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Find spin only magnetic moment of [Fe(H2O)5NO](2+)

  1. Feb 4, 2016 #1


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    1. The problem statement, all variables and given/known data
    Find spin only magnetic moment of ##[Fe(H2O)5NO]^{2+}##

    2. Relevant equations

    3. The attempt at a solution
    When I used Google, the magnetic moment in 3.87
    ##Fe## is in +1 oxidation state. So configuration is $$[Ar]3d^64s^1$$
    There are 5 unpaired electrons. How does magnetic moment become 3.87?
  2. jcsd
  3. Feb 5, 2016 #2
    the configuration readjusts to 3d7. And there are more problems: Fe is said to be in +3 and NO is in -1. Look it up in Google.
  4. Feb 5, 2016 #3


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    If thats the case, then there are 5 unpaired electrons. Magnetic moment won't be 3.87
  5. Feb 6, 2016 #4
    Yeah, I know. And it is a problem for me too, for my textbook says that it is Fe+ and NO+. But apparently that is 15 year old data. I'm not sure which one to follow, and as of now, I have no time to follow my research on the net up. I will look it up again and get back.
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