Question

Examine the following set of ionization energy values for a certain element. How many valence electrons does an atom of the neutral element possess? Ionization step Ionization energy (kJ/mol) Ei1 1012 Ei2 1903 Ei3 2912 Ei4 4956 Ei5 6273 Ei6 22233 Ei7 25997 Enter your answer numerically as an integer.

Answer 1

Concept and reason

The concept used is to identify the number of valence electrons present in the neutral atom of the certain element.

Fundamentals

Ionization energy is the amount of energy required to remove an electron from the neutral isolated gaseous molecule.

The ionization energy for the removal of valence electrons is relatively lower than the ionization energy for the removal of core electrons.

1.

Given,

$\begin{array}{l}\\{E_{i1}} = 1012{\rm{ kJ/mol}}\\\\{E_{i2}} = 1903{\rm{ kJ/mol}}\\\\{E_{i3}} = 2912{\rm{ kJ/mol}}\\\\{E_{i4}} = 4956{\rm{ kJ/mol}}\\\\{E_{i5}} = 6273{\rm{ kJ/mol}}\\\\{E_{i6}} = 22233{\rm{ kJ/mol}}\\\\{E_{i7}} = 25997{\rm{ kJ/mol}}\\\end{array}$

The difference between the first and second ionization energies is calculated as follows:

$\begin{array}{l}\\\Delta {E_{i1 - 2}} = 1903 - 1012{\rm{ kJ/mol}}\\\\{\rm{ = 891 kJ/mol}}\\\end{array}$

The difference between the second and third ionization energies is calculated as follows:

$\begin{array}{l}\\\Delta {E_{i2 - 3}} = 2912 - 1903{\rm{ kJ/mol}}\\\\{\rm{ = 1009 kJ/mol}}\\\end{array}$

The difference between the third and fourth ionization energies is calculated as follows:

$\begin{array}{l}\\\Delta {E_{i3 - 4}} = 4956 - 2912{\rm{ kJ/mol}}\\\\{\rm{ = 2044 kJ/mol}}\\\end{array}$

The difference between the fourth and fifth ionization energies is calculated as follows:

$\begin{array}{l}\\\Delta {E_{i4 - 5}} = 6273 - 4956{\rm{ kJ/mol}}\\\\{\rm{ = 1317 kJ/mol}}\\\end{array}$

The difference between the fifth and sixth ionization energies is calculated as follows:

$\begin{array}{l}\\\Delta {E_{i5 - 6}} = 22233 - 6273{\rm{ kJ/mol}}\\\\{\rm{ = 15960 kJ/mol}}\\\end{array}$

The difference between the sixth and seventh ionization energies is calculated as follows:

$\begin{array}{l}\\\Delta {E_{i6 - 7}} = 25997 - 22233{\rm{ kJ/mol}}\\\\{\rm{ = 3764 kJ/mol}}\\\end{array}$

The largest difference between two successive ionization energies is between the ${E_{i5}}$ and ${E_{i6}}$ .

Hence, there are five valence electrons.

Ans: Part 1

Therefore, there are five valence electrons in the atom of the given neutral element.