A) CO
B) \[\text{O}_{\text{2}}^{\text{-}}\]
C) \[\text{C}{{\text{N}}^{\text{-}}}\]
D) \[\text{N}{{\text{O}}^{+}}\]
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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