Question

For one element various successive ionization energies are given below:

Ionization energy	1st	2nd	3rd	4th	5th
In Kj/mol	577.5	1810	2750	11580	14820

The element is:
a.) Magnesium
b.) Aluminium
c.) Silicon
d.) Phosphorus

Answer 1

Hint: Ionization energy is the energy required to remove an electron from an atom present in an isolated gaseous state. Ionization energy is also represented as IE.

Complete Solution :

Ionization energy is the energy needed to remove a valence electron off of its molecular orbital of the atom in its neutral state. The ionization energy of an atom decreases as we move down the group and the ionization energy of an atom increases as we move along the period. Atoms which have the lowest ionization energy will easily remove valence electrons and form cations. Valence electrons are the electrons which are present in the valence shell of the electron.

\[X(g)\to {{X}^{+}}(g)+{{e}^{-}}\]

Given in the question:

First ionization energy is = 577.5

Second ionization energy is = 1810

Third ionization energy is = 2750

Fourth ionization energy is = 11580

Fifth ionization energy is = 14820

- We can see that the difference between the third and fourth ionization energy is much higher than the difference between the other ionization energies. This means that at that point the element either attained the half-filled configuration or fully filled electronic configuration.

- Hence, the number of valence electrons in the given element is 3. Among the following options only Aluminum has 3 electrons in the valence shell.

So, the correct answer is “Option B”.

Note: The ionization energy can also be predicted using the atom reactivity. The more reactive the compound, less will be the ionization energy of this compound. But ionization energy is mostly determined experimentally.