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 considerably 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 effect of the positively charged nucleus because the inner or core electrons will effectively 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 3s sub-shell in the case of Mg and a 3p 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 (Z_eff) 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.

Understanding Shielding and Effective Nuclear Charge (Period 3)

A simple model of shielding

We can build a simple model to show how shielding changes across any period in the periodic table. In this simplified model we will assume that the outer or valence electrons will shield or screen each other poorly, this means that the inner or core electrons will be responsible for partly shielding or screening some of the nuclear charge (Z) from the outer valence electrons. This means that the outer valence electrons will not feel the full attractive force from the nucleus but will instead feel only a partial attraction for the nucleus; we call this partial attraction the effective nuclear charge (Z_eff).

The activity below illustrates how to calculate the effective nuclear charge (Z_eff) that the valence electrons feel for the elements across period 3. The effective nuclear charge (Z_eff) is simply calculated by subtracting the number of shielding core electrons (S) from the number of positively charged protons in the nucleus. It should be noted that this is a simplification but does help in explaining the trend in the shielding across the period 3 elements and helps in explaining the trends in the ionisation energy across period 3. Simply select the element symbols in the right-hand bin to view the effective nuclear charge (Z_eff) for each element.

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⁺: 1s²2s²2p⁶3s²3p²

S: 1s²2s²2p⁶3s²3p⁴ → S⁺: 1s²2s²2p⁶3s²3p³

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 repulsion between them.

Self-check- trends in the ionisation energies of the period 3 elements

Click the period 3 element symbols and place them on the correct blue dot on the graph to show the trend in the first ionisation energy, press the check answer and reveal graph button when your done.

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 shield each 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 nuclear charge that the electrons feel will be increasing.


• One dip occurs when we leave the s-block and enter the p-block, moving from magnesium to aluminium. Here the outer valence electron moves from a 3s sub-level to a 3p sub-level, which is higher in energy and more shielded, so it requires less energy to remove.


• The second dip occurs when we go from a 3p³ to a 3p⁴ electronic configuration, moving from phosphorus to sulfur. In sulfur the 3p electrons begin to pair up in the 3p sub-levels, and the repulsion between paired electrons makes it easier to remove one, so the ionisation energy is lower than expected.

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