Energy in chemical reactions My question concerns how energy is used in chemical reactions. I am working with a reaction between magnesium and hydrochloric acid in a class (although this is not a homework question). What I've found is that the magnesium gives away its valence electrons to either the hydrogen or to the chloride. But in the second case, the chloride ions would first give their valence electrons to the hydrogen, and then they would accept the electrons from a magnesium atom. As I was thinking about which of the cases was more probable, I started thinking about the energy involved. For example, I would think that moving the electrons first from the chloride ions to the hydrogen, and then from the magnesium to the chloride, would use much more energy than the magnesium just giving its electrons to the hydrogen ions. 

My question is, as a general rule of thumb, although I wouldn't mind a detailed answer, are reactions that, like this one (as far as I know), have multiple "ways" of reacting going to take the way that uses the minimal amount of energy?  Or are the electrons free to move to whichever atom is closest?
Magnesium cannot give one or two electrons to the chloride ion. Because Chloride ions have the maximum number of 8 electrons in their external shells. There is no possibility for a new electron to stay on a Chloride ion. And the Chloride ion will be reluctant to accept or to loose one of its electrons, because it has the ideal conformation (8 outer electrons). So there are no "multiple ways" of reacting, as you imagine. The only thing that may happen is for magnesium to loose two electrons per atom, because at the end of this departure, it will have 8 outer electrons, and this seems to be to ultimate goal for all atoms. These electrons will be caught by $H^+$ ions, which are transformed into $H$ atoms and then into $H_2$ molecules.