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 :)
