Shapes Of Molecules A Level Help

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Chowderzzz
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I’ve seen Eliot Rintouls video about how to derive the shape of a molecule from its molecular formula. However, this method only seems to work when the outer shell electrons and electrons being bonded are even when added. How do I go about this sort of question when it’s odd?
Here’s a link to the video: https://youtu.be/Tz4gzoqxvCQ . I’m talking about the part at 15:30
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Kian Stevens
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This is what's known as VSEPR theory, so you can have a read up on it if you want

The total number electrons will always be even, and this is why when dealing with ions, electrons must be added/subtracted
As we all know from school, atoms 'want' to have 8 valence electrons and so that's why they form a certain number of bonds (this is a massive simplification, but oh well)
This is unless you have a central atom such phosphorous which has the ability to expand its octet in higher oxidation states, but you don't need to know how/why this works, all I'm trying to say is that the number of electrons would still be even and VSEPR would still work regardless (I'll show an example later)
One major thing that the video didn't clarify is what bonds are being formed, because it's not clear enough to just say how many electrons a certain bond will contribute

So in the case of NH3, we know that three N-H \sigma-bonds are formed because hydrogen is only capable of forming these bonds, hence a total of three electrons will be contributed from these bonds
After doing the algorithm shown in the video, yes we end up having four electron pairs, but it's clearer to see what these are: three of the pairs will be \sigma-bonding pairs, whilst the last one will be a lone pair
This lone pair is very important, because one thing that the video only brushed on is the relative repulsion of different electron pairs: all electron pairs repel to be as far apart as possible, but lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion
Thus, the lone pair will repel all the bonding pairs down slightly to give the trigonal pyramidal shape as shown in NH3; however, no matter what molecule you have, VSEPR theory assumes that any molecule with 3 bonding pairs and 1 lone pair will exhibit this shape

It gets a bit trickier when you do have something like a phosphorous compound, because like I said, it can expand its octet and actually have more than 8 valence electrons; PCl5 is a good example to consider
So, just like nitrogen, phosphorous in its ground-state has 5 valence electrons; however, if you apply VSEPR theory to PCl5 you realise that the central phosphorous atom doesn't have 8 valence electrons, it actually has 10... But this is absolutely fine, because phosphorous is able to do this due to what I mentioned previously
Since 5 P-Cl \sigma-bonds are formed, and there are 5 electron pairs in total, this means that all the electron pairs are P-Cl \sigma-bonding pairs (with no lone pairs present)
Thus, PCl5 will exhibit what's known as a trigonal bipyramidal shape, and like I mentioned earlier, VSEPR will assume that any molecule with this configuration will exhibit this shape

You don't need to know/understand the phosphorous example because I doubt you'll encounter anything like it at A-level, but I was just trying to demonstrate how VSEPR works regardless of the number of valence electrons, which will always be even
Last edited by Kian Stevens; 1 year ago
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