• # 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

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 ${{H}_{2}}>L{{i}_{2}}>{{B}_{2}}$