If you stick to atoms, and use the atomic orbital approximation, it's easy to work out the electronic spin by working out what atomic orbitals are populated. The atomic orbitals are populated in a slightly strange order that's given in http://en.wikipedia.org/wiki/Electron_configuration#Aufbau_principle_and_Madelung_rule. You have to remember that the p orbitals are really 3 orbitals of equal energy, the d orbitals are 5 orbitals of equal energy and if you get as high as the f orbitals there are 7 of these.
So hydrogen with 1 electron is 1s$^1$ and has spin half. helium is 1s$^2$ and the spins pair so you get spin zero. Likewise Lithium is 1s$^2$2s$^1$ with spin half and Beryllium is 1s$^2$2s$^2$ with spin zero. But take Carbon, which is 1s$^2$2s$^2$2p$^2$. It looks as if the spin should be zero because you have an even number of electrons. However the two p electrons will not be in the same p orbital because there are three p orbitals with equal energy. Because the p electrons are in different orbitals their spins are not necessarily paired, so the spin could be zero or it could be one. As a general guide Hund's rule tells you the higher spin is favoured, so the electronic spin of the carbon would be one (actually I can't remember if this is true or not).
Neon is 1s$^2$2s$^2$2p$^6$ and has spin zero because there are two electrons in each of the 1s, 2s and all three 2p orbitals, so the spins are all paired.
For molecules you proceed in a similar manner but you have to fill the molecular orbitals. It's rarer to get degenerate molecular orbitals since molecules don't have the spherical symmetry of atoms.
I'm not sure why Boris is telling you to ignore nuclear spin. True, it doesn't play any role in chemistry, but it is important in the spectra of atoms. In fact interactions between the electronic and nuclear spin are responsible for the Hydrogen 21cm radiation, and there's a lot of that in the universe!