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Formal charge on these two atoms

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data

    find formal charge on P in PBr3 and N in NBrI2

    2. Relevant equations

    Formal charge = number of valence electrons-number of lone pairs- 1/2 number of bonding electrons

    3. The attempt at a solution

    FC on P = 5 - 1 - 3 = 1
    FC on N = 5 - 1 - 3 = 1

    these are wrong for some reason and I can't figure out. I know I have the lewis diagram correct for these molecules(I checked)
     
  2. jcsd
  3. Dec 9, 2012 #2

    chemisttree

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    How does the Lewis diagram help you here? I always calculated what I knew first and then balanced the molecule's charge by assigning the oxidation number of the unknown. In these examples, I would have assumed the halogens were -1 as you have and the P or N would have an equal but opposite charge.
     
  4. Dec 10, 2012 #3

    symbolipoint

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    NBrI2

    Halides usually have charge of -1. The compound given is as neutral charge.
    N, u
    Br, -1
    I, -1
    I, -1
    Total, 0.

    Find u.
     
  5. Dec 21, 2012 #4
    The formal charges in both the cases is zero.
    In your above formula for FC you wrote the second term as " number of electron pairs ",it should be " number of non-bonded electrons ".
     
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