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Ionisation energy is a measure of the ease in which atoms lose electrons and become positive ions.
The first ionisation energy is the energy required to remove one electron from each atom of a mole of gaseous atoms.
M(g) - e- → M+(g)
Further electrons may be removed giving successive i.e.:
M+(g) - e- → M2+(g)
This energy is usually quoted in units of kilojoules per mole (kJ mol-1).
Energy is required to remove an electron from any atom because there is an attractive force between the nucleus and the electron being removed which has to be overcome.
The value of the first ionisation energy depends upon:
- The effective nuclear charge
- The distance between the electron and the nucleus
- The 'shielding' produced by lower energy levels
Shielding involves the repulsion between electrons in inner, filled orbitals and electron being removed from the outer orbital.
The graphs of atomic number against first ionisation number show that across each period there is an increase in ionisation energy.
Beryllium (Group II) has an extra electron and proton compared with lithium. The extra electron goes into the same 2s orbital. The increase in ionisation energy (I.E.) can be attributed to the increased nuclear charge.
The ionisation energy of Boron is less than that of Beryllium because in Boron there is a complete 2s orbital. The increased shielding of the 2s orbital reduces the ionisation energy.
Similarly, the I.E. of Oxygen is less than that of Nitrogen because the extra electron is shielded by the half-filled 2p orbital.
The break in the graph between N-O can be explained by the increased repulsion produced when two electrons are in the same orbital. The latter seems to be preferred by examiners!
Within a group the first I.E. decreases down the group as the outer electron becomes progressively further from the nucleus. Also there is more shielding because of the extra filled orbitals.
The graph below shows the successive I.E. for sodium:
The electronic structure for sodium is 1s2 2s2 2p6 3s1. The energy required to remove the first electron is relatively low. This corresponds to the loss of one 3s electron.
To remove the second electron needs a much greater energy because this electron is closer to the nucleus in a 2p orbital.
There is a steady increase in energy required as electrons are removed from 2p and then 2s orbitals.
The removal of the tenth and eleventh electrons requires much greater amounts of energy, because these electrons are closer to the nucleus in the 1s orbital.
Finally an alternative way of expressing electron configuration as s, p,d, and f is to use box notation as shown below for silicon:
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