pi$ bonds: 2 (C=O, one $\pi$ bond per double bond) = 2

Ratio $\sigma :\pi =2:2=1:1$

C: Benzene (C₆H₆)

Structure: Cyclic, alternating single and double bonds (delocalized).

$\sigma$ bonds: 6 (C-C) + 6 (C-H) = 12

$\pi$ bonds: 3 (from the double bonds) = 3

Ratio $\sigma :\pi =12:3=4:1$

D: 1,3-Butadiene (CH₂=CH-CH=CH₂)

Structure: Linear chain with two double bonds.

$\sigma$ bonds: 3 (C-C) + 6 (C-H) = 9

$\pi$ bonds: 2 (from the double bonds) = 2

Ratio $\sigma :\pi =9:2=4.5:1$

Comparing the ratios:

A: $1:1=1$

B: $1:1=1$

C: $4:1=4$

D: $4.5:1=4.5$

Order of $\sigma :\pi$ ratio: A = B < C < D

The final answer is $\boxed{A}$