ionisation energy

Trends in the ionisation energy for the period 3 elements

Ionisation energy

The first ionisation energy is the amount of energy required to remove 1 mole of electrons from an isolated atom in the gaseous state. It can be represented by the equation:

X(g) → X+(g) + e

This process will obviously be an endothermic one, energy will have to be provided to remove the electron from the attractive force it feels from the positively charged nucleus. The first ionisation energies varies considerable for different elements. The three factors that you must consider when discussing ionisation energy are:

  1. the size of the nuclear charge, the larger the number of positively charged protons present in the nucleus then the greater will be the attraction for the electrons.
  2. The further away the electrons are from the nucleus then the easier they will be to remove since the force of attraction from the positvely charged protons in the nucleus will decrease with distance.
  3. The last factor to consider is shielding. The electrons in the valence shell (outer shell) will not feel the full affect of the positively charged nucleus because the inner or core electrons will effective shield or screen the nucleus from them. This shielding effect will reduce the size of the attractive force from the nucleus that the electrons feel and so it will require less energy to remove them.

Trends in the ionisation energies for the period 3 elements

The ionisation energies for the period 3 elements Na-Ar are shown below in the diagram. There are a few observations worth making:

Trends in the ionisation energies of the period 3 elements

The graph below shows the trend in the ionisation energies across period 3.

graph to show the trends in the ionisation energy of the period 3 

There is also a drop in the ionisation energy as we go from the element phosphorus to sulfur. If we consider the electronic configuration of these two elements then we can offer an explanation as to why this drop happens:

P: 1s22s22p63s23p3 → P+: 1s22s22p63s2
S: 1s22s22p63s23p4 → S+: 1s22s22p63s3

In phosphorus the 3p-electrons all occupy separate p-orbitals, however in sulfur the electrons begin to pair up in the p-orbitals. This pairing up of electrons will introduce some repulsion between the electrons, this means that a filled orbital will be slightly higher in energy than a half-filled orbital so less energy will be needed to remove this one electron. Recall Hund's rule of maximum multiplicity, this rule will require the three p-electrons which remains in the orbitals all have parallel spins and occupy separate orbitals, so the one electron in the p-orbitals which has a spin in the opposite direction to the other 3, will be the one which is removed. This is shown in the diagram below:

 explanation of which electrons are removed when sulfur and 
phosphorus are ionised

Key Points

Practice questions

Check your understanding - Questions on ionisation energies