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How can some chemicals, if they have an equilimbrium of electrons, take away electrons from other chemicals? One example I believe is placing a small amount of gallium on top some alluminum and watching the alluminum melt. Why does the gallium, if it is at equilibrium state, need more electrons?

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One example I believe is placing a small amount of gallium on top some alluminum and watching the alluminum melt. Why does the gallium, if it is at equilibrium state, need more electrons?

No, gallium dissolving aluminium is not an example of electron exchange but of formation of an alloy.

Exchange of electrons between atoms in chemical reactions can roughly be explained by the Octet Rule. Atoms with less than 8 electrons in their outer electron shell (so called valence electrons) tend to shed or acquire electrons to make up a total of 8 (an octet), which happens to be a very stable configuration.

An example is the reaction involving aluminium foil and iodine crystals, which starts vigorously even at room temperature.

In it, $\mathrm{Al}$ loses three valence electrons, becoming an $\mathrm{Al^{3+}}$ cation, with the electron configuration of $\mathrm{Ne}$ which has a full octet.

The iodine atom $\mathrm{I}$ on the other hand acquires one electron, becoming an iodide $\mathrm{I^{-}}$ anion, with the electron configuration of $\mathrm{Xe}$ which also has a full octet.

The overall reaction is thus:

$$2\mathrm{Al}(s)+3 \mathrm{I_2}(s)\to 2\mathrm{AlI_3}(s)$$ Elements that tend to acquire electrons are called electronegative and are mostly found on the right hand side of the Periodic Table. Elements that tend to lose electrons are called electropostive and are mostly found on the left hand side of the Periodic Table.

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  • $\begingroup$ Gert, about octet and why there are sp-hybridisations introduced in chemistry I get an unusual thought About the distribution of electrons magnetic dipole moments in atom and the answer to symmetry in atomic orbitals with my comment. Could you please show why it should be wrong. $\endgroup$ – HolgerFiedler Nov 13 '16 at 6:29
  • $\begingroup$ Hybridisations serve to minimise electrostatic repulsion between MOs. I explain it well here: sciencemadness.org/talk/… and here: sciencemadness.org/talk/… $\endgroup$ – Gert Nov 13 '16 at 15:19
  • $\begingroup$ That is how we have to see it if we suppose that there is a s-orbital. It comes from the spherical harmonics we have choosen under the influence of our Cartesian coordinate thinking. See my comment under my answer from the second link. There are spherical harmonics through which you can describe the octet as 8 electrons, all indistinguishable except that 4 have spin down (magnetic dipole moments in one direction) and the other 4 have spin up (magnetic dipole moments in the opposite direction). Try something new. $\endgroup$ – HolgerFiedler Nov 13 '16 at 17:59
  • $\begingroup$ The SE for atoms isn't solved in Cartesian coordinates. The octet in $CH_4$ (e.g.) also describes it as "8 electrons, all indistinguishable except[...]" Try something new? If it ain't broke, don't fix it! $\endgroup$ – Gert Nov 13 '16 at 19:33
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There is no electron exchange when you put gallium on top of aluminum. The known observed reaction is that aluminum diffuses into the gallium because it has a very high solubility there.

The tendency of an atom/molecule to take electrons away from others is related to the concept of electronegativity. See Electronegativity. In the end it is due to the fact that the total energy of the reactants is lower after the electron transfer.

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