Question

The first four ionization values of an element are 120, 240, 520, and 6420 \[{\text{k}}{\text{cal}}\]. The number of valence electrons in the element is:

Answer 1

Hint: In simple language, ionization can be defined as the amount of energy which is required to remove an electron from its valence shell. The energy required to remove the first electron from the valence shell is called the first ionization energy, energy required to remove the second electron from the valence shell is called the second ionization energy, energy required to remove the third electron from the valence shell is called the third ionization energy and so on.

Complete step by step solution:
We see that the given the first ionization energy is 120 \[{\text{k}}{\text{cal}}\], second ionization energy is 240 \[{\text{k}}{\text{cal}}\], third ionization energy is 520 \[{\text{k}}{\text{cal}}\], and the fourth ionization energy is 6420 \[{\text{k}}{\text{cal}}\].
We see that the difference between the first, second and third ionization energy is less. But the difference between the third ionization energy and fourth ionization energy is higher. From this we can say that after losing three electrons the element has attained a noble gas configuration, hence the fourth ionization energy value of the element is higher, that is, 6420 \[{\text{k}}{\text{cal}}\]. This value is high because noble gas configuration is a stable configuration as there are no half-filled orbitals in this configuration.
Therefore, we can conclude that the number of valence electrons in the element must be three.

Note: In the periodic table, the elements placed to the left have a much lower ionization energy. The elements placed to the right have a higher value of ionization energy.