Question

The first, second, third, fourth and fifth ionisation potential values of an element are 6.11, 11.87, 51.21, 67.0, 84.39 \[eV\] respectively. The element is:
A.Calcium
B.Potassium
C.Aluminium
D.Carbon

Answer 1

Hint: to answer this question you need to know the number of ions present in the element and the shell structure of the element. Isomerism is the energy per mole necessary to remove electrons from atoms.

Step by step answer: Let’s first discuss what ionization energy is. Molar ionization energies is the energy per mole necessary to remove electrons from gaseous atoms or atomic ions and it is measured in \[kJmo{l^{ - 1}}\] .
The first ionization energy applies to the neutral atoms. The second, third etc. Molar energies apply to the further removal of electrons from a singly, doubly etc. charged ion. The ionization energies are measured in unit eV. The closer the outermost electrons to the nucleus of the atom, the higher is the atom’s ionization energy.
Now, from the values we can see that the first two values are low and the third value increases at a very high rate. The last two values have increased but at a normal rate. The third ionization potential of the given element is very large as compared to first and second ionization potential. Therefore, whatever the element is, it should have two electrons in the outermost shell. Thus, the given element is an alkaline earth metal. The answer to our question is Calcium i.e. option A. We will get to the answer by seeing the ions present in the molecules.

This is a diagram of calcium with ions present in each shell. We can see that in the last shell there are 2 electrons. These two electrons get easily removed as there are only two and they are far from the centre nucleus. But when we try to remove the third ion the potential value increases at a higher rate. This is because there are more ions in the third shell and they are closer to the nucleus thus making it harder to remove.

Note: You need to keep in mind the number of ions present in an element. When you know the structure of the element you will be able to solve the question easily. Also, the structure of the element can be memorized for saving time.