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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