IONISATION ENERGY

This page explains what first ionisation energy is, and then looks at the way it varies around the Periodic Table - across periods and down groups. It assumes that you know about simple atomic orbitals, and can write electronic structures for simple atoms. You will find a link at the bottom of the page to a similar description of successive ionisation energies (second, third and so on).

Important! If you aren't reasonable happy about atomic orbitals and electronic structures you should follow these links before you go any further.

Defining first ionisation energyDefinitionThe first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.This is more easily seen in symbol terms.

It is the energy needed to carry out this change per mole of X.

Worried about moles? Don't be! For now, just take it as a measure of a particular amount of a substance. It isn't worth worrying about at the moment.

Things to notice about the equationThe state symbols - (g) - are essential. When you are talking about ionisation energies, everything must be present in the gas state.Ionisation energies are measured in kJ mol-1 (kilojoules per mole). They vary in size from 381 (which you would consider very low) up to 2370 (which is very high).All elements have a first ionisation energy - even atoms which don't form positive ions in test tubes. The reason that helium (1st I.E. = 2370 kJ mol-1) doesn't normally form a positive ion is because of the huge amount of energy that would be needed to remove one of its electrons.

Patterns of first ionisation energies in the Periodic TableThe first 20 elements

First ionisation energy shows periodicity. That means that it varies in a repetitive way as you move through the Periodic Table. For example, look at the pattern from Li to Ne, and then compare it with the identical pattern from Na to Ar.These variations in first ionisation energy can all be explained in terms of the structures of the atoms involved.

Factors affecting the size of ionisation energyIonisation energy is a measure of the energy needed to pull a particular electron away from the attraction of the nucleus. A high value of ionisation energy shows a high attraction between the electron and the nucleus.The size of that attraction will be governed by:The charge on the nucleus.The more protons there are in the nucleus, the more positively charged the nucleus is, and the more strongly electrons are attracted to it.The distance of the electron from the nucleus.Attraction falls off very rapidly with distance. An electron close to the nucleus will be much more strongly attracted than one further away.The number of electrons between the outer electrons and the nucleus.Consider a sodium atom, with the electronic structure 2,8,1. (There's no reason why you can't use this notation if it's useful!)If the outer electron looks in towards the nucleus, it doesn't see the nucleus sharply. Between it and the nucleus there are the two layers of electrons in the first and second levels. The 11 protons in the sodium's nucleus have their effect cut down by the 10 inner electrons. The outer electron therefore only feels a net pull of approximately 1+ from the centre. This lessening of the pull of the nucleus by inner electrons is known as screening or shielding.

Warning! Electrons don't, of course, "look in" towards the nucleus - and they don't "see" anything either! But there's no reason why you can't imagine it in these terms if it helps you to visualise what's happening. Just don't use these terms in an exam! You may get an examiner who is upset by this sort of loose language.

Whether the electron is on its own in an orbital or paired with another electron.Two electrons in the same orbital experience a bit of repulsion from each other. This offsets the attraction of the nucleus, so that paired electrons are removed rather more easily than you might expect.

Explaining the pattern in the first few elementsHydrogen has an electronic structure of 1s1. It is a very small atom, and the single electron is close to the nucleus and therefore strongly attracted. There are no electrons screening it from the nucleus and so the ionisation energy is high (1310 kJ mol-1).Helium has a structure 1s2. The electron is being removed from the same orbital as in hydrogen's case. It is close to the nucleus and unscreened. The value of the ionisation energy (2370 kJ mol-1) is much higher than hydrogen, because the nucleus now has 2 protons attracting the electrons instead of 1.Lithium is 1s22s1. Its outer electron is in the second energy level, much more distant from the nucleus. You might argue that that would be offset by the additional proton in the nucleus, but the electron doesn't feel the full pull of the nucleus - it is screened by the 1s2 electrons.

You can think of the electron as feeling a net 1+ pull from the centre (3 protons offset by the two 1s2 electrons).If you compare lithium with hydrogen (instead of with helium), the hydrogen's electron also feels a 1+ pull from the nucleus, but the distance is much greater with lithium. Lithium's first ionisation energy drops to 519 kJ mol-1 whereas hydrogen's is 1310 kJ mol-1.

The patterns in periods 2 and 3Talking through the next 17 atoms one at a time would take ages. We can do it much more neatly by explaining the main trends in these periods, and then accounting for the exceptions to these trends.The first thing to realise is that the patterns in the two periods are identical - the difference being that the ionisation energies in period 3 are all lower than those in period 2.

