Question

Using the molecular orbital model to describe the bonding in${\mathit{F}}_{\mathbf{2}}^{\mathbf{+}}\mathbf{,}{\mathit{F}}_{\mathbf{2}}\mathbf{,}{\mathit{F}}_{\mathbf{2}}^{\mathbf{-}}$ predict the bond order and the relative bond length for these species. How many unpaired electrons are present in each species?

Short answer

The filling order for F is${\sigma }_{1s}{{\sigma }^{*}}_{1s}{\sigma }_{2s}{{\sigma }^{*}}_{2s}{\sigma }_{2pz}{\pi }_{2px}{\pi }_{2py}{{\pi }^{*}}_{2px}{{\pi }^{*}}_{2px}{{\sigma }_{2pz}}^{*}$

The bond order is$\frac{1}{2}(n_b-n_a)$

Where

$$\begin{gathered} n_b=bondingelectrons \\ {n}_a=noofantibondingelectron \end{gathered}$$

Step-by-step solution

Step 1

Filling the molecular orbitals

For$F_2^{+}\left(17e\right)$ is as follow${{\sigma }^2}_{1s}({{\sigma }^{*}}_{1s}{)}^2{{\sigma }^2}_{2s}({{\sigma }^{*}}_{2s}{)}^2{{\sigma }^2}_{2pz}{{\pi }^2}_{2px}{{\pi }^2}_{2py}({{\pi }^{*}}_{2px}{)}^2({{\pi }^{*}}_{2px}{)}^1$

Bond order is

$$\begin{gathered} B.O=\frac{1}{2}(10-7) \\ B.O=1.5 \end{gathered}$$

For$F_2$ there will be one extra electron so orbital filling is as follow

${{\sigma }^2}_{1s}({{\sigma }^{*}}_{1s}{)}^2{{\sigma }^2}_{2s}({{\sigma }^{*}}_{2s}{)}^2{{\sigma }^2}_{2pz}{{\pi }^2}_{2px}{{\pi }^2}_{2py}({{\pi }^{*}}_{2px}{)}^2({{\pi }^{*}}_{2px}{)}^2$

Bond order is

$$\begin{gathered} B.O=\frac{1}{2}(10-8) \\ B.O=1 \end{gathered}$$

For${F_2}^{-}$

The bond filling is as follow${{\sigma }^2}_{1s}({{\sigma }^{*}}_{1s}{)}^2{{\sigma }^2}_{2s}({{\sigma }^{*}}_{2s}{)}^2{{\sigma }^2}_{2pz}{{\pi }^2}_{2px}{{\pi }^2}_{2py}({{\pi }^{*}}_{2px}{)}^2({{\pi }^{*}}_{2px}{)}^2{\sigma }_{2pz}^{*}$

Bond order is

$$\begin{gathered} B.O=\frac{1}{2}(10-9) \\ B.O=0.5 \end{gathered}$$

Step 2

The bond order is inversely proportional to bond length.

Since bond order follows this order:${F^{+}}_2⟩F_2⟩F_2^{-}$

Hence bond length order will be${F^{+}}_2⟨F_2⟨F_2^{-}$

Step 3

The number of unpaired electrons is

$$\begin{gathered} {F^{+}}_2=1 \\ {F}_2=0 \\ {F}_2^{-}=1 \end{gathered}$$