Hi, This isn't really homework, but I'm trying to fully understand the reaction between sodium metal (Na) and water (H2O). Here's the equation: 2Na + 2H2O -> 2NaOH + H2 And here's the steps I think the process goes through: 1) Na has a single electron in its valence shell. This electron is quite a long way from the nucleus so it has a weak attraction to the Na atom. 2) H2O is a polar molecule with the O atom being slightly negative and the 2 H atoms being slightly positive. 3) I've read that when Na is placed in water it becomes a positive ion (Na+). I assume this is because it loses its valence electron, but why? What pulls the valence electron away? I guess its the slightly positive H atoms in the H2O molecule. If this is true, does this cause one of the H atoms in the H2O molecule to break free? If so, does the now free H atom have 0, 1 or 2 electrons? I'm thinking it has just the 1 electron leaving the OH molecule a negative ion (OH-). 4) Now at some point the Na+ ion bonds with OH to form NaOH. I'm thinking that when the H atom breaks free from the H2O molecule (point 3) it leaves OH-, which then bonds with the Na+ ion. When they bond you end up with a neutral NaOH molecule. 5) The H atom, which broke free in step 3, has a single electron and forms a bond with another H atom (H2). 6) The heat from the reaction is enough to ignite the hydrogen and oxygen mix that forms above the above. Does this sound correct or have I totally missed the point? Thanks for any guidance.