JEE Main 2013 — Chemical Bonding and Molecular Structure Question with Solution

The molecular orbital configuration of the given molecules is ${\mathrm{H}}_2=\sigma 1s^2$ (no electron anti-bonding) ${\mathrm{Li}}_2=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2$ (two anti-bonding electrons) $${\mathrm{B}}_2=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2$$ $$\left\{\pi 2p_y^1=\pi 2p_z^1\right\}$$ (4 anti-bonding electrons) Though the bond order of all the species are same $(\mathrm{B}.\mathrm{O}=1)$ 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 $${\mathrm{H}}_2>{\mathrm{Li}}_2>{\mathrm{B}}_2$$