How does an ionic bond form?

When you have a piece of Na metal and Chlorine gas in a closed box, how would a Chlorine molecule react with an Na atom to form a NaCl ionic bond? And how would these NaCl ions form a crystal?

• Explain in details as possible.

• Answer:

  The reaction will proceed very easily if the sodium metal is first heated. It is a strongly exothermic reaction which will produce a bright flame and a lot of heat as the chlorine oxidizes the sodium. But of course, the reaction will not be between a Cl2 molecule and a single Na atom - first it will be necessary to dissociate the Cl2 gas into two separate atoms, since the formation of NaCl2 is not energetically favorable - it simply costs too much energy to form an Na 2+ ion. The bond dissociation enthalpy for chlorine is about 121 kJ/mol. 12Cl2→Cl12Cl2→Cl\frac{1}{2} Cl^2  \rightarrow Cl It will also cost energy to cause the sodium metal to sublimate into the gas phase. This amounts to 107 kJ/mol. Then the gaseous Na atoms must be ionized. Na→Na++e−Na→Na++e−Na  \rightarrow Na^+ + e^- The first ionization energy of sodium is 496 kJ/mol. Then the chlorine atoms accept the electrons, forming an ion, which releases an energy of -349 kJ/mol, corresponding to the first electron affinity of chlorine. Cl+e−→Cl−Cl+e−→Cl−Cl + e^-  \rightarrow Cl^- Finally the Na+ and the Cl- ions can condense out of the gas phase into an ionically bonded crystal, releasing a lattice energy of -787 kJ/mol. So overall the reaction will release about -411 kJ/mol. In details the actual reaction kinetics will of course be more complicated, but this gives the basic idea. It is spectacular to watch - it can be done in a transparent vessel.