Chemistry - CHEMICAL BONDING Question with Solution | TestHub

To determine the number of $\pi$-bonds and $\sigma$-bonds, draw the Lewis structure for each species.

A. C₂N₂ (Cyanogen): N≡C-C≡N.

$\pi$-bonds = 4 (two in each C≡N triple bond)

$\sigma$-bonds = 3 (one C-C, two C-N)

Here, $\pi$-bonds (4) > $\sigma$-bonds (3).

B. C₃O₂ (Carbon suboxide): O=C=C=C=O.

$\pi$-bonds = 4

$\sigma$-bonds = 4

Here, $\pi$-bonds (4) = $\sigma$-bonds (4).

C. N₂O₃ (Dinitrogen trioxide): O=N-N(O)-O.

$\pi$-bonds = 1

$\sigma$-bonds = 4

Here, $\pi$-bonds (1) < $\sigma$-bonds (4).

D. (SCN)₂ (Thiocyanogen): N≡C-S-S-C≡N.

$\pi$-bonds = 4

$\sigma$-bonds = 5

Here, $\pi$-bonds (4) < $\sigma$-bonds (5).

Thus, C₂N₂ has more $\pi$-bonds than $\sigma$-bonds.