Question

Which of the following diatomic molecular species has only $\pi$ bonds according to molecular orbital theory?

• A. $O_2$
• B. $N_2$
• C. $C_2$
• D. $B{e}_2$

Answer 1

The correct option is C $C_2$

Molecular orbital configuration of $C_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2=\pi 2p_y^2$
$\text{Bond order}=\frac{1}{2}[N_b-N_a]$
where, $N_b$ and $N_a$ are number of electrons in bonding and antibonding molecular orbitals respectively.
Therefore, bond order of $C_2$ is 2 and both are $\pi$ bonds.
Thus, $C_2$ molecule contains only $\pi$- bonds according to MOT.

MO electronic configuration of $O_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$
Bond order = 2
1 $\sigma$ and 1 $\pi$-bond.

MO electronic configuration of $N_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2=\pi 2p_y^2,\sigma 2p_z^2$
Bond order = 3
1 $\sigma$ and 2 $\pi$-bond.

MO electronic configuration of $B{e}_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2$
Bond order = 0
Thus, $B{e}_2$ does not exist.

Hence, option (c) is the correct answer.