Free radicals are usually very reactive. If we bring two free radicals or a free radical and a molecule with electrons available for bonding together a bond will usually form. Yes, it does cost energy to bring those electrons close together. But look at the payoff, a new chemical bond is formed and this is usually quite exothermic (stabilizing), usually more than enough energy to offset the energy cost of bringing those electrons close enough to form a new bond. Because of the exothermicity of bond formation the reaction profile usually has a small activation energy. Translation: a small activation energy means a fast reaction. A caveat, at the outset I said that "free radicals are usually reactive". There are many known free radicals that are quite stable. This is usually due to steric factors (the reactive free radical center is surrounded by bulky groups and just not sterically accessible for reaction) or electronic factors (some free radicals exist in very large delocalized systems, hence the spin density at any one atom in the system is so small that reaction is unlikely). One further point, most of what I've said relates to free radicals, which was the title of your question, however you did mention "unpaired electrons" in the beginning of your post. Molecules with unpaired electrons (note the plural, electrons) are different than free radicals with (usually) just one unpaired electron. Some molecules like $\ce{O2}$ are quite stable even though they have two unpaired electrons.
• Well, you can draw a Lewis dot structure for $\ce{O2}$ that has two unpaired electrons, it just doesn't have an octet around each oxygen. Sometimes MO diagrams are a poor approximation of reality, but they are handy and quick. Same can be said for Lewis dot structures, and they're an even more primitive approximation methodology. It is important to know where these various approximations break down. So examples of failures ($\ce{O2}$ with Lewis dot structures) can be very helpful in this regard. – ron Jun 10 '14 at 23:53