Ionization energy (IE) is the energy required to remove the outer electron, also referred to as valence electron, from a gaseous atom in order to form a positive ion called a cation.
The First Ionization Energy (IE1)
The first ionization energy represents the energy required to remove the first valence electron from a given gaseous atom. For example, the first ionization energy equation for the gaseous sodium atom is written as:
The removal of an electron involves the breaking of its connection to the sodium atom. Breaking connections or bonds are an endergonic process that involves the absorption of energy. As a result, ionization energies will always be greater than zero.
Generally, ionization energy increases going from left to right across a period and increases going up any group.
However, there are exceptions that do exist for the first ionization energy of different gaseous atoms.
One exception that exists is when in the same row or period, Group 3A elements have lower ionization energy than elements in Group 3A. This is most notably seen when comparing beryllium and boron.
The electron configurations and orbital diagrams of beryllium and boron are written as:
After the first ionization energy we remove the outermost valence electron to obtain:
When it comes to an s orbital it is most stable when it is completely filled. For B+ we have a half-filled 2s orbital, which energetically is not favored. As a result, the boron atom will require more energy to remove its first valence electron than the beryllium atom.
A similar exception to the periodic trend for ionization energy can be observed when observing elements from Group 5A vs 6A.
The Other Ionization Energies (IE2 , IE3 , IE3…..)
With each additional valence electron you remove there will be an increase in ionization energy. Some of these increases in ionization energies can be observed in the following chart:
Based on the chart, the first ionization energy of aluminum can be observed as:
The second ionization energy is the energy required to remove an additional electron from Al+ to create Al2+.
The third ionization energy equation for Al2+ is shown as:
Notice how the energy more or less doubles or triples when going to the next ionization energy. This is the result of the number of repulsive electrons diminishing while the number of protons remaining constant. As a result, there is a greater attractive force holding onto the electrons and increasing the ionization energy.
Finally, we have the removal of the fourth electron to give the fourth ionization energy.
The noble gases are the most stable elements because they have a completely filled outer shell. When an element obtains the same number of electrons as a noble gas it reaches this new level of optimal stability and a disruption of it would result in a large investment of energy. The difference between the third and fourth ionization energy is so different because removing a fourth electron would cause Al3+ to no longer behave as a noble gas.
PRACTICE: Of the following atoms, which has the smallest second ionization energy?
Answer: “Smallest” means that the cost to remove that particular electron is low and therefore easy to do. Since it says “smallest” second ionization energy it means which element loses its second electron most easily. Because we have gone beyond the first ionization energy we must think in terms of stability.
An element from Group 2A would want to lose two electrons the most because doing so would allow them to obtain a noble gas status. This leaves Mg and Ca as the possible answers. They are both in the same group and remember going up a group causes an increase in ionization energy. Since Ca is positioned lower in Group 2A it would possess the “smallest” second ionization energy. Choice D will be the answer.
The Other Periodic Trends
Ionization Energy represents one of the major periodic trends when dealing with the electron structure of an atom. Other periodic trends involve electron affinity, effective nuclear charge, electronegativity, atomic radius and ionic radius.