Chemguide: Support for CIE A level Chemistry

Learning outcome 13

13.3: Shapes of organic molecules; σ and π bonds

Statement 13.3.1

Read the statement - it is self-explanatory.

A straight chain isn't, of course, literally straight. The diagram shows butane, C4H10.

Both of the models are exactly the same molecule because the carbons and hydrogens are still joined up in the same way. But there is rotation around carbon-carbon single bonds, and the molecule can take up all sorts of different shapes in space.

Even the straightest possible chain isn't actually straight, but a sort of zig-zag.

And here is another form of C4H10.

You can see that one of the carbon atoms has been taken off the end of the chain and re-attached in the middle. The right-hand model shows a branched chain.

You will also come across examples of carbon atoms being joined in rings. The molecules are described as cyclic.

Statements 13.3.2, 13.3.3 and 13.3.4

These statements cover the bonding and shapes of organic molecules with different types of hybridisation around the carbon atoms.

Before you go on, you should find and read the statements in your copy of the syllabus.

Much of this has already been covered in statement 3.4.2. Read this again before you start.

Bonding and shapes of methane and ethane - sp3 hybridisation

You should read the page bonding in methane which includes both methane and ethane

If you are asked to draw the shape of an ethane molecule, I suggest you use something like this:

Note:  If you have forgotten what the various ways of drawing bonds mean, re-read the page on how to draw organic molecules.

The bonding and shape of ethene - sp2 hybridisation

You will find everything you need on the page bonding in ethene.

By the end of this, you should understand that:

  • Ethene is a planar molecule.

  • The double bond in ethene consists of a sigma bond and a pi bond.

  • The pi bond lies above and below the plane of the molecule.

  • There is no rotation around a double bond. If you tried to twist the molecule, the pi bond would break, and that costs quite a lot of energy. (The pi bond would break because the unhybridised p orbitals would no longer be parallel to each other and touching sideways.)

The bonding and shape of ethyne - sp hybridisation

Ethyne (the first member of the alkyne homologous series) isn't specifically on the syllabus, but is the simplest example of a compound with sp hybridised carbon atoms.

If you followed the link to statement 3.4.2 above, you will already have read this, but refresh your memory by looking again at the formation of sp hybrid orbitals by reading the page about bonding in ethyne.

Predicting shapes of similar molecules

As a general guide, remember that:

  • If you have an sp3 carbon atom with four single bonds attached to it, the shape around that carbon atom will be tetrahedral, with bond angles of about 109.5°.

  • If you have an sp2 carbon atom with two single bonds and a double bond attached to it, the shape around that carbon atom will be planar, with bond angles of about 120°. That is true whether the double bond is attached to another carbon atom or to, say, an oxygen atom.

  • If you have an sp carbon atom with one single bond and a triple bond attached to it, the shape around that carbon atom will be linear, with bond angles of 180°.

Go to the Section 13 Menu . . .

To return to the list of learning outcomes in Section 13

Go to the CIE Main Menu . . .

To return to the list of all the CIE sections

Go to Chemguide Main Menu . . .

This will take you to the main part of Chemguide.

© Jim Clark 2020