
If the IP of ${\text{Na}}$ is$\,{\text{5}}{\text{.48 eV}}$, the ionization potential of ${\text{K}}$ will be:
A. Same as that of ${\text{Na}}$
B. $4.34\,{\text{eV}}$
C. $5.68\,{\text{eV}}$
D. $10.88\,{\text{eV}}$
Answer
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Hint: The ionization potential decreases down in a group whereas the ionization potential increases from left to right in a period.
Complete step by step answer:
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.
Due to both of these factors, (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.
The reason for decreases in ionization potential is as follows:
1. Decrease in effective nuclear charge
2. Increases in size.
Due to both of these factors, (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.
${\text{Na}}$ and ${\text{K}}$ both are the elements of the same group. ${\text{K}}$ comes next to ${\text{Na}}$ in group-I. Down in a group, ionization potential decreases, so IP of ${\text{K}}$ will be lower than IP of ${\text{Na}}$.
So, the IP of ${\text{K}}$is $4.34\,{\text{eV}}$.
Therefore, option (B) $4.34\,{\text{eV}}$,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.
Complete step by step answer:
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.
Due to both of these factors, (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.
The reason for decreases in ionization potential is as follows:
1. Decrease in effective nuclear charge
2. Increases in size.
Due to both of these factors, (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.
${\text{Na}}$ and ${\text{K}}$ both are the elements of the same group. ${\text{K}}$ comes next to ${\text{Na}}$ in group-I. Down in a group, ionization potential decreases, so IP of ${\text{K}}$ will be lower than IP of ${\text{Na}}$.
So, the IP of ${\text{K}}$is $4.34\,{\text{eV}}$.
Therefore, option (B) $4.34\,{\text{eV}}$,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.
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