Question

Match the atomic numbers 4, 14, 8, 15 and 19 with a gas of valency 2.
A) 4
B) 8
C) 14
D) 15

Answer 1

Valence electrons are the electrons which enter into the valence shell of an atom.
- The valence electrons are actively involved during the formation of bond, they share, gain or lose these valence electrons during the bond formation.
- The valency of an element is the number of electrons an element can lose or gain to obtain a stable condition.

Complete Solution :
So in the question we provided the atomic number of various elements and we have to identify that element which possess a valency of 2 and which is in gaseous phase too.
- We know that the valence electrons are those electrons present in the outermost shell of an atom and according to the outer electrons or valence electrons present each group has its own group valency.
- These valence electrons are those species which play an active role during the chemical reactions in which they share the valence electrons or lose or gain the valence electrons to form a bond and to obtain stable octet configuration.
- Most important point for solving the question is that we should know how to write the electronic configuration and to identify the element related to the atomic number.
- The increasing order of energy of the orbitals for filling the electrons in various orbitals present in each shell of the atom are:
$1s\langle 2s\langle 2p\langle 3s\langle 3p\langle 4s\langle 3d\langle 4p$…

Now let’s write the electronic configuration of the options given and identify the elements.
- So option (A) is 4, element with atomic number 4 is Be.
The electronic configuration of Be = $1{{s}^{2}}2{{s}^{2}}$
It has 2 valence electrons and the valency is 2, but Be is not a gas, it exists as solid in room temperature and it belongs to an alkaline earth metal group.
- Now let’s see option (B) which has an atomic number of 8 and the element is oxygen so it satisfies one of the criteria of the question since it is a gas. Now let’s write its electronic configuration and find the valency.
The electronic configuration of O = $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$
In this case the O atom needs two more electrons to complete the octet configuration and it gains two electrons to form the stable noble gas configuration and shows a valency 2.
So this option satisfies both the conditions of the question.
- Now if we see option C, the element with atomic number 14 is Silicon and it belongs to group 14 which possesses the group valency of +4.
The electronic configuration of Si = $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{2}}$
- Now let’s check the last option, option (D), element having atomic number 15 is P which belongs to group 15 and possess a group valency of +3.
And the electronic configuration of P = $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}$
So, the correct answer is “Option B”.

Note: The element with atomic number 19 is K, which belongs to alkali metal, the group-1 element having a valency of +1.
- While filling the electrons according to the increasing order of energy, each orbital could only accommodate two electrons with opposite spins, more than two electrons are not possible for an orbital.