A) \[{{H}_{2}}>L{{i}_{2}}>{{B}_{2}}\]
B) \[L{{i}_{2}}>{{H}_{2}}>{{B}_{2}}\]
C) \[L{{i}_{2}}>{{B}_{2}}>{{H}_{2}}\]
D) \[{{B}_{2}}>{{H}_{2}}>L{{i}_{2}}\]
Correct Answer: A
Solution :
[a] The molecular orbital configuration of the given molecules is \[{{H}_{2}}=\sigma {{(1s)}^{2}}\](no electron anti-bonding) \[L{{i}_{2}}=\sigma {{(1s)}^{2}}\sigma *{{(1s)}^{2}}\sigma {{(2s)}^{2}}\] (two anti-bonding electrons) \[{{B}_{2}}=\sigma {{(1s)}^{2}}\sigma *(1{{s}^{2}})\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\] \[\pi {{(2{{p}_{x}})}^{1}}=\pi {{(2{{p}_{y}})}^{1}}\] (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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