Chemguide: Support for CIE A level Chemistry
Learning outcomes 3.2(c) and 3.2(d)
These statements deals with the shapes of some simple molecules
Before you go on, you should find and read the statements in your copy of the syllabus.
You should read the page which looks at shapes of molecules and ions, and then follow the link at the bottom of the page to go on to look at a couple of double bonded examples as well.
In a note following the shape of the methane molecule, you will find a link to a page about drawing organic molecules. You should follow this and make sure that you fully understand about the use of lines, dotted lines and wedges in diagrams of this sort.
You don't need to read the whole of that organic chemistry page. About half way down, you will find a short section titled "How to draw structural formulae in 3-dimensions". Just read that - but be sure that you understand this before you go any further down the shapes page. It won't make sense otherwise!
On the page about the shapes of molecules containing double bonds, just read the bits about carbon dioxide and sulfur dioxide. Read the rest if you are interested, but you only need to know about CO2 and SO2.
If you compare the contents of my shapes pages with the syllabus, you will find that they have lots more examples than the syllabus asks for, but you should nevertheless read, and make sure you understand, everything on the first page, and at least as far as SO2 on the second.
In learning outcome 3.2(d), the syllabus says that it may ask you to explain other similar examples - in other words, it won't necessarily restrict questions to the molecules mentioned in this learning outcome, 3.2(c). That means that the more examples you have worked through, the better.
This statement allows the examiners to ask you about the shapes and bond angles of other molecules similar to the ones already mentioned previously in this section.
There is no short-cut to this! You need to understand how to work out shapes of molecules and their bond angles as described above.
There is no substitute for understanding this. If you try to learn examples parrot-fashion, you won't be able to cope with unfamiliar cases. And CIE like to use unfamiliar examples, because it is a good test of candidates' ability.
The best way of testing yourself is to look for examples from past question papers, and make sure that you can do them by looking at the mark schemes and examiner's reports.
However, don't then waste time trying to learn these extra examples because, next time, the CIE examiners will most probably have come up with something entirely different! As I said just now, there is no real alternative to understanding this.
Important: Don't read the next bit until you have read and understood the pages I have pointed you to. The rest of this page explains a tricky example which cropped up in a CIE exam question. You won't understand it until you are confident about the more straightforward examples.
At the time of writing (June 2010), they had already asked quite an awkward question, which you almost certainly wouldn't have come across as a part of your course.
You were asked for the shape of the chlorine dioxide molecule, ClO2, and told that the chlorine-oxygen bonds were both double.
The temptation is to think of this as being like carbon dioxide - linear. In fact, if you jump to that conclusion, you would be totally wrong. Incidentally, that is why it is important to have read about the sulfur dioxide case, even though it isn't specifically mentioned by the syllabus.
Let's work through it for chlorine dioxide . . .
Chlorine is in Group 7, and so has 7 electrons in its outer level. It is forming a total of 4 bonds to oxygens (2 double bonds), and so that adds another 4 electrons to the outer level - making a total of 11. It is a neutral molecule, not an ion, and so that's the final total of electrons.
11 electrons will be in five-and-a-half pairs. What that means, of course, is that there are 5 pairs, and an odd single electron. If you haven't come across molecules with unpaired electrons before, that might worry you - but such things are perfectly possible.
4 of the 5 pairs will be used to form the 2 double bonds.
So, there will be 2 double bond units, a lone pair and a single electron. These will arrange themselves in an approximately tetrahedral fashion to minimise the repulsions. When you think about the shape that the atoms take up, the molecule will be bent (or V-shaped or, as CIE tend to call it, "non-linear").
This question was only worth 2 marks - 1 for the shape, and 1 for the explanation. On occasions, CIE certainly make you work for your marks!
© Jim Clark 2010 (last modified March 2014)