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Question

(D) The bond order for all three molecules is $1$. The molecular orbital configurations are: $H_2: (\sigma 1s)^2$ ($0$ anti-bonding electrons) $Li_2:

Vedclass · Accepted Answer

$H_2 > Li_2 > B_2$

Vedclass · Answer

(D) The bond order for all three molecules is $1$. The molecular orbital configurations are: $H_2: (\sigma 1s)^2$ ($0$ anti-bonding electrons) $Li_2: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$ ($2$ anti-bonding electrons) $B_2: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_y)^1 (\pi 2p_z)^1$ ($4$ anti-bonding electrons) Stability is inversely proportional to the number of anti-bonding electrons. Therefore,the correct order of stability is $H_2 > Li_2 > B_2$.

Bond order normally gives an idea of the stability of a molecular species. All the molecules,namely $H_2$,$Li_2$,and $B_2$,have the same bond order,yet they are not equally stable. Their stability order is:

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