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The first four ionization energies of the element are given to be 738, 1451,7733 and 10541, (all values in KJ/mol).
It can be seen that the first and second ionization energies are quite low 738 KJ/mol and 1451 KJ/mol respectively while the third and fourth ionization energies are very high, 7733 KJ/mol and 10541 KJ/mol respectively.
The ionization energies give the group in the periodic table that the element belongs to as 2. In this group, elements have an atomic configuration that consists of a noble gas followed by 2 other electrons. After the element has lost the first and second electrons, the remaining electrons have the configuration of a noble gas which is a very stable configuration. Making a change in this requires a very large amount of energy.
The easiest way to find the group of an element given its ionization energies, is to find the first largest "jump" between two ionization energies.
The first four ionization energies are 738, 1451,7733 and 10541 (in KJ/mol) of an element. Find the difference between two consecutive numbers. The first largest "jump" or difference occurs between 1451 and 7733. Therefore the element belongs to group 2. Remember it should be the first largest "jump".
First, it is important to examine how the individual ionization energies. The first two ionization energies, 738 and 1451 kJ respectively, are relatively low numbers for ionization energies, especially as compared to 7733 and 10541 kJ. This means that the element can easily lose its first two electrons but requires a great amount of energy to lose the third electron.
Keep in mind the octet rule of elements that states that elements are most stable when they have 8 valence electrons, or 8 electrons in their outermost energy level. The ionization energies tell us that the element most likely has a filled octet after losing two electrons, pointing us to group 2 on the periodic table.
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