Question

When $N_2$ goes to $N_2^{+}$, and when $O_2$ goes to $O_2^{+}$. What happens to the $N-N$ bond distance and the $O-O$ bond distance respectively?

• A. Decreases, increases
• B. Increases, Decreases
• C. Increases, increases
• D. None of these

Answer 1

The correct option is B Increases, Decreases

When $N_2$ goes to $N_2^{+}$ , the $N-N$ bond distance increases, and when $O_2$ goes to $O_2^{+}$, the $O-O$ bond distance decreases.
Electronic configuration of $N_2=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2.$
Hence, the bond order of the nitrogen molecule is $3$.
If an electron from bonding $\pi$ MO is lost to form $N_2^{+}$. The bond order decreases to $2.5$. This decreases the strength of the bond. As the bond strength decreases, bond length increases.
Electronic configution of $O_2=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1{\pi }^{\ast }2p_y^1$.
The bond order of oxygen is 2. An electron is lost from antibonding $\pi$ MO to form $O_2^{+}$. The bond order increases to $2.5$. This increases the strength of the bond and the bond distance decreases.