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
E) Both\[{{O}_{2}},O_{2}^{+}\]are paramagnetic
Correct Answer: D
Solution :
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.
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