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Question

None of the given option is correct.

The molecular orbital configuration of the given molecules is 

$H_2 = \sigma 1s^2$ (no electron anti-b

Vedclass · Accepted Answer

None of these

Vedclass · Answer

None of the given option is correct.

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 

$H_2 > Li_2 >B_2$

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

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