Question

What charge would be needed on \({{\rm{F}}_{\rm{2}}}\) to generate an ion with a bond order of \({\rm{2}}\)?

Short answer

The bond order would be \({\rm{2}}\) when \({{\rm{F}}_{\rm{2}}}\) has a \({\rm{2 + }}\) charge on it.

Step-by-step solution

Ion

An ion is an electric charge-carrying atom or molecule. A superscript to represent the sign and size of an ion's electric charge is used to identify it.

Bond order

The bond order of a diatomic molecule can be determined by using the formula:

\(bond\,order = \frac{1}{2}\left[ {{n_{bonding\,electrons}} - {n_{antibonding\,electrons}}} \right]\)

Explanation

Configuration of electrons in\({{\rm{F}}_{\rm{2}}}\),

\({{\rm{(}}{{\rm{\sigma }}_{{\rm{2s}}}}{\rm{)}}^{\rm{2}}}{{\rm{(\sigma }}_{{\rm{2s}}}^{\rm{*}}{\rm{)}}^{\rm{2}}}{{\rm{(}}{{\rm{\sigma }}_{{\rm{2px}}}}{\rm{)}}^{\rm{2}}}{\left( {{{\rm{\pi }}_{{\rm{2py}}}}{{\rm{\pi }}_{{\rm{2pz}}}}} \right)^{\rm{4}}}{\left( {{\rm{\pi }}_{{\rm{2py}}}^{\rm{*}}{\rm{\pi }}_{{\rm{2pz}}}^{\rm{*}}} \right)^{\rm{4}}}\)

Order of the bonds:\({\rm{1}}\)

Configuration of electrons,\({\rm{F}}_{\rm{2}}^{{\rm{2 + }}}\)

\({{\rm{(}}{{\rm{\sigma }}_{{\rm{2s}}}}{\rm{)}}^{\rm{2}}}{{\rm{(\sigma }}_{{\rm{2s}}}^{\rm{*}}{\rm{)}}^{\rm{2}}}{{\rm{(}}{{\rm{\sigma }}_{{\rm{2px}}}}{\rm{)}}^{\rm{2}}}{\left( {{{\rm{\pi }}_{{\rm{2py}}}}{{\rm{\pi }}_{{\rm{2pz}}}}} \right)^{\rm{4}}}{\left( {{\rm{\pi }}_{{\rm{2py}}}^{\rm{*}}{\rm{\pi }}_{{\rm{2pz}}}^{\rm{*}}} \right)^{\rm{2}}}\)

Order of the bonds:\({\rm{2}}\)

Therefore, for the bond order to be \({\rm{2}}\), \({{\rm{F}}_{\rm{2}}}\) should have a \({\rm{2 + }}\) charge on it.