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Electron Confusion in Solids

  1. Aug 8, 2006 #1
    Since metals are neutral, and their electrons delocalized, is the number of electrons in a metallic solid equal to the number of protons in the solid? Also I am guessing that the valence shells of the metals are not complete?

    I am confused on this regard.

    Furthermore, I can see how the myriad molecular orbitals in metallic solids form a continuum for electrons to travel within throughout the solid, but how exactly do network solids such as silicon have conduction orbitals? Are the the MOs in silicon solid also as closely spaced as to form a continuous orbital for electrons to travel within? It just doesn't seem like the tetrahedral shape of a silicon bonds in the solid would provide MO's that are spaced closely enough to form the bands seen in metallic solids.
    Last edited: Aug 8, 2006
  2. jcsd
  3. Aug 9, 2006 #2


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    Perhaps a good explanation is:
    from http://www.schoolscience.co.uk/content/5/chemistry/steel/steelch1pg1.html [Broken]

    In a metal, and in most matter, the positive and negative charges are balanced, unless of course electrons are removed or added to the material. In a metal, the number of electrons equals the number of protons in order to maintain charge neutrality.

    Some commentary on Group 4 elements, which includes silicon, and there is a discussion on electrical conductivity.
    Last edited by a moderator: May 2, 2017
  4. Aug 9, 2006 #3


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    First of all, I moved this from the Chemistry forum because your questions are more relevant to solid state physics.

    If depends on how many electrons are "donated" to the conduction band per atom. Typically, each atom in the metal gives up an electron. If this is the case, then the number of electrons in the conduction band will equal to the number of atoms in the metal. This simplistic scenario, however, doesn't work all the time. Some metals might easily have a +2 valence, or even fractional valence such as +1.5, meaning half of the atoms give up 1 electron, while the other half give up 2 electrons. This gets even more complicated with you have other exotic metals.

    To be able to see this, you have to do rigorous band-structure calculations. You will need to account for the tight-binding structure, and see how far each local orbital overlap. This could include more than just the nearest neighbor. It is from such band structure calculation that you get the valence band, the band gap, and the conduction band for such semiconductor. You just don't see these things by looking at one, or a few molecular orbitals. It is why this is a many-body physics problem.

  5. Aug 9, 2006 #4


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    In exactly the same way.
    Yes (loosely speaking).
    Why not? How does one make a definitive judgement like that purely qualitatively? Tin - a metal - has the exact same bonding geometry.
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