There are three important pieces of information you will need to complete the Electron Configuration Worksheet from yesterday’s lesson.
Lets do the Electron Configuration for 57Ce. It would look like this:
Because ‘La‘ comes before ‘Ce‘, we have to add the 5d1 before we start counting the 4f orbital.
Now if we did the Electron Configuration for 72Hf, which is next to La, we have to go entirely through the 4f orbital before we go back to the 5d.
BUT! When we do this, it would not make any sense to write 5d1 4f14 5d2…
SO… we combine the 5d’s and place them after the 4f
Simple… if it’s a (-) then you add electrons to the final orbital.
9F is 1s2 2s2 2p5
9F-1 is 1s2 2s2 2p6
If its a (+) then you subtract electrons from the final orbital(s).
Na is 1s2 2s2 2p6 3s1
Na+ is 1s2 2s2 2p6 The 3s orbital has been lost!
This is the shortcut so you do not have to write out those huge-mungous (yes that is a word) electron configurations.
Lets do the electron configuration for Cl.
17Cl 1s2 2s2 2p6 3s2 3p5
The shorthand version uses the MOST RECENTLY PASSED NOBEL GAS (so when you pass gas use the shortcut… get it… c’mon that’s funny!) So for Cl, the most recently passed gas is Ne. So instead of starting at our ‘House’ (H), we’ll start at Ne and then count the orbitals from there.
17Cl [Ne] 3s2 3p5
Excellent sir so cool idea