A) \[\sigma \] orbital
B) \[\pi \] orbital
C) \[{{\sigma }^{*}}\] orbital
D) \[{{\pi }^{*}}\] orbital
Correct Answer: D
Solution :
MO electronic configuration of \[N{{O}^{+}}={{(\sigma 1s)}^{2}}{{(\overset{*}{\mathop{\sigma }}\,1s)}^{2}}{{(\sigma 2s)}^{2}}{{(\overset{*}{\mathop{\sigma }}\,2s)}^{2}}{{(\pi 2{{p}_{x}})}^{2}}\] \[{{(\pi 2{{p}_{y}})}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}\] \[NO={{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}\] \[{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{y}})}^{1}}\] Hence, electron is added to n* orbital.
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