Q45E - Chemistry (OpenStax)

Question

A friend tells you that the \({\rm{2s}}\)orbital for fluorine starts off at a much lower energy than the \({\rm{2s}}\) orbital for lithium, so the resulting \({{\rm{\sigma }}_{{\rm{2s}}}}\) molecular orbital in \({{\rm{F}}_{\rm{2}}}\) is more stable than in \({\rm{L}}{{\rm{i}}_{\rm{2}}}\). Do you agree?

Step-by-Step Solution

Answer

Yes, it is agreed.

Step 1: Definition of diatomic molecule

A diatomic molecule is a molecule that consists of only two atoms of the same or different chemical elements

Step 2: Evaluating \({{\bf{F}}_{\bf{2}}}\) is more stable than \({\bf{L}}{{\bf{i}}_{\bf{2}}}\)

 

Yes, i agree. The \(2s\) Fluorine orbits start at  much lower energies \(2s\) this is because   fluorine is smaller than lithium and is orbital lithium \(2s\) The orbit of fluorine is closer to the nucleus  than  lithium and is more stable. Therefore, the result is \({\sigma _{2s}}\)molecular orbital in \({F_2}\)is more stable than in \({\rm{L}}{{\rm{i}}_{\rm{2}}}\).