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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    The electronic configuration of valence shell of nitrogen molecule in ground state is \[(\sigma 2{{s}^{2}}),({{\sigma }^{*}}2{{s}^{2}}),(\pi 2{{p}^{4}}),(\sigma 2{{p}^{2}})\]. Hence, the bond order in nitrogen molecule is

    A)  3                                                      

    B)  0

    C)  1                

    D)  2

    Correct Answer: A

    Solution :

     Bond order \[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] where,\[{{N}_{b}}=\]Number of bonding electrons \[{{N}_{a}}=\]Number of antibonding electrons \[{{N}_{2}}(14)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},\pi 2p_{y}^{2}\] \[\approx \pi 2p_{z}^{2},\sigma 2p_{x}^{2}\] Bond order of\[{{N}_{2}}=\frac{10-4}{2}=3\]


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