A) \[{{N}_{2}}\]
B) \[{{C}_{2}}\]
C) \[N_{2}^{+}\]
D) \[O_{2}^{2-}\]
Correct Answer: C
Solution :
\[{{C}_{2}}\to \]the molecular orbital configuration(12) is \[\to \]\[{{C}_{2}}=[{{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}]\] \[{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{2}}]\] \[{{C}_{2}}\]molecule is diamagnetic as it does not contain any unpaired electron \[{{N}_{2}}\to \]the molecular orbital configuration (14) is \[{{N}_{2}}=[(\sigma 1{{s}^{2}}){{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}\] \[{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{2}}{{(\sigma 2{{p}_{x}})}^{1}}]\] \[{{N}_{2}}\]molecule is diamagnetic because it does not contain any unpaired electron. \[N_{2}^{+}\to \]the molecular orbital configuration (13) is \[N_{2}^{+}=[(\sigma 1{{s}^{2}}){{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}\] \[{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{2}}{{(\sigma 2{{p}_{x}})}^{2}}]\] \[{{N}_{2}}\]molecule is paramagnetic because it contain one unpaired electron. \[O_{2}^{2-}\to \]the molecular orbital configuration of \[O_{2}^{2-}\] (18) is \[O_{2}^{2-}=[{{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}\] \[{{(\sigma 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{2}}{{({{\pi }^{*}}2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{z}})}^{2}}]\] It does not contain any unpaired electron so, it is diamagnetic
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