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Can we picture metallic bonding as an equilibrium between electrons and cations?

Suppose:

$$\ce{Al^3+ + 3e- <=> Al}$$

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  • $\begingroup$ I mean isn't $\ce{Al^3+}$ stable, why would it need the extra electrons since its outer shell is stable? $\endgroup$
    – Asker123
    Mar 29, 2015 at 16:56
  • $\begingroup$ Related although not a direct duplicate. chemistry.stackexchange.com/questions/25003/… $\endgroup$
    – bon
    Mar 29, 2015 at 16:56

1 Answer 1

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In metals, electrons are non-localized, forming a "sea" of electrons, rather than having them localized, as in the $\ce{Na+Cl-}$ lattice of crystalline salt. See Metallic bonding for a more complete description.

It is, of course, a matter of degree, as covalent, ionic and metallic bonding can "blend" from one to the other. A bond can be considered partially ionic and covalent, for example; see these helpful graphics

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