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4.14.1: example magnetic moment data and their interpretation

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    36221
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    Summary of applicable formulae

    1) Spin-Only magnetic moment
    μs.o. = √{4S(S+1)} B.M.

    2) For A and E ground terms
    μeff = μs.o. (1-α λ /Δ) B.M.
    Do not expect Temperature dependence.

    3) For T ground terms with orbital angular momentum contribution
    μS+L = √{4S(S+1) + L(L+1)} B.M.
    T terms generally show marked Temperature dependence.

    The examples that follow are arranged showing the experimentally observed values, the theoretical "spin-only" value and possible variations expected.

    A number of the examples involve "alums" where the central Transition Metal ion can be considered to be octahedrally coordinated by water molecules.


    d1

    VCl4
    V(IV) tetrahedral

          80K       300K             μs.o. /B.M.
          1.6       1.6                   1.73             
    

    2E ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For less than a half-filled d shell, the sign of λ is positive so the effect on μ should be that μeff < μs.o.

    VCl62-
    V(IV) octahedral

          80K       300K             μs.o. /B.M.
          1.4       1.8                   1.73             
    

    2T2g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
    Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
    For a full S+L contribution this would give μS+L = 3 B.M. which is clearly much higher than the 1.8 B.M. found at 300K. So, μs.o. < μobs < μS+L
    showing that the magnetic moment is partially quenched.

    d2

    V3+ in (NH4)V(SO4)2.12H2O (an alum)
    V(III) octahedral

          80K       300K             μs.o. /B.M.
          2.7       2.7                   2.83             
    

    3T1g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
    Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
    For a full S+L contribution this would give μS+L = √(20) = 4.47 B.M. which is clearly much higher than the 2.7 B.M. found at 300K.
    So, μobs < μs.o. < μS+L
    showing that the magnetic moment is significantly quenched.
    In this case, there is no observed Temperature variation between 80 and 300K and it may require much lower temperatures to see the effect?

    d3

    Cr3+ in KCr(SO4)2.12H2O (an alum)
    Cr(III) octahedral

          80K       300K             μs.o. /B.M.
          3.8       3.8                   3.87             
    

    4A2g ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For less than a half-filled d shell the sign of λ is positive so the effect on μ should be that μeff < μs.o.

    d4

    CrSO4.6H2O
    Cr(II) octahedral

          80K       300K             μs.o. /B.M.
          4.8       4.8                   4.9             
    

    5Eg ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For less than a half-filled d shell the sign of λ is positive so the effect on μ should be that μeff < μs.o.

    K3Mn(CN)6
    Mn(III) low-spin octahedral

          80K       300K             μs.o. /B.M.
          3.1       3.2                   2.83             
    

    3T1g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
    Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
    For a full S+L contribution this would give μS+L = √(20) = 4.47 B.M. which is clearly much higher than the 3.2 B.M. found at 300K.
    So, μs.o. < μobs < μS+L
    showing that the magnetic moment is partially quenched.
    In this case, there is a small Temperature variation observed between 80 and 300K.

    d5

    K2Mn(SO4)2.6H2O (an alum)
    Mn(II) high-spin octahedral

          80K       300K             μs.o. /B.M.
          5.9       5.9                   5.92             
    

    6A1g ground term - hence do not expect Temperature dependence and L=0 so no orbital contribution possible.
    Expect μeff = μs.o.

    K3Fe(CN)6
    Fe(III) low-spin octahedral

          80K       300K             μs.o. /B.M.
          2.2       2.4                   1.73             
    

    2T2g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
    Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
    For a full S+L contribution this would give μS+L = √(9) = 3 B.M. which is clearly much higher than the 2.4 B.M. found at 300K.
    So, μs.o. < μobs < μS+L
    showing that the magnetic moment is partially quenched.

    d6

    Fe2+ in (NH4)2Fe(SO4)2.6H2O (an alum)
    Fe(II) high-spin octahedral

          80K       300K             μs.o. /B.M.
          5.4       5.5                   4.9             
    

    5T2g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
    Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
    For a full S+L contribution this would give μS+L = √(30) = 5.48 B.M. which is close to the 5.5 B.M. found at 300K.
    So, μs.o. < μobs ~ μS+L
    showing that the magnetic moment is not showing much quenching.

    d7

    Cs2CoCl4
    Co(II) tetrahedral

          80K       300K             μs.o. /B.M.
          4.5       4.6                   3.87             
    

    4A2 ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For more than a half-filled d shell the sign of λ is negative so the effect on μ should be that μeff > μs.o.
    The observed values are somewhat bigger than expected for the small (0.2 B.M.) variation due to equation 2) so other factors must be affecting the magnetic moment. These effects will not be covered in this course!

    Co2+ in (NH4)2Co(SO4)2.6H2O (an alum)
    Co(II) high-spin octahedral

          80K       300K             μs.o. /B.M.
          4.6       5.1                   3.88             
    

    4T1g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
    Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
    For a full S+L contribution this would give μS+L = √(27) = 5.2 B.M. which is close to the 5.1 B.M. found at 300K.
    So, μs.o. < μobs ~ μS+L
    showing that the magnetic moment is not showing much quenching.

    d8

    Ni2+ in (NH4)2Ni(SO4)2.6H2O (an alum)
    Ni(II) octahedral

          80K       300K             μs.o. /B.M.
          3.3       3.3                   2.83             
    

    3A2g ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For more than a half-filled d shell the sign of λ is negative so the effect on μ should be that μeff > μs.o.
    The observed values are somewhat bigger than expected for the small (0.2 B.M.) variation due to equation 2) so other factors must be affecting the magnetic moment. These effects will not be covered in this course!

    (Et4N)2NiCl4
    Ni(II) tetrahedral

          80K       300K             μs.o. /B.M.
          3.2       3.8                   2.83             
    

    3T2 ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
    Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
    For a full S+L contribution this would give μS+L = √(20) = 4.47 B.M. which is higher than the 3.8 B.M. found at 300K.
    So, μs.o. < μobs < μS+L
    showing that the magnetic moment is partially quenched.

    d9

    Cu2+ in (NH4)2Cu(SO4)2.6H2O (an alum)
    Cu(II) octahedral

          80K       300K             μs.o. /B.M.
          1.9       1.9                   1.73             
    

    2Eg ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For more than a half-filled d shell the sign of λ is negative so the effect on μ should be that μeff > μs.o.

    Contributors and Attributions


    4.14.1: example magnetic moment data and their interpretation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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