Question

The pair of species with the same bond order is:

• A. $O_2^{+},N{O}^{+}$
• B. NO, CO
• C. $N_2,O_2$
• D. $O_2^{2-},B_2$

Answer 1

Answer: (D)

$O_2^{2-}:\left(8+8+2=18\right)$
$O_2^{2-}:{\left({\sigma }_{1s}\right)}^2{\left({\sigma }_{1s}^{\cdot }\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{\cdot }\right)}^2{\left({\sigma }_{2pz}\right)}^2{\left({\pi }_{{}_{2}px}\right)}^2={\left({\pi }_{2py}\right)}^2{\left({\pi }_{2px}^{\cdot }\right)}^2={\left({\pi }_{2px}^{\cdot }\right)}^2$
Bond order = $\frac{N_{b-N_a}}{2}=\frac{10-8}{2}=1$
$B_2:(5+5=10)$
$B_2:{\left({\sigma }_{1s}\right)}^2{\left({\sigma }_{1s}^{\cdot }\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{\cdot }\right)}^2{\left({\pi }_{2px}\right)}^1={\left({\pi }_{2py}\right)}^1$
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.