Question

The successive ionization energy values of an element are $7.1, \, 14.3, \, 34.5, \, 46.8$ and $162.2 \, \text{e}\text{V}$ respectively. The element is likely to be:

• A. $Na$
• B. $Si$
• C. $F$
• D. $Ca$

Answer 1

Answer: (B)

The $IE_{1},IE_{2},IE_{3},IE_{4}$ and $IE_{5}$ of an element are $7.1,14.3,34.5,46.8$ and $162.2eV$ respectively. Therefore, We can see that there is a regular very close increment of ionization energy till the fourth ionization energy. After the fourth ionization energy, there is a sudden high jump in energy. It confirms that after losing four electrons, that element will achieve the noble gas configuration or stable configuration.

$_{11}^{}Na \rightarrow 1s^{2}2s^{2}2p^{6}3s^{1}$

$_{14}^{}Si \rightarrow 1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}$

$_{9}^{}F \rightarrow 1s^{2}2s^{2}2p^{5}$

We can see the configuration of the elements and can identify that silicon has this tendency to lose the four electrons and gets the noble gas configuration.

the valency of the element is $4$ and it is most likely to be silicon.