Question

How many valence electrons would be found in an atom of Ba?

Answer 1

Hint The number of electrons present in Ba is 56. The valence electrons are those electrons which enter the outermost shell of the atom.

Complete step by step solution:
So in the question, it is asked how many valence electrons will be present in a Ba atom. From the time we study about the periodic table and the elements in them, we are more concerned about the position of atoms which gives the number of shells present in them and about the valence shell and the valence electrons which accounts for various chemical and physical properties of the element.
So before going into the solution part, let’s briefly discuss the valence shell and valence electrons in the atoms.
The valence shells are those shells which are present in the outermost shell of the atom in which the valence electrons get filled. The valence electrons are responsible for many chemical properties as these are the electrons which take part in the chemical combinations. During the chemical combinations, the valence electrons are either shared or lost and in some cases the atoms gain electrons to complete its octet configuration and these electrons are accommodated in the valence shells.
Now let's discuss the approach to find the valence electrons in an element.
One method is, if we know the atomic number of the element, we will write the electronic configuration of the element and add up the number of electrons present in various orbitals of the outermost shell which gives the number of valence electrons of the element.
Let’s write the electronic configuration of Ba.
In the hint part it is given that there are 56 electrons in Ba, so the atomic number is 56 and let’s distribute the electrons in various energy level orbitals,
$\text{Electronic}\text{Configuration}\text{of}\text{Ba=1}{\text{s}}^{\text{2}}\text{2}{\text{s}}^{\text{2}}\text{2}{\text{p}}^{\text{6}}\text{3}{\text{s}}^{\text{2}}\text{3}{\text{p}}^{\text{6}}\text{3}{\text{d}}^{\text{10}}\text{4}{\text{s}}^{\text{2}}\text{4}{\text{p}}^{\text{6}}\text{4}{\text{d}}^{\text{10}}\text{5}{\text{s}}^{\text{2}}\text{5}{\text{p}}^{\text{6}}\text{6}{\text{s}}^{\text{2}}$
$\text{Electronic}\text{Configuration}\text{of}\text{Ba=}\left[Xe\right]\text{6}{\text{s}}^{\text{2}}$
Hence from the electronic configuration we know the outermost or valence shell is the sixth shell and there are two electrons in them.

Therefore the total number of valence electrons in Ba is 2 and it has a tendency to lose two electrons and attain the stable noble gas configuration of Xe.

Note: Another method to find the valence number is, through the group number of the element. If we know to which group the elements belong to then we could predict the valence electrons since the main group numbering is according to the number of valence electrons. The Ba belongs to group 2 elements i.e. the alkaline earth metals, since the Ba is a group 2 element, the number of valence electrons will be 2.