Results 1 to 2 of 2

Thread: Metallic bonding

  1. #1 Metallic bonding 
    Moderator Moderator AlexP's Avatar
    Join Date
    Jul 2006
    Are the electrons in a sample of metal not even in their orbitals...? I looked it up and it sounded as if they're not, but that just doesn't seem right to me... What happens when some sample of pure metal is exposed to some pure atomic nonmetal, does the nonmetal take valence electrons or does it just take any? Some clarification on this subject would be nice, as its obvious that i'm a bit confused.

    Reply With Quote  


  3. #2  
    Forum Freshman César's Avatar
    Join Date
    Sep 2006
    Spain, Europe
    The main concept to grasp here is , from my point of view, delocalized electron. Have you studied organic chemistry and aromatic rings? In this rings (for example benzene) you have six carbons forming the ring with a double bond every other carbon; as these double bonds can occur between any two adjacent carbons, from a quantum mechanics point of view, this doble bonds are shared by all the atoms, id est, you have an overall pi orbital with six electrons. This is the reason why benzene is usually represented by an hexagon with a circle in it.

    The same applies to metals but in this case we are talking about orbitals d and s. You have a mass of metallic ions sourrounded by a sea of electrons.

    Another key concept is that of resonance. I recommend you to read that entry on Wikipedia.

    Regarding your last question, keep in mind that all chemical processes huge amonts of atoms are implied everytime we are talking about any reaction, therefore 2. law of large numbers (see Wikipedia) apply.

    Best regards,


    Reply With Quote  

Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts