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? $\endgroup$– Asker123Commented Mar 29, 2015 at 16:56
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$\begingroup$ Related although not a direct duplicate. chemistry.stackexchange.com/questions/25003/… $\endgroup$– bonCommented Mar 29, 2015 at 16:56
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1 Answer
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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
answered Mar 29, 2015 at 21:29