JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer Bond order normally gives idea of stability of a molecular species. All the molecules viz. \[{{\operatorname{H}}_{2}},\] \[\operatorname{L}{{\operatorname{i}}_{2}}\] and \[{{\operatorname{B}}_{2}}\] have the same bond order yet they are not equally stable. The Their stability order is:     JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[{{\text{H}}_{\text{2}}}>{{\text{B}}_{\text{2}}}>\text{L}{{\text{i}}_{\text{2}}}\]      

    B)  \[\text{L}{{\text{i}}_{\text{2}}}\text{}{{\text{H}}_{\text{2}}}>{{\text{B}}_{\text{2}}}\]

    C)  \[{{\operatorname{Li}}_{2}}>{{\text{B}}_{\text{2}}}>{{\operatorname{H}}_{2}}\]  

    D)  \[{{\text{B}}_{\text{2}}}>{{\operatorname{H}}_{2}}>{{\operatorname{Li}}_{2}}\]

    E) (e) None of these

    Correct Answer: E

    Solution :

    None of the given option is correct.         The molecular orbital configuration of the given molecules is \[{{H}_{2}}=\sigma 1{{s}^{2}}\] (no electron anti-bonding) \[L{{i}_{2}}=\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}\](two anti-bonding electrons) \[{{B}_{2}}=\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\left\{ \pi 2p_{y}^{1}=\pi 2p_{z}^{1} \right\}\]  (4 anti-bonding electrons) Though the bond order of all the species are same (B.O = 1) but stability is different. This is due to difference in the presence of no. of anti-bonding electron. Higher the no. of anti-bonding electron lower is the stability hence the correct order is \[{{H}_{2}}>L{{i}_{2}}>{{B}_{2}}\]


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