A) \[{{O}_{2}}\]
B) \[{{O}_{2}}^{2-}\]
C) \[O_{2}^{-}\]
D) \[{{O}_{2}}^{+}\]
Correct Answer: B
Solution :
Key Idea If a species does not have any unpaired electron, then it is called diamagnetic and if a species has unpaired electron, then it is called paramagnetic. Oxygen molecule \[({{O}_{2}})\] has following electronic configuration. \[{{O}_{2}}\] K K \[{{(\sigma 2s)}^{2}}{{(\overset{*}{\mathop{\sigma }}\,\,2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}\] \[{{(\overset{*}{\mathop{\pi }}\,2{{p}_{x}})}^{1}}{{(\overset{*}{\mathop{\pi }}\,2{{p}_{y}})}^{1}}\] This is paramagnetic due to presence of two unpaired electrons. Peroxide ion \[(O_{2}^{2-})\] has following electronic configuration. \[O_{2}^{2-}\] K K \[{{(\sigma 2s)}^{2}}{{(\overset{*}{\mathop{\sigma }}\,2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}){{(\pi 2{{p}_{y}})}^{2}}\] \[{{(\overset{*}{\mathop{\pi }}\,2{{p}_{x}})}^{2}}{{(\overset{*}{\mathop{\pi }}\,2{{p}_{y}})}^{2}}\] There is no unpaired electron in \[O_{2}^{2-}\] ion. So, it is diamagnetic. Superoxide ion \[(O_{2}^{-})\] has following electronic configuration. \[O_{2}^{-}\] KK \[{{(\sigma 2s)}^{2}}{{(\overset{*}{\mathop{\sigma }}\,2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}\] \[{{(\overset{*}{\mathop{\pi }}\,2{{p}_{x}})}^{2}}{{(\overset{*}{\mathop{\pi }}\,2{{p}_{y}})}^{1}}\] There is one unpaired electron in \[O_{2}^{-}\]. So, it is paramagnetic. Oxygen molecule ion \[(O_{2}^{+})\] has following electronic configuration. \[O_{2}^{+}\] K K \[{{(\sigma 2s)}^{2}}{{(\overset{*}{\mathop{\sigma }}\,2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}\] \[{{(\overset{*}{\mathop{\pi }}\,2{{p}_{x}})}^{1}}\] There is one unpaired electron in \[O_{2}^{+}\]. So, it is paramagnetic.
You need to login to perform this action.
You will be redirected in
3 sec
