Dear researchers,
Can I suggest that if you just look at the relationship between the numbers 2, 8, 18 and 32. ...and its expansion. You will find a progression pattern between them, that will be the START to the solution. Its not very complicated at all, easy enough for any of us to understand.
Actually, the non-mathematic/simplistic explanation is as follows:
Electrons in an atom have to be in an orbital, which is a mathematically-defined volume of space. Each orbital is described by three quantum numbers- n (which determines the size of the orbital, and the number of nodes = areas which have no electron density), l (which determines the shape of the orbital, and the number of planar nodes), and m
l (which determines where in the x,y,z directions the orbital is pointing). No two orbitals can have the same set of three quantum numbers; no two orbitals can be identical, and each orbital can hold two electrons.
In the first shell, n = 1, and there are no nodes (how does it smell? Terrible!), so there is no planar nodes, and no variation in shape or direction. There can only be one orbital in the first shell. It is spherical and is called the 1s orbital (s doesn't actually stand for sphere, but it's an easy way to remember it). Therefore the first shell can have two and only two electrons.
In the second shell, n=2, and there is one node. This could be a radial node, which makes another spherical orbital (the 2s orbital), which has no variation in direction. We could also have a planar node, and this plane could lie in the xy plane, the xz plane, or the yz plane. We have three orbitals differing in their orientation in space, so we have the 2px, the 2py, and the 2pz orbitals. Grand total of 4 orbitals, so eight electrons.
Why do the 2p orbitals appear as dumbell shapes? If you take a spherical balloon and twist it so that there's no air in the centre, you now have the basic shape of something with a planar node. A dumbell.
In the third shell, n = 3, so we have two nodes.
Two radial nodes gives us the 3s orbital.
One radial and one planar gives us three 3p orbitals.
Two planar nodes give us the d orbitals. The shape of most of these is given by taking your balloon, twisting it in the centre to give a planar node, and then twisting it again along another axis (you can do it if your balloon isn't inflated very much)- you get a cloverleaf shape. The odd one out (the donut + dumbell shape) is made by taking the two nodal planes and mathematically adding them to give a nodal double-cone. The odd shape is simply what's left of a spheroid when you subtract a double cone from it.
Why are there 5 d orbitals in a shell? It's because of the possible values of m
l. l can be anywhere from zero to one less than n; m
l can be from -l to +1 (integer values only). So if n = 3, l can be 0 (3s), 1 (3p), or 2 (3d). When l = 2, m
l can be -2, -1, 0, +1, +2. Five possible values = 5 solutions to the Schroedinger Eq'n = 5 orbitals.
So the third shell has 1 + 3 + 5 = 9 orbitals, and thus can hold 18 electrons.
The fourth shell has one 4s orbital, three 4p orbitals, five 4d orbitals, and seven of the next set, the 4f orbitals (which look like you've twisted any of the d orbitals to give another planar node in the balloon). 1 + 3 + 5 + 7 = 16, so 32 electrons can fit in the fourth shell.
The number of orbitals in a shell is always equal to n^2, so the number of electrons that can fit in the shell is always twice a perfect square, which is the amazing pattern you were pointing out earlier.
If you want to see the shapes of orbitals, I recommend the Orbitron Gallery:
http://winter.group.shef.ac.uk/orbitron/index.html
Any more chemistry-related numerology?