Explaining the general trend across periods 2 and 3The general trend is for ionisation energies to increase across a period.In the whole of period 2, the outer electrons are in 2-level orbitals - 2s or 2p. These are all the same sort of distances from the nucleus, and are screened by the same 1s2 electrons.The major difference is the increasing number of protons in the nucleus as you go from lithium to neon. That causes greater attraction between the nucleus and the electrons and so increases the ionisation energies. In fact the increasing nuclear charge also drags the outer electrons in closer to the nucleus. That increases ionisation energies still more as you go across the period.

Note: Factors affecting atomic radius are covered on a separate page.

In period 3, the trend is exactly the same. This time, all the electrons being removed are in the third level and are screened by the 1s22s22p6 electrons. They all have the same sort of environment, but there is an increasing nuclear charge.Why the drop between groups 2 and 3 (Be-B and Mg-Al)?The explanation lies with the structures of boron and aluminium. The outer electron is removed more easily from these atoms than the general trend in their period would suggest.Be1s22s21st I.E. = 900 kJ mol-1

B1s22s22px11st I.E. = 799 kJ mol-1

You might expect the boron value to be more than the beryllium value because of the extra proton. Offsetting that is the fact that boron's outer electron is in a 2p orbital rather than a 2s. 2p orbitals have a slightly higher energy than the 2s orbital, and the electron is, on average, to be found further from the nucleus. This has two effects. The increased distance results in a reduced attraction and so a reduced ionisation energy. The 2p orbital is screened not only by the 1s2 electrons but, to some extent, by the 2s2 electrons as well. That also reduces the pull from the nucleus and so lowers the ionisation energy.The explanation for the drop between magnesium and aluminium is the same, except that everything is happening at the 3-level rather than the 2-level.Mg1s22s22p63s21st I.E. = 736 kJ mol-1

Al1s22s22p63s23px11st I.E. = 577 kJ mol-1

The 3p electron in aluminium is slightly more distant from the nucleus than the 3s, and partially screened by the 3s2 electrons as well as the inner electrons. Both of these factors offset the effect of the extra proton.

Warning! You might possibly come across a text book which describes the drop between group 2 and group 3 by saying that a full s2 orbital is in some way especially stable and that makes the electron more difficult to remove. In other words, that the fluctuation is because the group 2 value for ionisation energy is abnormally high. This is quite simply wrong! The reason for the fluctuation is because the group 3 value is lower than you might expect for the reasons we've looked at.

Why the drop between groups 5 and 6 (N-O and P-S)?Once again, you might expect the ionisation energy of the group 6 element to be higher than that of group 5 because of the extra proton. What is offsetting it this time?N1s22s22px12py12pz11st I.E. = 1400 kJ mol-1

O1s22s22px22py12pz11st I.E. = 1310 kJ mol-1

The screening is identical (from the 1s2 and, to some extent, from the 2s2 electrons), and the electron is being removed from an identical orbital.The difference is that in the oxygen case the electron being removed is one of the 2px2 pair. The repulsion between the two electrons in the same orbital means that the electron is easier to remove than it would otherwise be.The drop in ionisation energy at sulphur is accounted for in the same way.

Trends in ionisation energy down a groupAs you go down a group in the Periodic Table ionisation energies generally fall. You have already seen evidence of this in the fact that the ionisation energies in period 3 are all less than those in period 2.Taking Group 1 as a typical example:

Why is the sodium value less than that of lithium?There are 11 protons in a sodium atom but only 3 in a lithium atom, so the nuclear charge is much greater. You might have expected a much larger ionisation energy in sodium, but offsetting the nuclear charge is a greater distance from the nucleus and more screening.Li1s22s11st I.E. = 519 kJ mol-1

Na1s22s22p63s11st I.E. = 494 kJ mol-1

Lithium's outer electron is in the second level, and only has the 1s2 electrons to screen it. The 2s1 electron feels the pull of 3 protons screened by 2 electrons - a net pull from the centre of 1+.The sodium's outer electron is in the third level, and is screened from the 11 protons in the nucleus by a total of 10 inner electrons. The 3s1 electron also feels a net pull of 1+ from the centre of the atom. In other words, the effect of the extra protons is compensated for by the effect of the extra screening electrons. The only factor left is the extra distance between the outer electron and the nucleus in sodium's case. That lowers the ionisation energy.Similar explanations hold as you go down the rest of this group - or, indeed, any other group.

