A) both paramagnetic character and bond order increase
B) bond order decreases
C) paramagnetic character increases
D) paramagnetic character decreases and the bond order increases
Correct Answer: D
Solution :
(i) \[{{O}_{2}}-{{(\sigma 1s)}^{2}}{{(\sigma *1s)}^{2}}{{(\sigma 2s)}^{2}}{{(\sigma *2s)}^{2}}\] \[{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{(\pi *2{{p}_{x}})}^{1}}\] \[{{(\pi *2{{p}_{y}})}^{1}}\] Bond order \[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] \[=\frac{8-4}{2}\] \[=2\] \[{{O}_{2}}\] molecule having 2 unpaired electron. (ii) \[O_{2}^{+}-{{(\sigma 1s)}^{2}}{{(\sigma *1s)}^{2}}{{(\sigma 2s)}^{2}}{{(\sigma *2s)}^{2}}\] \[{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{(\pi *2{{p}_{x}})}^{1}}\] \[{{(\pi *2{{p}_{y}})}^{1}}\] Bond order \[=\frac{8-3}{2}\] \[=2.5\] \[O_{2}^{+}\] ion having only 1 unpaired electron. Hence, when \[{{O}_{2}}\] is converted into \[O_{2}^{+}\] paramagnetic character decrease and the bond order increases.
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