The molecular orbital configuration of the given molecules is
$H_2 = \sigma 1s^2$ (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 specie 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
$H_2 > Li_2 >B_2$
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Consider the above reaction and identify the missing reagent/chemical.