Arrange the members of the following set of elements in order of increasing first ionization energies?

Question

(A) O(B) B(C) C(D) Li(E) Ne(F) N

Answer 1

Li < B < C < O < N < Ne

Notice that this trend follows the trend from left to right across the periodic table with one very important exception. The first ionization energy of O is less than N.

For O, the first ionization energy = 13.62 eV
For N, the first ionization energy = 14.53 eV
(eV is electronvolts)

The reason for this exception to the general trend is the stability of electron configuration in the valance orbitals of nitrogen. The valence orbitals for both N and O are 2px, 2py, and 2pz. Nitrogen has 3 valence electrons, one in each of the valence orbitals. This is a very stable configuration which makes the removal of one of these electrons (1st ionization) somewhat difficult.

However, oxygen has 4 valence electrons. Two of the valence orbitals has one electron and one of the valence orbitals has 2 electrons. Removal of one of these electrons (1st ionization) from the doubly occupied valence orbital results in a configuration having a single electron in each valence orbital (a very stable configuration).

Nitrogen has a very stable valence configuration which makes the first ionization energy greater than for oxygen which has one electron beyond the stable half-filled configuration.

Answer 2

a. Alkaline Earth Metals: Ra, Ba, Sr, Ca, Mg, Be b. Cs, H, Br, B, F Ionization skill is the minimum skill required to get rid of an electron at floor state. Ionization skill will advance Up and to the right in the time of the Periodic table. So Mg has a larger ionization skill then ok, and C has a larger ionization skill then Zn for instance.

Answer 3

Li < B < C < N < O < Ne
Reason? All of them in the second period (in the periodic table). So the effective attraction force between the nuclear and the electron is increasing.