Trends in ionisation energy in a transition series

Apart from zinc at the end, the other ionisation energies are all much the same.All of these elements have an electronic structure [Ar]3dn4s2 (or 4s1 in the cases of chromium and copper). The electron being lost always comes from the 4s orbital.

Note: The 4s orbital has a higher energy than the 3d in the transition elements. That means that it is a 4s electron which is lost from the atom when it forms an ion. It also means that the 3d orbitals are slightly closer to the nucleus than the 4s - and so offer some screening.Confusingly, this is inconsistent with what we say when we use the Aufbau Principle to work out the electronic structures of atoms. I have discussed this in detail in the page about the order of filling 3d and 4s orbitals.If you are a teacher or a very confident student then you might like to follow this link.If you aren't so confident, or are coming at this for the first time, I suggest that you ignore it. Remember that the Aufbau Principle (which uses the assumption that the 3d orbitals fill after the 4s) is just a useful way of working out the structures of atoms, but that in real transition metal atoms the 4s is actually the outer, higher energy orbital.

As you go from one atom to the next in the series, the number of protons in the nucleus increases, but so also does the number of 3d electrons. The 3d electrons have some screening effect, and the extra proton and the extra 3d electron more or less cancel each other out as far as attraction from the centre of the atom is concerned.The rise at zinc is easy to explain.Cu[Ar]3d104s11st I.E. = 745 kJ mol-1

Zn[Ar]3d104s21st I.E. = 908 kJ mol-1

In each case, the electron is coming from the same orbital, with identical screening, but the zinc has one extra proton in the nucleus and so the attraction is greater. There will be a degree of repulsion between the paired up electrons in the 4s orbital, but in this case it obviously isn't enough to outweigh the effect of the extra proton.

Note: This is actually very similar to the increase from, say, sodium to magnesium in the third period. In that case, the outer electronic structure is going from 3s1 to 3s2. Despite the pairing-up of the electrons, the ionisation energy increases because of the extra proton in the nucleus. The repulsion between the 3s electrons obviously isn't enough to outweigh this either.I don't know why the repulsion between the paired electrons matters less for electrons in s orbitals than in p orbitals (I don't even know whether you can make that generalisation!). I suspect that it has to do with orbital shape and possibly the greater penetration of s electrons towards the nucleus, but I haven't been able to find any reference to this anywhere. In fact, I haven't been able to find anyone who even mentions repulsion in the context of paired s electrons!If you have any hard information on this, could you contact me via the address on the about this site page.

Ionisation energies and reactivityThe lower the ionisation energy, the more easily this change happens:

You can explain the increase in reactivity of the Group 1 metals (Li, Na, K, Rb, Cs) as you go down the group in terms of the fall in ionisation energy. Whatever these metals react with, they have to form positive ions in the process, and so the lower the ionisation energy, the more easily those ions will form.The danger with this approach is that the formation of the positive ion is only one stage in a multi-step process.For example, you wouldn't be starting with gaseous atoms; nor would you end up with gaseous positive ions - you would end up with ions in a solid or in solution. The energy changes in these processes also vary from element to element. Ideally you need to consider the whole picture and not just one small part of it.However, the ionisation energies of the elements are going to be major contributing factors towards the activation energy of the reactions. Remember that activation energy is the minimum energy needed before a reaction will take place. The lower the activation energy, the faster the reaction will be - irrespective of what the overall energy changes in the reaction are.The fall in ionisation energy as you go down a group will lead to lower activation energies and therefore faster reactions.

THE ORDER OF FILLING 3d AND 4s ORBITALS

This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. I am grateful to Dr Eric Scerri from UCLA, who pointed these problems out to me and provided me with some useful academic papers I wouldn't otherwise have been able to get hold of.The way that the order of filling of orbitals is normally taught gives you an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements.This page takes a closer look at this, and offers a more accurate explanation which avoids the problems.

The commonly taught versionThis section is just a summary of the way this is currently taught. It is taken from things you have probably already read elsewhere on Chemguide. You shouldn't find anything unfamiliar in it.The order of filling orbitalsElectrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. Where there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible.The diagram (not to scale) summarises the energies of the orbitals up to the 4p level.

The oddity is the position of the 3d orbitals.They are shown at a slightly higher level than the 4s - and so it is the 4s orbital which will fill first, followed by all the 3d orbitals and then the 4p orbitals. Similar confusion occurs at higher levels, with so much overlap between the energy levels that the 4f orbitals don't fill until after the 6s, for example.

