IE is related to the energy required to remove an electron from an atom.
If an element has more than one electron to be removed, it will have more than one ionization Energy (IE)
The first ionization energy is defined as the energy required to remove the outer most electron from a neutral atom in the gas phase. An equation can be written to show this definition,
M(g) -> M+(g) + 1e–
The second ionization energy is the energy required to remove the next outer electron from the singly charged ion.
M+(g) -> M2+(g) + 1e–
Subsequent removal of electrons can also be written and determined.
The general trend is for the ionization energy to increase going across a period. This makes sense if we recall the atomic radius decreases going across the period. If the atom gets smaller it means the electrons feel a greater attraction to the nucleus and it is more difficult to remove the electron.
As you would expect each successive removal of an electron requires more energy.
As electrons are removed the remaining electrons experience a greater attraction to the nucleus. As shown in the table the ionization energies increase with each successive removal. Notice that there occur major changes as successive electrons are removed.
For magnesium, 1s2 2s2 2p6 3s2 there is a large change in IE between the second and third electron removal. Why? (Consider full vs partially full orbitals and stable octets)
Magnesium has the electron configuration of 1s22s22p63s2. The first two electrons are removed from the third level. The third electron is removed from the second level. Electrons in lower levels feel a greater attraction to the nucleus and are more difficult to remove. Also the second level is stable octet with a full 2s and a full 2p6 orbital.
A similar pattern occurs with Aluminum. There is a large jump in IE for aluminum between the third and fourth electron removal
aluminium: 1s2 2s2 2p6 3s2 3p1 ( easier to remove the level 3 valence electrons and not easy to
remove an electron from a stable 2p6 orbital or energy level 2 with 8
electrons)
The IE for silicon jumps between the fourth and fifth.
Looking at the table suggests the valence electrons, the outer most electrons, require considerably less energy to remove. This suggests that the outer electrons are much easier to remove from the atom compared to the inner level electrons. The inner level electrons are too tightly bound to be involved in chemical bonding.
TABLE Successive values of ionization energies (I) for the elements sodium through argon (kJ/mol)a
| Element | I1 | I2 | I3 | I4 | I5 | I6 | I7 |
| Na | 490 | 4560 | |||||
| Mg | 735 | 1445 | 7730 | ||||
| Al | 580 | 1815 | 2740 | 11,600 | |||
| Si | 780 | 1575 | 3220 | 4350 | 16,100 | ||
| P | 1060 | 1890 | 2905 | 4950 | 6270 | 21,200 |
When we look at first and second ionization energy for sodium there is a significant difference. The second ionization is almost ten times larger than the first. What is going on? To explain this all we need to do is consider which electrons are being ionized and what level are those electrons coming from. So lets look at the electron configuration for sodium,
1s22s22p63s1
The first electron ionized it the electron in the highest energy level…the 3s. That electron requires some amount of energy to remove. After removing the 3s electron we have Na+, which has the electron configuration of 1s22s22p6
When the second electron is removed it comes from the 2p sublevel, as that is the sublevel with the next highest energy.
Why does it take so much more energy to remove the second electron in sodium? It has to do with the fact that the second electron feels a greater attraction to the nucleus compared to the first electron removed. There is more proton pull per electron .
The other general trend is that ionization energy decreases as you go down a group. This is because the electrons enter a new energy level as one moves down a block on the periodic table. Also there are more inner electrons to shield the valence electrons from the nuclear pull. Thus, it requires less energy to remove electrons.
