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 \[CO(6+8=14)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\] \[\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] (All the electrons are paired so it is diamagnetic.) \[O_{2}^{-}(8+8+1=17)\] \[=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\] \[=\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] 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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