There are three factors which decide how easily an electron can be removed, in this order of priority:
- Number of shells (distance from the nucleus in effect)
- Effect of shielding
- Nuclear attractive force from protons
First ionisation
Both $"K"$ (potassium) and $"Ca"$ (calcium) are in Period 4.
To ionise once, we must remove one electron from these two metals.
$"K"$ is in Group 1. It, therefore, has one electron in its outer shell (in the $s$ orbital). $"Ca"$ is in Group 2, and has two electrons in its outer shell (again both in the $s$ orbital).
$"Ca"$ has the same number of shells as $"K"$, a similar amount of shielding, but more protons in the nucleus. This means there is a stronger attraction between the nucleus and the electron to be removed, meaning more energy is required to remove it (moving it to a potential of $"0 eV"$).
$"Ca"$ has a higher first ionisation energy than $"K"$.
In the case of $"K"$, this leaves it with a full third shell, as a $"K"^+$ ion. For $"Ca"$, it becomes a $"Ca"^+$ ion, with one electron still remaining in the 4th shell.
Second ionisation
In this case, we are ionising the $"K"^+$ ion and the $"Ca"^+$ ion. These are formed after the first ionisation.
$"K"^+$ has the electron configuration $["Ar"]$. This means it has three full shells. $"Ca"^+$ has the same electron configuration as an atom of $"K"$, i.e. $["Ar"]4s^1$.
$"Ca"^+$ has more shells and shielding than $"K"^+$. This outweighs the fact that $"Ca"^+$ has more protons in the nucleus, meaning means there is a weaker attraction between the nucleus and the electron to be removed, meaning less energy is required to remove it.
$"K"^+$ has a higher ionisation energy than $"Ca"^+$, so $"K"$ has a higher second ionisation energy than $"Ca"$.
Hope this helped :)
