Question

The electronic configuration 2, 8, 8, 2 represents the element:
A. Argon
B. Potassium
C. Calcium
D. Chlorine

Answer 1

Hint: Think about the capacity of each orbital to accommodate electrons based on how far it is from the nucleus. Also, take into consideration the sub-orbitals and how they are related according to energy.

Complete answer:
From the configuration, we can see that the electrons are distributed among 4 orbits. So, the element is present in the fourth period of the periodic table.
We know that the first orbital has a capacity of 2 since only one $1s$ orbital is present, and $s$ are the spherical orbitals that can accommodate only 2 electrons. So, this orbital is complete.
The next orbital has a total capacity 8 electrons including the 2 electrons from the $2s$ orbital and 6 electrons from the $2p$ orbital. The $p$ orbital is a dumbbell-shaped orbital has the capacity of 6 electrons. We can see that this orbital is completely filled.
Before moving on to the next orbital, let us check what the energy level of each sub-orbitals is relative to each other. The orbitals having an increasing amount of energy are:
\[1s<2s<2p<3s<3p<4s<3d\]
Here, we can see that the $4s$ orbital has lower energy than the $3d$ orbital and the orbitals with lower energy will be filled first. So, the third orbital will accommodate only 8 electrons instead of the total capacity of 18 and then move on to filling the $4s$ orbital. After filling the $4s$ orbital the $3d$ orbital will be filled.
In this atom, we can see that all orbitals up to the $4s$ orbital are filled, the next election will be placed in the $3d$ orbital. The last electron is placed in the $4s$ orbital, and it contains 2 electrons. So, we can deduce that the element we are asked about is an alkaline earth metal which is present in the fourth period.
The total number of electrons present is 20, we know that the atomic number of the element calcium is 20.

Hence, the correct answer to this question is ‘C. Calcium’

Note: Remember that an easy way to verify whether we have reached the correct conclusion or not by checking where the last electron of the atom is placed. Here, it is placed in an $s$ orbital. So, this element will be an s-block element i.e. one of the alkali metals or the alkaline earth metals.