Chemguide: Support for CIE A level Chemistry
Learning outcome 9.2(b) and part of 9.2(f)
This statement - 9.2(b) - deals with the oxidation numbers (oxidation states) of the elements in Period 3 in their oxides and chlorides. 9.2(f) is involved in the explanation for this.
Before you go on, you should find and read the statements in your copy of the syllabus.
Oxidation state (oxidation number)
The terms "oxidation number" and "oxidation state" are used interchangeably.
I tend use the term "oxidation state", and you will find that used on the page I will point you to in a moment. But for the rest of this page, I will stick to "oxidation number", because that is what the CIE syllabus uses.
You may have met oxidation numbers in Section 6 of the syllabus; if not, the first thing you have to do is to get comfortable with the idea of oxidation number.
You will find this on the page about oxidation states (oxidation numbers).
Don't try to rush this - spend an hour or so on it if necessary. You won't make sense of the rest of this part of the syllabus unless you understand how to use oxidation numbers.
Variation of the oxidation numbers in the chlorides
As in the earlier statement 9.2(a), I am only going to look at the chlorides of the elements from sodium to phosphorus.
In each case, the chlorine in the compound is the more electronegative element, and so the chlorine always has a negative oxidation number - in each case, -1.
Before you read on, work out the oxidation number of the other element in each of the following compounds.
NaCl, MgCl2, Al2Cl6, SiCl4, and PCl5
Don't waste time trying to learn these oxidation numbers - just quickly work them out if you need them. It is important that you can do this. I will discuss a CIE question further down the page where you would get completely confused if you tried to learn this rather than understand it.
Let's look at the most complicated-looking one - Al2Cl6. You can then apply exactly the same principles to the other simpler ones.
All of these are electrically neutral compounds, and so the sum of all the oxidation numbers is zero. If the oxidation number of the aluminium is "x", and the chlorine is -1, then:
2x + 6(-1) = 0
2x = +6
x = +3
So the oxidation number of the aluminium is +3.
You would then go through the same simple process for the other chlorides, and find the following oxidation numbers:
In each case the oxidation number is the same as the number of electrons in the outer energy level of the atom in question - what the syllabus describes as "their valance shell electrons".
These compounds all involve the element using all of its outer electrons in the bonding - either giving them away to form positive ions, or sharing them to form covalent bonds. In each of these cases, there is a movement of electrons away from the element towards the more electronegative chlorine. That is why all of these oxidation numbers are positive.
Although it isn't specifically mentioned by the syllabus, it is worth exploring the phosphorus case a little bit further, because phosphorus forms two chlorides, PCl3 and PCl5.
You will find the bonding in these discussed towards the bottom the page on covalent bonding.
The oxidation number of the phosphorus in PCl3 is +3. Work it out!
The point I am trying to make is that the maximum possible oxidation number of an element in Period 3 is going to be the same as the number of electrons in its outer energy level, but that smaller oxidation numbers are also possible where not all of the outer electrons are used.
You will see another example of this in the oxides of sulphur (discussed below).
Variation of the oxidation numbers in the oxides
In this case, we are concerned with the oxides of the elements from sodium to sulphur.
The oxygen in the compound is always the more electronegative element, and so always has a negative oxidation number - in each case, its usual -2.
So work out the oxidation numbers of the other element in each of the following:
Na2O, MgO, Al2O3, SiO2, P4O10, SO2 and SO3.
Don't just read this next bit - work them out!
You can see that the pattern is much the same as before. The maximum oxidation state shown by the elements in this period is the same as the number of outer electrons.
in the case of sulphur, the maximum oxidation state is +6. All of sulphur's 6 outer electrons are being used to make the bonds in sulphur trioxide. But that doesn't have to be the case.
In sulphur dioxide (where sulphur has an oxidation state of +4), only 4 of its outer electrons are involved in bonding.
You might wonder why sulphur wasn't included amongst the list of chlorides in the previous section. I suspect it is because it would break the tidy pattern. In compounds between sulphur and chlorine, the highest oxidation state shown by the sulphur is +4, not +6. In other words, you can have SCl4, but not SCl6.
There is, however, a compound SF6. So sulphur only shows its maximum oxidation state when it is combined with oxygen or fluorine. The reason for this isn't all that difficult, but isn't required by the syllabus.
What about the oxides of chlorine (not asked in the syllabus)? Chlorine has 7 valence electrons, and so you would expect its maximum oxidation state to be +7. It is - in the oxide Cl2O7, although chlorine also forms several other oxides with smaller oxidation numbers.
A look at what CIE actually ask in exams
You must expect exam questions to go beyond what is actually asked in the syllabus. Examiners can justify this provided you could work the answer out from things which are on the syllabus.
I am going to explore an example to show you the sort of problems that arise. It comes from November 2009 Paper 22. The syllabus statements on the previous syllabus covering that exam were almost the same as the current ones.
As a part of the question, they gave a list of chlorides of the elements from sodium to sulphur and asked you to give the oxidation numbers of the elements, and then explain the variation across the period.
Here is the list of compounds, and I have worked out the oxidation numbers for you:
Notice that the last two chlorides in the question (phosphorus and sulphur) don't have the element in its highest possible oxidation state.
This means that if you had simply learnt that the maximum oxidation number shown by the elements from sodium to phosphorus was the same as the number of outer electrons, you couldn't have done this question.
And if you can't work out the correct oxidation numbers, you obviously can't explain the variation across the period.
You must be able to understand what you are doing.
I have a real problem with the answers to the other part of this question asking you to explain the variation across the period!
The Examiner's Report says about part (a)(ii):
. . . and I take issue with that!
There is no mention in the syllabus of explaining this in terms of noble gas structures. In fact, for the chlorides actually mentioned in the syllabus you can't explain this in terms of noble gas structures, because the chloride of phosphorus and the oxides of phosphorus and sulphur quoted in the syllabus don't involve noble gas structures in the phosphorus or sulphur.
In this instance, it also isn't necessarily true that the aluminium in AlCl3 has a neon structure. That is only true if the AlCl3 is ionic. But in the vapour at a high enough temperature, AlCl3 is a simple molecule, covalently bound - and the aluminium has 6 electrons in its outer level. Even in solid AlCl3, the bonding is ionic with a high degree of covalent character.
What the positive sign in the oxidation numbers actually shows is some loss of electrons in each case, either their full removal to make ions, or their movement towards the chlorine in a polar covalent bond.
What the size of the oxidation numbers (+1, +2, +3, +4) shows is the number of electrons actually involved in this process.
It doesn't surprise me in the slightest that students struggled with this question. What it does make clear is that you must get hold of (and go through very carefully) as many past papers, mark schemes, and examiner's reports as you can. You have to know exactly what the examiners expect for any question - you can't assume that it is the same as you have learnt from a teacher, a textbook or from Chemguide.
You will find the syllabuses part of the CIE site by following this link. Unfortunately, CIE don't make all their past papers available to students, but teachers can access them from a password-protected part of the site. Pester your teacher to get these for you!
© Jim Clark 2010 (last modified May 2014)