I just want to focus on the fourth period.

The beginning of the fourth periodEverything is straightforward up to this point, but the 3-level orbitals aren't all full - the 3d levels haven't been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills first.K1s22s22p63s23p64s1

Ca1s22s22p63s23p64s2

d-block elements

d-block elements are thought of as elements in which the last electron to be added to the atom is in a d orbital. (Actually, that turns out not to be true! We will come back to that in detail later.)The electronic structures of the d-block elements are shown in the table below. Each additional electron you add usually goes into a 3d orbital. Most chemistry books and chemistry teachers try to explain the breaks in the pattern at chromium and copper - but not very convincingly. I will come back to that later as well.To make the table look less complicated, I am using [Ar] to represent 1s22s22p63s23p6.Sc[Ar] 3d14s2

Ti[Ar] 3d24s2

V[Ar] 3d34s2

Cr[Ar] 3d54s1

Mn[Ar] 3d54s2

Fe[Ar] 3d6 4s2

Co[Ar] 3d74s2

Ni[Ar] 3d84s2

Cu[Ar] 3d104s1

Zn[Ar] 3d104s2

Making positive ions from the d-block ionsThis is probably the most unsatisfactory thing about this approach to the electronic structures of the d-block elementsIn all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. The reversed order of the 3d and 4s orbitals only seems to apply to building the atom up in the first place. In all other respects, the 4s electrons are always the electrons you need to think about first.When d-block elements form ions, the 4s electrons are lost first.

So, for example:To write the electronic structure for Fe3+:Fe1s22s22p63s23p63d64s2

Fe3+1s22s22p63s23p63d5

The 4s electrons are lost first followed by one of the 3d electrons.

What is wrong with this version?Evidence from the formation of ionsThis last bit about the formation of the ions is clearly unsatisfactory. We say that the 4s orbitals have a lower energy than the 3d, and so the 4s orbitals are filled first. We know that the 4s electrons are lost first during ionisation. The electrons lost first will come from the highest energy level, furthest from the influence of the nucleus. So the 4s orbital must have a higher energy than the 3d orbitals.Those statements are directly opposed to each other. They can't both be right.When you talk about ionisation energies for these elements, you talk in terms of the 4s electrons as the outer electrons being shielded from the nucleus by the inner 3d levels.We say that the first ionisation energies don't change much across the transition series, because each additional 3d electron more or less screens the 4s electrons from the extra proton in the nucleus.The explanations around ionisation energies are based on the 4s electrons having the higher energy, and so being removed first.

Where is the flaw in what is usually taught?The usual way of teaching this is an easy way of working out what the electronic structure of any atom is - with a few odd cases to learn like chromium or copper.The problems arise when you try to take it too literally. It is way of working out structures - no more than that.The flaw lies in the diagram we started with:

We draw this diagram, and then assume that it works for all atoms. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram.If you stop and think about it, that has got to be wrong. As you move from element to element across the Periodic Table, you are adding extra protons to the nucleus, and extra electrons around the nucleus.The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals.That means that we need to rethink this on the basis that what we drew above isn't likely to look the same for all elements.

The solutionThe elements up to argonThere is no problem with these. The general pattern that we drew in the diagram above works well.

Potassium and calciumThe pattern is still working here. The 4s orbital has a lower energy than the 3d, and so fills next. That entirely fits with the chemistry of potassium and calcium.

