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 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 a negatively charged 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 the size of 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 them 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:

• The general trend in ionisation energy across period 3 elements is that it increases with increasing nuclear charge. There are 2 "dips" in this trend; these are with the elements aluminium and sulfur.


• The 2 dips in the ionisation energies, one as we go from Mg to Al and the other as we go from the element P to S. The drop in ionisation energy from Mg to Al is due to the fact that the electron being removed is from a 2s sub-shell in the case of Mg and a p sub-shell in the case of Al. Since the 3p sub-shell is further from the nucleus it will be higher in energy and also the electron in the 3p sub-shell is shielded more than an electron in the 3s sub-level, so it requires less energy to remove. That is the 3p electron in aluminium will experience a smaller effective nuclear charge than the 3s electron in magnesium because it will be shielded more effectively by the inner core electrons. We can show this as:

Mg: 1s²2s²2p⁶3s² → Mg⁺: 1s²2s²2p⁶3s¹

Al: 1s²2s²2p⁶3s²3p¹ → Al⁺: 1s²2s²2p⁶3s²

Since the 3p electron is further from the nucleus than the 3s electron and it is also shielded more by the inner core electrons and so it will require less energy to remove it.

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: 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 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:

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

• The general trend in ionisation energies across a period is that it increases due to increasing nuclear charge.


• The size of atoms across a period decreases, the main reason for this is increasing nuclear charge and the fact that additional electrons are being added to the same principal energy level, so as we cross a period the atoms also get smaller and with increasing nuclear charge the ionisation energies will generally increase.


• As we cross a period additional electrons are being added to the same principal energy level and electrons in the same shell or principal energy level do not shieldeach other effectively from the nuclear charge. This means that as we cross a period the amount of shielding present does not change much so the effective nucler charge that the electrons feel will be increasing.


• There are 2 dips in the first ionisation in period 3, one when we leave the s-block and enter the p-block, this happens in the move from magnesium to aluminium, here the outer valence electron moves from a 3s-orbital to 3p-orbital which is higher in energy and so requires less energy to remove the electron since there is an increase in nuclear charge and an increase in shielding.


• The second dip in the first ionisation energy in period 3 occurs when we go from a 3p³ to a 3p⁴ electronic configuration, this happens when we move from phosphorus to sulfur. The pairing up of the electrons in the p-sub shell raises its energy and so makes it easier to remove an electron.

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