Successive Ionization Energies

Remember that the first ionization energy (IE1) is the energy required to remove the most loosely bound electron from a neutral atom and the second ionization energy (IE2) is the energy required to remove the second most loosely bound electron. The energy required to remove the third electron is the third ionization energy (IE3) and so on.

Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in the table below, there is a large increase in the ionization energies for each element. This jump corresponds to the removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization.

ElementIE1IE2IE3IE4IE5IE6IE7
K418.83051.84419.65876.97975.59590.611343
Ca589.81145.44912.46490.68153.010495.712272.9
Sc633.11235.02388.77090.68842.910679.013315.0
Ga578.81979.42964.661808298.710873.913594.8
Ge762.21537.53302.14410.69021.4Not availableNot available
As944.51793.62735.54836.86042.912311.5Not available

Ranking Ionization Energies

Predict the order of increasing energy for the following processes:

  • IE1 for Al
  • IE1 for In
  • IE2 for Na
  • IE3 for Al
Solution

Both IE2 for Na and IE3 for Al are removing the last core electron from the atom. Since these processes will both begin from a cationic state, the electrons will be more difficult to remove. Al will be Al2+ and Na will be Na+ — the Al cation is more highly charged, and the electron will be more difficult to remove from it. IE2 for Na < IE3 for Al.

Both of these second-or-higher ionization energies will be larger than either IE1. We can compare these based on atomic radius (Zeff is the same for Al and In). Since In is a larger atom, its valence electrons are further from the nucleus and more loosely held (easier to remove): IE1 for Al > IE1 for In.

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All together and in increasing order: IE1 for In < IE1 for Al < IE2 for Na < IE3 for Al