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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Among the following species, identify the pair having same bond order\[C{{N}^{-}},O_{2}^{-},N{{O}^{+}},C{{N}^{+}}\]

    A) \[C{{N}^{-}}\]and\[O_{2}^{-}\]                 

    B) \[O_{2}^{-}\]and\[N{{O}^{+}}\]

    C) \[C{{N}^{-}}\]and \[N{{O}^{+}}\]  

    D)        \[C{{N}^{-}}\]and \[C{{N}^{+}}\]

    E) \[N{{O}^{+}}\]and\[C{{N}^{+}}\]

    Correct Answer: C

    Solution :

    Species having the same number of electrons, have same bond order.
    Species Number of electrons
    \[C{{N}^{-}}\] \[6+7+1=14\]
    \[O_{2}^{-}\] \[8+8+1=17\]
    \[N{{O}^{+}}\] \[7+8-1=14\]
    \[C{{N}^{+}}\] \[6+7-1=12\]
    Since,\[C{{N}^{-}}\]and\[N{{O}^{-}}\]have same number of electrons, they have same bond order, ie, 3. \[C{{N}^{-}}\]or \[N{{O}^{+}}=14=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\]                                                 \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2}\] Bond order\[=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-4}{2}=3\]


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