RAJASTHAN PMT Rajasthan - PMT Solved Paper-2006

  • question_answer
    The bond order of \[O_{2}^{+}\] is the same as in:

    A) \[N_{2}^{+}\]                                   

    B)        \[C{{N}^{-}}\]

    C)        \[CO\]                                  

    D)        \[N{{O}^{+}}\]

    Correct Answer: A

    Solution :

    \[O_{2}^{+}(15{{e}^{-}})=K\,\,K*{{(\sigma 2s)}^{2}}{{(\sigma *2s)}^{2}}\]                                 \[{{(\sigma 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{2}}\]                                           \[{{(\pi *2{{p}_{y}})}^{1}}{{(\pi *2{{p}_{z}})}^{0}}\] Hence, bond order\[=\frac{1}{2}(10-5)=2.5\] \[N_{2}^{+}(13{{e}^{-}})=K\,\,K*{{(\sigma 2s)}^{2}}{{(\sigma *2s)}^{2}}{{(\sigma *2{{p}_{x}})}^{2}}\]                                                 \[{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{1}}\] Hence, bond order\[=\frac{1}{2}(9-4)=5\]


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