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 $NN$ bond is weakened
• B. In $O_{2}^{+},$ the bond order increases
• C. In $O_{2}^{+},$ paramagnetism decreases
• D. $N_{2}^{+}$ becomes diamagnetic

Answer 1

Answer: (D)

$N_{2}^{+}$ becomes diamagnetic

Answer 2

MO configuration of ${{N}_{2}}$ molecule is ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\pi 2{{p}_{x}})}^{2}}$ ${{(\pi 2{{p}_{y}})}^{2}}{{(\sigma 2{{p}_{z}})}^{1}}$ $\therefore$ Bond order of ${{N}_{2}}=\frac{10-4}{2}=3$ MO configuration of $N_{2}^{+}$ is ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\pi 2{{p}_{x}})}^{2}}$ ${{(\pi 2{{p}_{y}})}^{2}}{{(\sigma 2{{p}_{z}})}^{1}}$ Thus, $N_{2}^{+}$ becomes paramagnetic. $\because$ Bond order of $N_{2}^{+}=\frac{9-4}{2}=2.5$ and $bond\text{ }order\propto bond\text{ }energy$ $\therefore$ In $N_{2}^{+},$ the $NN$ bond is weakened. MO configuration of ${{O}_{2}}$ is ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}$ ${{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{1}}{{({{\pi }^{*}}2{{p}_{y}})}^{1}}$ $\therefore$ Bond order $=\frac{10-6}{2}=2$ MO configuration of $O_{2}^{+}$ is ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}$ ${{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{1}}$ $\therefore$ Bond order $=\frac{10-5}{2}=2.5$ Hence, ${{O}_{2}}$ and $O_{2}^{+}$ both are paramagnetic, but in $O_{2}^{+}$ paramagnetism decreases.