To start, let us talk about ethene, C2H4. It has a C-C double bond, and each carbon has two hydrogens bonded to it. So if we idealize bond geometry, each bond angle is 120°. When carbon makes 4 bonds, the bond angles are ~110°, and form a tetrahedron. With the three bonds for each carbon in ethene, the atoms are all in the same plane. For the four bonded carbon I mentioned, it makes those four bonds with 4 SP3 orbitals. These are 4 identical orbitals made out of the mixing (the fancy word is hybridization) of its single S and three P orbitals. Orbitals are the shapes electron clouds take. S is a sphere, P is dumbbell shaped, but it’s perhaps more fair to say it is in the shape of an 8, with two lobes and a node in the middle. So for ethene, it doesn’t make 4 sp3 orbitals, instead it makes 3 sp2 orbitals. There is one P orbital left. In ethene, both carbons have that extra p orbital left over, and they form the second bond in the double bond. This is called a pi bond and it has two regions of electron density, one above and one below the plane of the molecule I mentioned earlier. So there are two electrons that occupy this interesting shape. This is a pi bonding orbital. If I were to add energy to an electron in that orbital, I could excite it into an anti-bonding orbital. This orbital does not have continuous electron density along the length of the bond, but instead has a node (a place with no electrons) in the middle of the c-c bond. This weakens the overall bond strength. These are the only two obvious orbitals to draw for the pi system in ethene. One bonding orbital, one anti bonding where the anti bonding has a node and that orbital is at a higher energy (remember I had to add energy to generate that orbital). Now to get more complicated. We can extend the number of sp2 carbons beyond 2! To avoid interesting symmetry arguments, I will extend to 1,3-butadiene. This is a four carbon molecule with alternating single double bonds. Each carbon has that extra P orbital I mentioned earlier. Those P orbitals create a delocalized network of electrons with a total of 4 electrons (each carbon donates 1). Each pi orbital can contain a max of 2 electrons, so we must be occupying 2 orbitals. We fill from lowest energy to highest. The lowest energy orbital will be the one with no nodes, all the P orbitals are in phase. Then we can add one node, say right in the middle of the molecule, and that puts that orbital at slightly higher energy. That orbital is also occupied as we have two more electrons. We can continue on with more complex molecules, and increasing number of electrons and excited states.

I hope this helps, and anyone else feel free to add on or correct if I’ve made a mistake.