Question

The ratio of number of $\sigma$ - bonds to $\pi$ - bonds in $C_2(CN{)}_4$ molecule

Answer 1

No. of $\sigma$ - bonds$\left(n_1\right)$ $\to$ 9
No. of $\pi$ - bonds$\left(n_2\right)$ $\to$ 9
$\frac{n_1}{n_2}$ = $\frac{9}{9}$ = 1