Question

A: tetracyanomethane B: Carbon dioxide C: Benzene D: 1,3-buta-di-ene. Ratio of $\sigma$ and $\pi$ bonds is in order:

• A. $A=B<C<D$
• B. $A=B<D<C$
• C. $A=B=C=D$
• D. $C<D<A<B$

Answer 1

The correct option is A $A=B<C<D$

Tetracyanomethane: $C(CN{)}_4$ has 8 sigma and 8 pi bond so ration is 1.
$C{O}_2$ has 2 sigma and 2 pi bond so ration is 1.

Benzene has 6 sigma bonds and 3 pi bonds. The ratio is 2.

$1,3$-buta-di-ene has 9 sigma bonds and 2 pi bonds. The ratio is 9/2. Hence, the order:$A=B<C<D$