What happens to the number of valence electrons in atoms of elements as we go down a group in the periodic table?
Answer
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Hint: Solve this question by taking an example. Take a random element from the periodic table and the subsequent elements in the same group. Now determine if there is any change in the number of valence electrons. Based on this answer the question mentioned above.
Complete answer:
As suggested in the hint above, we will take an element from the periodic table, say sodium (Na) and the elements of the respective group i.e potassium (K), Rubidium (Rb), Caesium (Cs).
We will now write down the electronic configuration of the elements of the first group of the periodic table.
Sodium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$
Potassium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{1}}$
Rubidium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{1}}$
Caesium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{2}}4{{d}^{10}}5{{p}^{6}}6{{s}^{1}}$
In the electronic configuration of these elements, we observe that the number of electrons in the last shell remains the same i.e. 1.
Valency of an element is the number of electrons in the outer shell of the atom.
Therefore, we can conclude that the number of valence electrons in atoms of elements as we go down a group in the periodic table remains the same.
The correct answer is option (C).
Note:
In the above set of elements taken, we did not consider the last element of the group i.e. francium. This is because it demonstrates radioactivity. The nuclei is highly unstable and is thus not under consideration.
Complete answer:
As suggested in the hint above, we will take an element from the periodic table, say sodium (Na) and the elements of the respective group i.e potassium (K), Rubidium (Rb), Caesium (Cs).
We will now write down the electronic configuration of the elements of the first group of the periodic table.
Sodium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$
Potassium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{1}}$
Rubidium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{1}}$
Caesium : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{2}}4{{d}^{10}}5{{p}^{6}}6{{s}^{1}}$
In the electronic configuration of these elements, we observe that the number of electrons in the last shell remains the same i.e. 1.
Valency of an element is the number of electrons in the outer shell of the atom.
Therefore, we can conclude that the number of valence electrons in atoms of elements as we go down a group in the periodic table remains the same.
The correct answer is option (C).
Note:
In the above set of elements taken, we did not consider the last element of the group i.e. francium. This is because it demonstrates radioactivity. The nuclei is highly unstable and is thus not under consideration.
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