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Question on bonding and formal charge

  1. Nov 9, 2012 #1
    Hello! I was wondering why, when calculating a formal charge, you actually use that formula of (valence electrons in ground state) - lone electrons after bonding - 1/2(bonded electrons)

    Lets take H2O.

    H - O - H


    The O has 6-4-2 = 0 formal charge.
    However, the O originally had 6 electrons, and now it has 8, so shouldn't that mean it has a charge of -2 because it has 2 more electrons than before. Also, when that happens, doesn't that mean the number of electrons doesn't match with the number of protons, meaning the atom is now unstable? I would appreciate it if anyone could help to clear this doubt.
     
  2. jcsd
  3. Nov 10, 2012 #2

    AGNuke

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    Gold Member

    That is the "Formal" Charge, not an actual charge. Oxygen has 6 electrons, 2 of which them is "shared" with other atoms. They do not fully belongs to O-atom and the other 2 electrons also are shared, so it is not like that 6-proton 8-electron system.
     
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