Chemguide: Support for CIE A level Chemistry
Learning outcome 1.4
Learning outcomes 1.4.1 to 1.4.8
These statements deals with ionisation energy and the way ionisation energy varies across and down the Periodic Table. I am dealing with them together because the order in Chemguide isn't quite the same as the order in the syllabus.
Before you go on, you should find and read the statements in your copy of the syllabus.
First ionisation energy
You will find everything that you need to know on this page about first ionisation energy.
You will find this really difficult if you aren't already confident about electronic structures using s and p notation. Don't even think about looking at this page until you are happy about that.
Successive ionisation energies
You should then go on to the page about successive ionisation energies.
More about learning outcomes 1.4.7 and 1.4.8
These two statements look confusingly similar. Learning outcome 1.4.8 seems to me to be the easier one, so we will have a quick look at that first.
Learning outcome 1.4.8
In this case, you are likely to be given a graph or a set of data for the various ionisation energies of an element, and be asked which Group of the Periodic Table it is in. This is covered on the Chemguide page above, but here is another example.
The first eight ionisation energies of an element are:
1310 3390 5320 7450 11000 13300 71000 84100
Which Group of the Periodic Table is it in?
Look for the big jump. In this case, the values are increasing by a couple of thousand at a time until you get to the gap between the sixth and seventh values, where the jump is huge.
So, six electrons are relatively easy to remove, but the seventh is far more difficult. That is because it is being removed from an inner level.
The element is in Group 6 of the Periodic Table.
If you can get hold of a copy, there is a multi-choice question using a different example (this time with the values on a graph rather than as numbers) on November 2007 Paper 1 Q4. The answer is C.
Learning outcome 1.4.7
You will find a multi-choice question related to this statement on June 2009 Paper 1 Q3, which relates to the same data as above. The difference this time is that you are asked for the outer electronic configuration rather than the Group number.
Basically, all you need to do is to work out what Group the element is in, and then work out the arrangement of the outer electrons from that.
We already know that it is in Group 6. Group 6 elements will have the electronic structure ns2np4.
In the question, you were told that the element was between lithium and neon in the Periodic Table, and so the outer structure would be 2s22p4.
If you had to write down an answer instead of choosing one from a list, you might also have written it in a more detailed form as 2s22px22py12pz1.
© Jim Clark 2019