Question

What charge would be needed on ${\mathrm{F}}_2$ to generate an ion with a bond order of 2?

Answer 1

Step 1: Bond Order:

• "The bond order (BO) of a bond in a covalent molecule is defined as one-half the difference of the number of electrons in the bonding and anti-bonding orbitals.
• It can be expressed as:

$\mathrm{BO}=\frac{\left(\mathrm{Number}\mathrm{of}\mathrm{bonding}{\mathrm{e}}^{-}\right)-\left(\mathrm{Number}\mathrm{of}\mathrm{antibonding}{\mathrm{e}}^{-}\right)}{2}$

Step 2: Calculation for bond order of ${\mathrm{F}}_2$ molecule:

• The total number of valence electrons in ${\mathrm{F}}_2$ are 18.
• The electronic configuration of ${\mathrm{F}}_2$ molecule is $\mathrm{K}2{\mathrm{\sigma s}}^2,2\mathrm{\sigma }{*}^2,\mathrm{\sigma }2{\mathrm{pz}}^2,\mathrm{\pi }2{\mathrm{px}}^2,\mathrm{\pi }2{\mathrm{py}}^2,\mathrm{\pi px}{*}^2,\mathrm{\pi }2\mathrm{py}{*}^2$
• Thus, the bond order will be:

$\mathrm{BO}=\frac{8-6}{2}=\frac{2}{2}=1$

There are $8$ bonding electrons and $6$ anti-bonding electrons, resulting in a BO of $1.0$.

Step 3: Calculation charge on Fluorine with bond order 2:

• Now, since the BO has to be increased to $2.0$ , hence electron(s) must be removed from the antibonding orbital(s).
• The below calculation shows that BO equals $2.0$ is possible when there are $4$ antibonding electrons. Thus, one electron is removed from each of the HOMO orbitals, giving a BO of $2$.

$$\begin{gathered} 2=\frac{\left(\mathrm{Number}\mathrm{of}\mathrm{bonding}{\mathrm{e}}^{-}\right)-\left(\mathrm{Number}\mathrm{of}\mathrm{antibonding}{\mathrm{e}}^{-}\right)}{2} \\ \Rightarrow 2=\frac{\left(8\right)-\left(\mathrm{Number}\mathrm{of}\mathrm{antibonding}{\mathrm{e}}^{-}\right)}{2} \\ \Rightarrow 4=\left(8\right)-\left(\mathrm{Number}\mathrm{of}\mathrm{antibonding}{\mathrm{e}}^{-}\right) \\ \Rightarrow \mathrm{Number}\mathrm{of}\mathrm{antibonding}{\mathrm{e}}^{-}=4 \end{gathered}$$

• Now, since there are two electrons short of the total nuclear charge, hence, the overall charge carried by the ion will be $+2$ and each $\mathrm{F}$ atom will carry charge $+1$.