Chapter 6, Problem 89A

Interpretation Introduction

Interpretation:

The given element from the given table needs to be identified.

Concept introduction:

Amount of energy needed to remove most loosely bound electron from isolated gaseous atom of elements in its lowest energy state to produce a cation is known as ionization energy.

Answer to Problem 89A

The element is sulfur.

Explanation of Solution

Period 3 elements are s-block elements and p-block elements. The elements can potentially lose six valence electrons in the given table of ionization energy. Element is unlikely to lose six valence electrons thus there is a large difference between fifth and sixth valence electron. This is because 6th electron is to be removed from 3s which become half filled after removal of 5 electrons and close to the nucleus. Sulfur has the potential to lose six valence electrons and it is period 3 elements. The electronic configuration of sulfur is $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}$. After lose six electrons in the outermost shell sulfur gets noble gas configuration,  [Ne]. It is difficult to remove one electron from the element with full valence shell.

Conclusion

The electronic configuration of sulfur is $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}$. After lose six electrons in the outermost shell sulfur gets noble gas configuration. Thus the element is sulfur.