The successive ionization energies of a certain element are I1= 577.9 kJ/mol, I2 = 1820 kJ/mol, I3= 2750 kJ/mol, I4 = 11,600 kJ/mol, and I5 = 14,800 kJ/mol. This pattern of ionization energies suggests that the unknown element is .

Given question is incomplete. The complete question is as follows.

The successive ionization energies of a certain third-period element are I1 = 577.9KJ/mol, I2 + 1820 KJ/mol, I3 = 2750 KJ/mol, I4 = 11600 KJ/mol, I5 = 14800 KJ/mol. what element do these ionization energies suggest? Explain your reasoning.

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

Here, given ionization energies belong to an element present in third period.

We know that second ionization energy will always be greater than third ionization energy.

For the given ionization energies, there is a huge  difference between third and fourth ionization energy. This means that there are three valence electrons present in the element.

This is because after losing three electrons it is difficult for the given element to lose fourth electron. Hence, is high as compared to .

Hence, this element has 3 valence electrons and it belongs to 3A group of the periodic table.

Thus, we can conclude that the given unknown element is aluminium (Al).

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