Examine the following set of ionization energy values for a certain element. How many valence electrons…

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,

Ei1=1012kJ/molEi2=1903kJ/molEi3=2912kJ/molEi4=4956kJ/molEi5=6273kJ/molEi6=22233kJ/molEi7=25997kJ/mol\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:

ΔEi12=19031012kJ/mol=891kJ/mol\begin{array}{l}\\\Delta {E_{i1 – 2}} = updating{\rm{ kJ/mol}}\\\\{\rm{ = 891 kJ/mol}}\\\end{array}

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

ΔEi23=29121903kJ/mol=1009kJ/mol\begin{array}{l}\\\Delta {E_{i2 – 3}} = updating{\rm{ kJ/mol}}\\\\{\rm{ = 1009 kJ/mol}}\\\end{array}

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

ΔEi34=49562912kJ/mol=2044kJ/mol\begin{array}{l}\\\Delta {E_{i3 – 4}} = updating{\rm{ kJ/mol}}\\\\{\rm{ = 2044 kJ/mol}}\\\end{array}

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

ΔEi45=62734956kJ/mol=1317kJ/mol\begin{array}{l}\\\Delta {E_{i4 – 5}} = updating{\rm{ kJ/mol}}\\\\{\rm{ = 1317 kJ/mol}}\\\end{array}

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

ΔEi56=222336273kJ/mol=15960kJ/mol\begin{array}{l}\\\Delta {E_{i5 – 6}} = updating{\rm{ kJ/mol}}\\\\{\rm{ = 15960 kJ/mol}}\\\end{array}

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

ΔEi67=2599722233kJ/mol=3764kJ/mol\begin{array}{l}\\\Delta {E_{i6 – 7}} = updating{\rm{ kJ/mol}}\\\\{\rm{ = 3764 kJ/mol}}\\\end{array}

The largest difference between two successive ionization energies is between the

Ei5{E_{i5}}

and

Ei6{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.

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