Question

The maximum tendency to form unipositive ion is for the element which has the following electronic configuration:
A.${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}$
B.${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{1}}}$
C.${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}$
D.${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^3}$

Answer 1

Hint: The unipositive means a positive charge. To form a unipositive ion, the metal will lose one electron. The tendency of the formation of unipositive ions is based upon the stability of the unipositive ion.

Complete Step by step answer: Every metal loses electrons to gain a stable electronic configuration like noble metal because the Noble metals are the most stable. So, the fully-filled or partially-filled electronic configurations are the most stable.
To remove an electron the energy is required. Ionization potential is defined as the energy required for the removal of an electron from the outermost shell of an isolated gaseous atom. It is also known as ionization energy.
Now we will remove one electron to form the unipositive ion of each electronic configuration and check its stability as follows:
-The electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}$ is of Mg atom.
-The electronic configuration of the unipositive ion of Mg is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^1}$. So, the electronic configuration of Mg was stable because it has fully-filled orbitals whereas the unipositive ion of Mg is unstable.
-The electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{1}}}$ is of Al atom.
-The electronic configuration of the unipositive ion of Al is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}$. So, the electronic configuration of Al was unstable due to the presence of a single electron. After the formation of the unipositive ion Al gains the fully-filled electronic configuration. So, the unipositive ion of Al is stable. So, Al will form a unipositive ion. So, option (B) is correct.
-The electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}$is of Ne atom.
-The electronic configuration of the unipositive ion of Ne is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^5}$. So, the electronic configuration of Ne was stable because it has fully-filled orbitals whereas the unipositive ion of Ne is unstable.
-The electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^3}$ is of P atom.
The electronic configuration of unipositive ion of P is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^2}$. So, the electronic configuration of P was stable because it has half-filled orbitals whereas the unipositive ion of P is unstable.

Therefore, option (B) ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^1}$ is correct.

Note: On going left to right in a period, the ionization potential increases. Due to an increase in effective nuclear charge and decreases in size the stabilization of the outermost shell increases, so it becomes hard to remove an electron from the outermost shell on going left to right in a period. On going down in a group, the ionization potential decreases. Due to a decrease in effective nuclear charge and increases in size the stabilization of the outermost shell decreases, so it becomes easy to remove an electron from the outermost shell on going down in a group.