Question

How many electrons will a calcium atom (Ca) lose in order to get a noble gas configuration?

Answer 1

In order to find the number of electrons lost by calcium to form a noble gas configuration, we should know the electronic configuration of the calcium atom and the electronic configuration of the nearest noble gas. By losing the electrons, atoms form cation.

Complete step by step answer: The calcium is the chemical element present in group 2 and period 4 of the periodic table. The calcium atom is an s-block element and an alkaline earth metal.
The elements that are present in the last column of the periodic table are the group known as noble gases. The noble gas configuration of an atom contains an elemental symbol of the noble gas written prior to the atom which followed by the electronic configuration of the atom.
The noble gases are the stable atoms which follow octet rule. The octet rule is the tendency of an atom to have eight electrons in the valence electronic configuration.
The atomic number of calcium is 20 which means that the calcium atom contains total 20 number of electrons.
The electronic configuration of calcium is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}$.
The neutral calcium atom loses two electrons to form calcium cation $C{a^{2 + }}$. The electronic configuration of calcium ions is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$.
The atomic number of neutral argon atoms is 18. The total electrons present in argon is 18. The electronic configuration of argon is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$ which is same as the electronic configuration of calcium cation $C{a^{2 + }}$.
So, the calcium atom will lose its two electrons in order to get a noble gas configuration.

Note:
By gaining electrons, the atom forms anion. As the valence electronic configuration of calcium contains two electrons in s orbital, it will be difficult for calcium to gain 6 electrons to form stable configuration than losing its two electrons. So the metal calcium usually forms cation not an anion.