Metals tend to lose electrons to obtain the stable noble gas configuration of 8 valence electrons.

Why do they want to obtain this configuration, and how does the strength of their "desire" to obtain this configuration compare with the "desire" to maintain neutral charge. If the answer depends on the chemical, I'm happy for you to provide some examples.


  • $\begingroup$ The key point here is that overall, in a compound electrical neutrality is maintained, and you have additional stability added to the mix because of the bond. I may write up an answer on this later :) $\endgroup$ – ManishEarth Feb 11 '13 at 14:31

Firstly, atoms "want" to achieve the noble gas configuration of 8 valence electrons because it is the most stable form. All that means is that it doesn't tend to react under normal conditions that we experience on Earth, therefore it will stay in that configuration for quite a while and are less likely to react. There is a more complex quantum physical answer for that but you'll have to go elsewhere for than.

The main force that keeps electrons in atoms is the electrical attraction between the electrons and the protons in the nucleus and so, if it is more energetically favourable to lose that electron in order to form a bond, then that is what will happen.

Focusing on the Alkali metals as an example, as you move down the group, they get more and more reactive. This is because of two main reasons that are a result of the electrons being further away from the nucleus:

  1. Because they're further away, the attraction between the protons and the outer most electron is less
  2. Secondly, taking Rubidium as an example, it has 37 electrons and 37 protons. From the perspective of the outer-most electron, there are 36 electrons repelling it, and 37 protons attracting it, therefore acting as a net charge of 1. However, if you take into account the first point, the repulsion of the closer electrons is stronger than the attraction of the protons so it could even be less than one
  • $\begingroup$ Do you know which specific energies are involved? Bond formation energies, activation energies etc.? $\endgroup$ – Brian Feb 11 '13 at 18:49
  • $\begingroup$ I have a vague conceptual knowledge, yes, but I'm still on my second year of my GCSEs $\endgroup$ – Kian Feb 11 '13 at 19:05

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