Question

Potassium forms ${{\text{K}}^{\text{ + }}}$ion but it does not form ${{\text{K}}^{{\text{2 +
}}}}$because of
A. very low value of ${\left( {{\text{I}}{\text{.E}}{\text{.}}} \right)_{\text{1}}}$and ${\left(
{{\text{I}}{\text{.E}}{\text{.}}} \right)_2}$
B. very high value of ${\left( {{\text{I}}{\text{.E}}{\text{.}}} \right)_{\text{1}}}$and ${\left(
{{\text{I}}{\text{.E}}{\text{.}}} \right)_2}$
C. low value of ${\left( {{\text{I}}{\text{.E}}{\text{.}}} \right)_{\text{1}}}$and low value of${\left(
{{\text{I}}{\text{.E}}{\text{.}}} \right)_2}$
D. low value of ${\left( {{\text{I}}{\text{.E}}{\text{.}}} \right)_{\text{1}}}$and high value of${\left(
{{\text{I}}{\text{.E}}{\text{.}}} \right)_2}$

Answer 1

Hint:The ionization potential decreases down in a group whereas the ionization potential increases from left to right in a period.Ionization potential depends upon the electronic configuration.The filled configurations are more stable so the atom requires high ionization energy.

Complete step-by-step solution: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.

On going left to right in a period, the ionization potential increases. The reason for increases in ionization potential is as follows:
(1) Increase in effective nuclear charge
(2) Decreases in size.

On going down in a group, the ionization potential decreases. The reason for decreases in ionization potential is as follows:
(1) Decrease in effective nuclear charge
(2) Increases in size.

Potassium is the group-I element its electronic configuration is as follows:
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}
}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{1}}}$

Potassium has one electron in outermost s-orbitals. After losing this one electron it obtains a stable configuration, so it forms ${{\text{K}}^{\text{ + }}}$ion.

The electronic configuration of ${{\text{K}}^{\text{ 2+ }}}$ion is as follows:

${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}
}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}$

To remove the second electron from a very stable configuration, a high amount of energy is required. So, potassium does not form ${{\text{K}}^{{\text{2 + }}}}$.

Therefore, option (D) low value of ${\left( {{\text{I}}{\text{.E}}{\text{.}}} \right)_{\text{1}}}$and high the value of${\left( {{\text{I}}{\text{.E}}{\text{.}}} \right)_2}$, is correct.

Note:On going left to right in a period, the electrons are added in the same shell and the proton number also increases, so the attraction of the nucleus on the outermost shell increases. So, the size of the atom decreases. On-going down in a group, the electrons get added in the next higher shell so, the attraction of the nucleus on the outermost shell decreases so, the size of the atom increases. Down in the group, ionization decreases so, IE of ${\text{K}}$ will be lower than IE of ${\text{Na}}$.