When atoms lose more than one electron, the ionization energy to remove the second electron is always more than the ionization energy to remove the first. Similarly, the ionization energy to remove the third electron is more than the second and so on. However, the increase in ionization energy upon the removal of subsequent electrons is not necessarily uniform. For example, consider the first three ionization energies of magnesium:
First ionization energy 738 kJ/mol
Second ionization energy 1450 kJ/ mol
Third ionization energy 7730 kJ/ mol
The second ionization energy is roughly twice the first ionization energy, but then the third ionization energy is over five times the second. Use the electron configuration of magnesium to explain why this is so. Would you expect the same behavior in sodium? Why or why not?
Ionization energy is the amount of energy required to remove an electron from an atom to form an ion in a gaseous state. Ionization energy always increases as we move from left to right across a period of the periodic table. This is because of the increase in effective nuclear charge with increasing atomic number. Thus Mg has higher IE than Na.
Thus, there is huge difference between the 1st, 2nd and 3rd IE of Na whereas for magnesium, the 2nd ionization energy is simply double to 1st second, but then a huge difference between the second and third ionization energies. This can be explained by the electronic configuration of Na and Mg.
Na : [Ne]3s1
Therefore, 1st IE for Na involves the removing of 3s valence electron which is loosely bind with the nucleus. After removing the 3s electron from Na, the resulting ion attains a noble gas configuration which is very stable.
Na - e-----> Na+ IE1st = 496 kJ/mol,
The second IE of sodium, generally participates to removing a core electron from an ion with a noble gas...