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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 :
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The correct answer is:
None of these
None of these
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
$$
\mathrm{H}_2>\mathrm{Li}_2>\mathrm{B}_2
$$
$\mathrm{H}_2=\sigma 1 s^2$ (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
$$
\mathrm{H}_2>\mathrm{Li}_2>\mathrm{B}_2
$$
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