Question

$N_2$ and $O_2$ are converted to monopositive cations $N_2^{+}$ and $O_2^{+}$ respectively. Which is incorrect?

• A. In $N_2^{+}$, the $N-N$ bond is weakened
• B. In $O_2^{+}$, the bond order increases
• C. In $O_2^{+}$, the paramagnetism decreases
• D. $N_2^{+}$ becomes diamagnetic

Answer 1

The correct option is D $N_2^{+}$ becomes diamagnetic

Molecular orbital configuration of $N_2$:
$\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\pi 2p_x{)}^2\right(\pi 2p_y{)}^2(\sigma 2p_z{)}^2,B.O.=\frac{10-4}{2}=3$
Molecular orbital configuration of $N_2^{+}$:
$\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\pi 2p_x{)}^2\right(\pi 2p_y{)}^2(\sigma 2p_z{)}^{1}B.O.=\frac{9-4}{2}=2.5$
Molecular orbital configuration of $O_2$ :
$\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x{)}^2\left(\pi 2p_y{)}^2\right({\pi }^{\ast }2p_x{)}^1({\pi }^{\ast }2p_y{)}^{1}B.O.=\frac{10-6}{2}=2$
Molecular orbital configuration of $O_2^{+}$:
$\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x{)}^2\left(\pi 2p_y{)}^2\right({\pi }^{\ast }2p_x{)}^1({\pi }^{\ast }2p_y{)}^{0}B.O.=\frac{10-5}{2}=2.5$
$\text{Bond order}\propto \text{bond energy}\propto \text{strength of the bond}$
Out of all, option d is incorrect, as it contains unpaired electron in $\sigma 2p_z^1$ orbital. Hence, $N_2^{+}$ is paramagnetic.