1. Tardigrade
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  3. Chemistry
  4. The pair of species with the same bond order is:

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Q. The pair of species with the same bond order is:

Solution:

$O^{2-}_{2}: \left(8+8+2=18\right)$
$ O^{2-}_{2 } : \left(\sigma_{1s}\right)^{2}\left(\sigma^{\cdot}_{1s}\right)^{2}\left(\sigma_{2s}\right)^{2}\left(\sigma^{\cdot}_{2s}\right)^{2}\left(\sigma_{2pz}\right)^{2}\left(\pi_{_2px}\right)^{2 } = \left(\pi_{2py}\right)^{2 }\left(\pi^{\cdot}_{2px}\right)^{2} = \left(\pi^{\cdot}_{2px}\right)^{2} $
Bond order = $ \frac{N_{b - N_a}}{2} = \frac{10-8}{2}=1 $
$B_2 : (5 + 5 = 10)$
$B_{2} : \left(\sigma_{1s}\right)^{2}\left(\sigma_{1s}^{\cdot}\right)^{2}\left(\sigma_{2s}\right)^{2}\left(\sigma^{\cdot}_{2s}\right)^{2}\left(\pi_{2px}\right)^{1} = \left(\pi_{2py}\right)^{1} $
Bond order = $\frac{N_{b}-N_{a}}{2}=\frac{6-4}{2}=1 $
$\therefore \, O^{2-}_{2}$ and $B_2$ have the same bond order.