Bond order normally gives idea of stability of a molecular species. All the molecules viz. $\mathrm{H}_2,…

Bond order normally gives idea of stability of a molecular species. All the molecules viz. $\mathrm{H}_2, \mathrm{Li}_2$ and $\mathrm{B}_2$ have the same bond order yet they are not equally stable. Their stability order is :
  1. $\mathrm{H}_2>\mathrm{B}_2>\mathrm{Li}_2$
  2. $\mathrm{Li}_2>\mathrm{H}_2>\mathrm{B}_2$
  3. $\mathrm{Li}_2>\mathrm{B}_2>\mathrm{H}_2$
  4. None of these

Solution

The molecular orbital configuration of the given molecules is $\mathrm{H}_2=\sigma 1 s^2$ (no electron anti-bonding) $\mathrm{Li}_2=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2$ (two anti-bonding electrons) $ \mathrm{B}_2=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 $ $ \left\{\pi 2 p_y^1=\pi 2 p_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 $

Asked in: JEE Main 2013 (22 Apr Online)

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