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

The Correct Option is D

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.