MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    From elementry molecular orbital theory we can give the electronic configuration of the singly positive nitrogen molecular ion,\[N_{2}^{+}\]as

    A) \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{p}^{4}},\sigma 2{{p}^{1}}\]

    B) \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\pi 2{{p}^{3}}\]

    C) \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2{{p}^{3}},\pi 2{{p}^{2}}\]               

    D)  \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2{{p}^{2}},\pi 2{{p}^{4}}\]

    Correct Answer: A

    Solution :

                     \[N_{2}^{+}=(7+7-1=13)\] \[=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\pi 2p_{x}^{2}\] \[\approx \pi 2p_{y}^{2},\sigma 2P_{z}^{1}\]


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