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WA.As-Salaamu 'Alaykum.
Someone please explain to me the Electron-Pair Repulsion Theory; I don't get it at all. How do we determine what the bond angles and shapes of molecules are? Jazak Allah.
Please give this a read, it helped me quite a bit. Also, let me try to explain with an example, CH4.
The central atom is C and it is surrounded by 4 hydrogens. First find the number of electrons being shared. In this case, it is 4 from carbon and 1 from each hydrogen, totaling up to 8 electrons. Now, each bond uses 2 electrons. There are 4 bonds, and therefore no lone pairs. So CH4 has 4 bonds and no lone pairs.
Another one: NH3. The central atom is N and it makes 2 bonds with hydrogen. The total number of electrons being shared are 5 from Nitrogen and 3 from the hydrogens. So total electrons = 8. Nitrogen makes 3 bonds (which use 6 electrons), so 2 electrons are left making 1 lone pair. So NH3 has 3 bonds and 1 lone pair.
When you have the bonds and lone pairs decided, learn this:
2 bonds, 0 lone pair = linear and 180 degrees (e.g. CO2)
2 bonds, 1 lone pair = angular and 117 degrees. (e.g. SO2)
2 bonds, 2 lone pair = angular and 104.5 degrees. (e.g. H2O)
3 bonds, 0 lone pair = trigonal planar and 120 degrees. (e.g. AlCl3)
3 bonds, 1 lone pair = trigonal pyramidal = 107 degrees. (e.g. NH3 as I did above)
4 bonds, 0 lone pair = tetrahedral = 109.5 degrees. (CH4 as I did above)
5 bonds, 0 lone pair = trigonal bipyramidal = 90 and 120 degrees. (PCl5)
6 bonds, 0 lone pair = octahedral and 90 degrees. (SF6)
The theory is basically that the electron pairs arrange themselves around the central atom to minimize the amount of repulson, so they try to be as far apart as possible. Also, you need to know that lone pair - lone pair repulsion > lone pair - bond repulsion > bond-bond repulsion. Lone pair repulsions are stronger because they are closer to the atom and therefore exert more pressure.