Final answer:
The N, O, and F atoms have higher first ionization energies compared to their molecular forms N₂, O₂, and F₂ respectively, because the electrons are more tightly bound in the atomic orbitals. The same is true for the Li atom when compared to the Li₂ molecule.
Step-by-step explanation:
Identifying the member with the highest first ionization energy in each pair in the gas phase involves comparing the atomic and molecular orbital diagrams for the nitrogen atom (N) and the nitrogen molecule (N₂), the oxygen atom (O) and the oxygen molecule (O₂), the fluorine atom (F) and the fluorine molecule (F₂), and the lithium atom (Li) and the lithium molecule (Li₂).
The first ionization energy refers to the energy required to remove the most loosely bound electron from an atom or molecule in the gas phase.
For nitrogen, N atom has a higher first ionization energy than the N₂ molecule because the atomic electron is more tightly bound in the atom than in the molecular orbital of N₂.
Considering oxygen, the O atom also has a higher first ionization energy compared to the O₂ molecule. This is because in O₂, unpaired electrons in the degenerate π2p orbitals are less tightly bound as compared to the paired electrons in atomic oxygen.
For fluorine, the F atom has a higher first ionization energy than the F₂ molecule since electrons are removed from a non-bonding antibonding orbital in F₂ which is easier than removing one from the isolated F atom.
Lastly, the Li atom will have a higher first ionization energy than the Li₂ molecule because antibonding orbitals in Li₂ make it easier to remove an electron compared to the atomic case.
To conclude, in each of the cases above, the atoms have higher first ionization energies compared to their molecular counterparts.