A) \[CO\]
B) \[O_{2}^{-}\]
C) \[C{{N}^{-}}\]
D) \[N{{O}^{+}}\]
Correct Answer: B
Solution :
| Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration. |
| Molecular orbital configuration of the given species is as \[CO(6+8=14)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\]\[\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\pi 2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] |
| (All the electrons are paired so it is diamagnetic.) \[O_{2}^{-}(8+8+1=17)\] |
| \[=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\overset{{}}{\mathop{\sigma }}\,2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,{{s}^{2}},\overset{{}}{\mathop{\sigma 2p_{z}^{2}\pi 2p_{x}^{2}}}\,\] |
| \[\overset{{}}{\mathop{\pi }}\,2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] |
| It contains one unpaired electron so it is paramagnetic. |
| \[C{{N}^{-}}(6+7+1=14)=\]same as CO |
| \[N{{O}^{+}}(7+8-1)=\]same as CO |
| Thus, among the given species only \[O_{2}^{-}\]is paramagnetic. |
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