Question

Which of the following diatomic molecular species has only $\mathrm{\pi }$ bonds according to Molecular orbital Theory?

• A. ${\mathrm{O}}_2$
• B. ${\mathrm{N}}_2$
• C. ${\mathrm{C}}_2$
• D. ${\mathrm{Be}}_2$

Answer 1

The correct option is C

${\mathrm{C}}_2$

The correct option is C.

Explanation for correct option:

1. MO configuration ${\mathrm{C}}_2\text{is}\sigma 1{\mathrm{s}}^2,{\sigma }^{*}1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,{\sigma }^{*}2{\mathrm{s}}^2,\pi 2{\mathrm{p}}_{\mathrm{x}}^2=\pi 2{\mathrm{p}}_{\mathrm{y}}^2$​
2. So, ${\mathrm{C}}_2$​ molecule contains only $\mathrm{\pi }$− bonds according to MOT.

Explanation for incorrect options:

Option A:

1. Each oxygen atom possesses eight electrons, resulting in a total of 16 electrons in the ${\mathrm{O}}_2$ molecule.

Option B:

1. In the nitrogen molecule, there are a total of $6$ electrons involved in bond formation.
2. The nitrogen molecule has three bonds (one sigma and two $\mathrm{\pi }$bonds).

Option D:

1. There are no singly full atomic orbitals in beryllium, as evidenced by the electronic arrangement.
2. Because overlapping is impossible without the half-filled orbital, the ${\mathrm{Be}}_2$ molecule does not exist.

Final answer: ${\mathrm{C}}_2$ is the diatomic molecular species that has Only $\mathrm{\pi }$ bonds according to molecular orbital theory.