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Shapes of Molecules part 1
VSEPR (Valence Shell Electron Pair Repulsion Theory)
You may not have heard of the VSEPR abbreviation but it is worth knowing, as it is a good reminder of the theory behind this topic: valence shell = outer shell and therefore outer electrons. And everything is based on electrons repelling each other.
the reason any molecule adopts its’ shape is to minimise repulsion
Classic exam question….why does a molecule adopt a particular shape? The electrons position themselves as far apart as possible to minimize repulsion → the shapes shown below. The repulsion is between electrons in bonds (bonding pairs), between lone pairs or between lone pairs and bonding pairs.
Basic Shapes
I have listed the 7 basic shapes below plus their bond angles. You have to know these. You must know these….you just HAVE to know these! This is the starting point to the whole topic and makes life so much easier if you can recall them on demand. They use a lot of these examples in multiple choice questions, so you can gain some really cheap marks. How many times have they used NH 3!
✓ These are the common examples, of course there are others you could use instead. But I would stick
to these as they are straight out the marking schemes.
Linear: BeCl 2
Trigonal Planar: BCl 3 Tetrahedral: CH 4 Trigonal Pyramidal
(or pyramidal): NH 3
Bent or non-linear: H 2 O Trigonal Bipyramidal:^ PCl 5 Octahedral:^ SF^6
Click here for shapes part 2
✓ In most of examples we are trying to get 8 electrons around the central atom to satisfy the “octet”
rule. But look out for boron which only has 6 electrons around it (very unusual) and also larger elements like phosphorus which has 10 electrons around it or sulphur with 12.
Using the list
We can make the list a bit more user friendly. And this is will be the basis for working out shapes of molecules that are not on the list. The aim is to work out:
CH 4
We firstly look at what group the central atom is in (the atom in the middle) i.e. carbon …..group 4 → 4 outer electrons. This means there are 4 electrons on carbon available to form bonds.
✓ I am ignoring hydrogen. Assume that the atoms on the “outside” contribute one electron each to a
bond no matter what group it’s in.
✓ If you are unclear on any of this, just think of dot and cross diagrams. You take one electron from
carbon and one from hydrogen to form a bond. That’s all we are doing here. Now, how many bonds is carbon actually making? The formula tells you that! The H 4 part gives it away…it is making 4 bonds. Therefore all the 4 electrons are used up making the 4 bonds….and the important part…. there are no lone pairs as all the electrons from carbon are used up making the 4 bonds:
✓ This is what you are aiming to get to every single time you do these questions.
NH 3
Nitrogen is in group 5 → 5 outer electrons. We can see from the formula that there are only 3 bonds. So only 3 of the nitrogen electrons are being used in bonds, therefore two electrons must be left over …..and 2 electrons = 1 pair….so we have one lone pair. how many bonding pairs (bonds) and how many lone pairs there are in a molecule 4 bonding pairs and 0 lone pairs
H 2 S
Exactly the same as we did above….sulphur is group 6 → 6 outer electrons. We can see it is only making 2 bonds. Therefore we must have 4 electrons remaining → 2 lone pairs. Now look at the table above…which shape has 2 bonding pairs and 2 lone pairs? Bent…therefore H 2 S is the same shape as H 2 O. Easy! Reallllllyyy reaallllyyy easy.
PCl 3
Phosphorus is in group 5 → 5 outer electrons. It is making 3 bonds , therefore we must have one lone pair: So again, look at the table above…it’s pyramidal. It becomes really boring after a while. Same old shapes!
Reduction in bond angles
An extremely common question is to ask about the progression in bond angles from CH 4 → NH 3 → H 2 O…which is 109.5° → 107° → 104.5°.
✓ If you call it 109 or 109.5, it’s fine. Doesn’t have to be THAT accurate.
Why? It’s the introduction of lone pairs. The lone pairs cause more repulsion and forces the electrons in bonds to move further away from the lone pairs → reduction in bond angle. compare the number of bonding pairs and lone pairs to those in the table above → shape name 2 bonding pairs and 2 lone pairs 4 bonding pairs 0 lone pairs 3 bonding pairs 1 lone pair 2 bonding pairs 2 lone pairs 3 bonding pairs and 1 lone pair
Tetrahedral is the “big daddy” in this progression. 109.5° is the starting point. The NH 3 and H 2 O shapes are based on tetrahedral as the have “4 things” in total (see table further up the page). But the lone pairs as I mentioned cause this angle to reduce slightly. Alongside this, the name must change. One step further…CH 4 only has repulsion between electrons in bonds. NH 3 has the extra repulsion of the lone pair and the electrons in bonds. H 2 O has even more repulsion as the lone pairs repel each other as well. Lone pair:lone pair repulsion > lone pair:bonding pair repulsion > bonding pair:bonding pair repulsion This could apply to more complicated examples as well, which we will look at in part 2. It’s not always just the above progression. The names come from the bond angle. As soon as the angle changes, the name changes Click here for shapes part 2
