Question

A new element is discovered which has the following ionization energies: IE1 127 kJ/mol IE2 287 kJ/mol IE3 625 kJ/mol IE4 2651 kJ/mol IE5 3540 kJ/mol Which group do you suppose this new element belongs in

Answer 1

Answer:

This hypothetical element is likely to be found in IUPAC group 3 (or IIIA in the previous version of the IUPAC system.)

Explanation:

Note the trend in the consecutive ionization energies of this element. The first three ionization energies are relatively small (under $\rm 1000\; kJ \cdot mol^{-1}$) while ionization energies after the third are much larger.

The main energy levels are likely responsible for this trend. The first three electrons are in the valence shell. They have the greatest distance from the nucleus, which means that the nucleus has the weakest hold on them. Besides, there's the shielding effect: inner-shell electrons push these valence electrons away from the nucleus. That further reduces the amount of energy it takes to remove the first three electrons and "ionize" the atom.

In comparison, inner shell electrons are much harder to remove. That explains the jump in the ionization energy trend.

In a modern IUPAC periodic table, the group number of an element is the same as the number of valence electrons in its atom. Based on the trend in ionization energy, there are three valence electrons in each atom of this hypothetical element. As a result, the IUPAC group number of this hypothetical element would be 3. That's the same as IIIA in the previous version of the IUPAC system.