Question

what is bond order of F2 & H2+

Answer 1

Dear Student,


$$\begin{gathered} \mathrm{Bond}\mathrm{order}=\frac{\mathrm{Bonding}\mathrm{electrons}-\mathrm{Antibonding}\mathrm{electrons}}{2} \\ \mathrm{MO}\mathrm{configuration}\mathrm{of}{\mathrm{F}}_2=\left(\mathrm{\sigma }1\mathrm{s}{)}^2\right(\mathrm{\sigma }*1\mathrm{s}{)}^2\left(\mathrm{\sigma }2\mathrm{s}{)}^2\right(\mathrm{\sigma }*2\mathrm{s}{)}^2\left(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{z}}{)}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}{)}^2\right(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}{)}^2(\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}{)}^2(\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}}{)}^2 \\ \mathrm{Bond}\mathrm{order}=\frac{10-8}{2}=\frac{2}{2}=1 \\ \mathrm{MO}\mathrm{configuration}\mathrm{of}{{\mathrm{H}}_2}^{+}=(\mathrm{\sigma }1\mathrm{s}{)}^1 \\ \mathrm{Bond}\mathrm{order}=\frac{1-0}{2}=\frac{1}{2} \end{gathered}$$