Find the odd-electron molecules from the following. $$ \begin{array}{ccccc} \mathrm{C}_2 \mathrm{H}_2 & \mathrm{SCl}_2 & \mathrm{NO} & \mathrm{N}_2 & \mathrm{NO}_2 \\ \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } & \text { (v) } \end{array} $$

• A

  i, iii, iv
• B

  ii, iii
• C

  i, iv
• Correct answer
  D

  iii, v

Solution

$\mathrm{NO}$ and $\mathrm{NO}_2$ species are having odd electron and disobey the octet rule. $\ddot{N}=\ddot{O}$ is a stable free radical.
Seven electrons around nitrogen. Formal charges and the molecule's resonance structures are indicate.

Hence, the correct answer is option: 'D'.