Metals contain free moving electrons, in both the solid and liquid state. Does this happen when it is in vapor form?


1 Answer 1


Interesting question. There are free electrons and ions in a metal's gas phase, but even at rather high temperatures, they are relatively rare. The formation of free electrons is given by the reaction:

$$\ce {M_{(g)} → M^+_{(g)} + e^{-}_{(g)}}$$

This is an endothermic reaction for all atoms and molecules, and the energy involved receives a special name: the (first) ionization energy. Comparing a chart of ionization energies for the elements, one finds that almost all elements require over $ 500\ kJ/mol$ (or about $5\ eV/atom$) to ionize. Even at $5750\ K$, the temperature of the surface of the Sun, the average thermal energy of a particle is about only about $0.5\ eV$.

Also, a large part of the electrons that did manage to free themselves from their parent neutral atoms would actually be consumed by other neutral atoms, in a different reaction:

$$\ce {M_{(g)} + e^{-}_{(g)} → M^{-}_{(g)}}$$

This reaction is slightly exothermic for most elements and molecules, and the energy associated with it also receives a special name: the (first) electron affinity.

Calculating the equilibrium concentrations of each species at a given temperature is possible but is a bit of work. However, for most metals, you can expect to find significantly less than one free electron per ten thousand neutral atoms at $5750\ K$, and the relative amount of free electrons decreases exponentially with temperature.

  • $\begingroup$ You are confusing de-localisation with ionisation. Its significant- its present in traditional fluorescent tubes where mercury is vaporised and shortcircuits to give arcs which produce UV- in turn producing fluorescence. $\endgroup$ Dec 9, 2013 at 6:11
  • $\begingroup$ @user2617804 I'm considering a thermal plasma in equilibrium, with no applied voltage. Also I'm not sure what you mean about delocalisation. $\endgroup$ Dec 9, 2013 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.