A: Tetracyanomethane, B: Carbon dioxide, C: Benzene, and D: 1,3-Butadiene. Determine the correct order of the ratio of sigma (σ) and pi (π) bonds:
(A) A = B < C < D
(B) A = B < D < C
(C) A = B = C = D
(D) C < D < B < A

To determine the order of the ratio of σ and π bonds for the given compounds, let's analyze each compound:

A: tetracyanomethane

Tetracyanomethane has a central carbon atom bonded to four cyanide groups (−C≡N). Each carbon-nitrogen bond consists of one σ bond and two π bonds, resulting in a total of four σ bonds and eight π bonds.
B: carbon dioxide

Carbon dioxide has two double bonds between the central carbon atom and the oxygen atoms. Each bond consists of one σ bond and one π bond, resulting in a total of two σ bonds and two π bonds.
C: benzene

Benzene has a ring structure with alternating single and double bonds. Each bond consists of one σ bond and one π bond. In benzene, there are six σ bonds and six π bonds.
D: 1,3-butadiene

1,3-butadiene has two double bonds and two single bonds. Each double bond consists of one σ bond and two π bonds, while each single bond consists of one σ bond. So, there are two σ bonds and four π bonds in 1,3-butadiene.
Now, let's compare the ratios of σ and π bonds for each compound:

A: 4σ/8π
B: 2σ/2π
C: 6σ/6π
D: 2σ/4π

From the analysis, we can see that:

Compound A (tetracyanomethane) has the highest ratio of σ to π bonds.
Compound B (carbon dioxide) has the lowest ratio of σ to π bonds.
Compounds C (benzene) and D (1,3-butadiene) have equal ratios of σ to π bonds.
So, the correct order is:
(D) C < D < B < A