ionisation energy

Trends in the ionisation energy for the period 3 elements

Ionisation energy

Image to explain the trends in ionisation energy across a period in the period table. The first ionisation energy is the amount of energy required to remove 1 mole of electrons from 1 mole of isolated atoms in the gaseous state. It can be represented by the equation:

X(g) X+(g) + e(g)

This process will obviously be an endothermic one since 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 vary 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 positively 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. 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 in the graph below. There are a few observations worth making: graph to show the trends in the ionisation energy of the period 3 
elements

Trends in the ionisation energies of the period 3 elements

The graph opposite shows the trend in the ionisation energies across 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 easily offer an explanation as to why this drop happens:

P: 1s22s22p63s23p3P+: 1s22s22p63s2
S: 1s22s22p63s23p4S+: 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 paired 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. Now recall Hund's rule of maximum multiplicity, this rule will require the three p-electrons which remains in the p-orbitals to all have parallel spins and occupy separate orbitals, so the one electron in the sulfur p-orbitals which has a spin in the opposite direction to the other 3 electrons, 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

In the sulfur atom the electrons in the p-orbitals begin to pair up. This pairing up will introduce some repulsion between the two electrons in this orbital. This means that when we ionise the sulfur atom and remove one of the electrons in the 3p-orbitals then less energy than expected will be required to remove it because the other electron in the p-orbital will give it a bit of “a push” and help it leave due to this replusion between them.

Key Points

Practice questions

Check your understanding - Questions on ionisation energies

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