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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
${H_2} > {B_2} > L{i_2}$
$L{i_2} > {H_2} > {B_2}$
$L{i_2} > {B_2} > {H_2}$
None of these
Solution
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
$H_2 > Li_2 >B_2$
Similar Questions
Match List$-I$ with List$-II.$
| List$-I$ | List$-II$ |
| $(a)$ $Ne _{2}$ | $(i)$ $1$ |
| $(b)$ $N _{2}$ | $(ii)$ $2$ |
| $(c)$ $F _{2}$ | $(iii)$ $0$ |
| $(d)$ $O _{2}$ | $(iv)$ $3$ |
Choose the correct answer from the options given below:
