Question

How do I calculate the bond order for ${{\mathrm{H}}_2}^{+}$ and ${{\mathrm{H}}_2}^{-}$?

Answer 1

Bond order:

• Bond order is defined as the half of the difference between bonding electrons and anti-bonding electrons.
• It is calculated by the following formula:

$\mathrm{Bond}\mathrm{order}=\frac{1}{2}[\mathrm{Number}\mathrm{of}\mathrm{Bonding}\mathrm{electrons}({\mathrm{N}}_{\mathrm{b}})-\mathrm{Number}\mathrm{of}\mathrm{Antibonding}\mathrm{electrons}({\mathrm{N}}_{\mathrm{a}}\left)\right]$

Calculation of Bond order of ${{\mathrm{H}}_2}^{+}$ and ${{\mathrm{H}}_2}^{-}$:

• The number of electrons present in Hydrogen $({\mathrm{H}}_2)$ atom are two which involves in bonding and there are no anti-bonding electrons.
• The number of electrons present in ${{\mathrm{H}}_2}^{+}$ is only one.
• Symbol (+) indicates loss of one electron.
• There is only bonding electrons present in ${{\mathrm{H}}_2}^{+}$.
• Bond order of ${{\mathrm{H}}_2}^{+}$ is

$\mathrm{Bond}\mathrm{order}=\frac{1}{2}[\mathrm{Number}\mathrm{of}\mathrm{Bonding}\mathrm{electrons}({\mathrm{N}}_{\mathrm{b}})-\mathrm{Number}\mathrm{of}\mathrm{Antibonding}\mathrm{electrons}({\mathrm{N}}_{\mathrm{a}}\left)\right]$

$$\begin{gathered} \mathrm{Bond}\mathrm{order}\mathrm{of}{{\mathrm{H}}_2}^{+}=\frac{1}{2}[1-0] \\ \Rightarrow \mathrm{Bond}\mathrm{order}\mathrm{of}{{\mathrm{H}}_2}^{+}=\frac{1}{2}\left(1\right) \\ \Rightarrow \mathrm{Bond}\mathrm{order}\mathrm{of}{{\mathrm{H}}_2}^{+}=\frac{1}{2} \\ \Rightarrow \mathrm{Bond}\mathrm{order}\mathrm{of}{{\mathrm{H}}_2}^{+}=0.5 \end{gathered}$$

• The number of electrons present in ${{\mathrm{H}}_2}^{-}$ are 3.
• Symbol (-) indicates gain of electrons.
• Here in ${{\mathrm{H}}_2}^{-}$, there are 2 bonding electrons and one anti-bonding electron.
• Bond order of ${{\mathrm{H}}_2}^{-}$ is

$\mathrm{Bond}\mathrm{order}=\frac{1}{2}[\mathrm{Number}\mathrm{of}\mathrm{Bonding}\mathrm{electrons}({\mathrm{N}}_{\mathrm{b}})-\mathrm{Number}\mathrm{of}\mathrm{Antibonding}\mathrm{electrons}({\mathrm{N}}_{\mathrm{a}}\left)\right]$

$$\begin{gathered} \mathrm{Bond}\mathrm{order}\mathrm{of}{{\mathrm{H}}_2}^{-}=\frac{1}{2}[3-1] \\ \Rightarrow \mathrm{Bond}\mathrm{order}\mathrm{of}{{\mathrm{H}}_2}^{-}=\frac{1}{2}\left(2\right) \\ \Rightarrow \mathrm{Bond}\mathrm{order}\mathrm{of}{{\mathrm{H}}_2}^{-}=1 \end{gathered}$$

Therefore the bond order of ${{\mathrm{H}}_2}^{+}$ is 0.5 and that of ${{\mathrm{H}}_2}^{-}$ is 1.