The d-block elementsAnd now it all goes wrong!For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series.So rather than working out the electronic structure of scandium by imagining that you just throw another electron into a calcium atom, with the electron going into a 3d orbital because the 4s is already full, you really need to look more carefully at it.Remember that, in reality, for Sc through to Zn the 3d orbitals have the lower energy - not the 4s.So why isn't scandium [Ar] 3d3 rather than [Ar] 3d14s2?Making Sc3+Imagine you are building a scandium atom from boxes of protons, neutrons and electrons. You have built the nucleus from 21 protons and 24 neutrons, and are now adding electrons around the outside.So far you have added 18 electrons to fill all the levels up as far as 3p. Essentially you have made the ion Sc3+.Making Sc2+Now you are going to add the next electron to make Sc2+. Where will the electron go?The 3d orbitals at scandium have a lower energy than the 4s, and so the next electron will go into a 3d orbital. The structure is [Ar] 3d1.Making Sc+You might expect the next electron to go into a lower energy 3d orbital as well, to give [Ar] 3d2. But it doesn't. You have something else to think about here as well.If you add another electron to any atom, you are bound to increase the amount of repulsion. Repulsion raises the energy of the system, making it less energetically stable. It obviously helps if this effect can be kept to a minimum.The 3d orbitals are quite compactly arranged around the nucleus. Introducing a second electron into a 3d orbital produces more repulsion than if the next electron went into the 4s orbital.There isn't a very big gap between the energies of the 3d and 4s orbitals. The reduction in repulsion more than compensates for the energy needed to do this.The energetically most stable structure for Sc+ is therefore [Ar] 3d14s1. Making ScPutting the final electron in, to make a neutral scandium atom, needs the same sort of discussion. In this case, the lowest energy solution is the one where the last electron also goes into the 4s level, to give the familiar [Ar] 3d14s2 structure.SummaryIn each of these cases we have looked at, the 3d orbitals have the lowest energy, but as we add electrons, repulsion can push some of them out into the higher energy 4s level.If you build up the scandium atom from scratch, the last electrons to go in are the two 4s electrons. These are the electrons in the highest energy level, and so it is logical that they will be removed first when the scandium forms ions. And that's what happens.The 4s electrons are also clearly the outermost electrons, and so will define the radius of the atom. The lower energy 3d orbitals are inside them, and will contribute to the screening. There is no longer any conflict between these properties and the order of orbital filling.

The electronic structures of two more d-block elementsThe difficulty with this approach is that you can't use it to predict the structures of the rest of the elements in the transition series.In fact, what you have to do is to look at the actual electronic structure of a particular element and its ions, and then work out what must be happening in terms of the energy gap between the 3d and 4s orbitals and the repulsions between the electrons.

Note: You will probably find this really frustrating. The common way of teaching this (based on the wrong order of filling of the 3d and 4s orbitals for transition metals) gives a method which lets you predict the electronic structure of an atom correctly most of the time.The better way of looking at it from a theoretical point of view no longer lets you do that.You can get around this, of course. If you want to work out a structure, use the old method. But remember that it is based on a false idea, and don't try to use it for anything else - like working out which electrons will be lost first from a transition element, for example.

Thinking about the other elements in the series in the same way as we did with scandium, in each case the 3d orbitals will take the first electron(s). Then at some point repulsion will push the next ones into the 4s orbital. When this happens varies from element to element.VanadiumVanadium has two more electrons than scandium, and two more protons as well, of course. Think about building up a vanadium atom in exactly the same way that we did scandium.We have the nucleus complete and now we are adding electrons. When we have added 18 electrons to give the argon structure, we have then built a V5+ ion.Now look at what happens when you add the next 5 electrons.V5+[Ar]

V4+[Ar]3d1

V3+[Ar]3d2

V2+[Ar]3d3

V+[Ar]3d4

V[Ar]3d34s2

The energy gap between the 3d and 4s levels has widened. In this case, it isn't energetically profitable to promote any electrons to the 4s level until the very end. In the ions, all the electrons have gone into the 3d orbitals. You couldn't predict this just by looking at it.ChromiumWhy is the electronic structure of chromium [Ar]3d54s1 instead of [Ar]3d44s2?Because that is the structure in which the balance of repulsions and the size of the energy gap between the 3d and 4s orbitals happens to produce the lowest energy for the system.Many chemistry textbooks and teachers try to explain this by saying that the half-filled orbitals minimise repulsions, but that is a flawed, incomplete argument. You aren't taking into account the size of the energy gap between the lower energy 3d orbitals and the higher energy 4s orbital.Two rows directly underneath chromium in the Periodic Table is tungsten. Tungsten has exactly the same number of outer electrons as chromium, but its outer structure is 5d46s2, NOT 5d56s1.In this case, the most energetically stable structure isn't the one where the orbitals are half-full. You can't make generalisations like this!

In conclusionThe current method of teaching students to work out electronic structures is fine as long as you realise that that is all it is - a way of working out the overall electronic structures, but not the order of filling.You can say that for potassium and calcium, the 3d orbitals have a higher energy than the 4s, and so for these elements, the 4s levels fill before than the 3d. That, of course, is entirely true! Then you can say that, looking at the structures of the next 10 elements of the transition series, the 3d orbitals gradually fill with electrons (with some complications like chromium and copper). That is also true.What is not right is to imply that the 3d levels across these 10 elements have higher energies than the 4s. That is definitely not true, and causes the sort of problems we have been discussing.