Question

5. The first ionization energies of some relevant species are given below. IE1(N) = 14.5 eV

   IE1(N2)= 15.6 eV IE1(F) = 17.4 eV IE1(F2) = 15.7 eV

   1. Explain the observation that IE1(N2) > IE1(N). Draw the relevant MO diagram.

   2. Explain the observation that IE1(F2) < IE1(F). Draw the relevant MO diagram.

   3. Do you expect the IE1 of NF to be greater than or less than the IE1 of N? Explain

      your answer, and draw the relevant MO diagram.

Answer 1

The ionization energy: It is the energy required to remove an electron from the outermost orbit of a neutral gaseous atom or ion.

The ionization energy of N₂ is greater than N because it requires more energy to remove an electron from the nitrogen molecule than from the nitrogen atom; the electron, therefore, has lower energy in the molecule. To separate the atoms, the energy of the electron must be more. Therefore, energy is required to break the bond, and the molecule is bound. This is the basic principle of bond formation.

Whereas The ionization energy of F₂ is smaller than F this is like a contradiction but the electron with the highest energy requires the least energy to remove from the molecule or atom. The molecular orbital energy clearly shows that the highest energy electrons in F₂ are in anti-bonding orbitals. Therefore, one of these electrons is easier to remove than an electron in an atomic 2p orbital, because the energy of an antibonding orbital is higher than that of the atomic orbitals, Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine.

It is theoretically expected that IE1 of NF to be greater than or less than the IE1 of N. Here the reason will be the same as that of the second one.

A standard Molecular Orbital Diagram of N₂

Relatively Simpler Representations.

Relevant MO diagram for NF