Is there any way of comparing the magnetic moments of $\ce{Co^2+}$ and $\ce{Cr^3+}$ gas phase ions by just going through their electronic configuration?Or is it just experimental data?

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    $\begingroup$ What are the complexes you are considering? Are the ions in water, or are they chelated or something else? You can make a good guess of the magnetic moment but they are not always right. $\endgroup$
    – S R Maiti
    Commented Jun 12, 2021 at 18:36
  • $\begingroup$ Both of them are in gaseous form this is the only information provided. $\endgroup$ Commented Jun 12, 2021 at 20:10
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    $\begingroup$ For spin only magnetic moment, you can just do $u_{spin} = \sqrt{n(n+2)}$ where n is the number of unpaired electron. Experimental data will differ though because this is based on spin only I believe. $\endgroup$
    – M.L
    Commented Jun 12, 2021 at 22:18
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    $\begingroup$ If they are in gaseous form, then yes, the use the formula in the above comment, and take n as the total number of unpaired electrons. $\endgroup$
    – S R Maiti
    Commented Jun 13, 2021 at 7:37
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    $\begingroup$ Hmm I think what I wrote before is actually wrong. Because in the gaseous state, without any ligand, the contribution of the orbital angular momentum (l) would not be cancelled, so you cannot use the spin only formula $\endgroup$
    – S R Maiti
    Commented Jun 15, 2021 at 13:28

1 Answer 1


I think the implicit question here is:

Given the same S and L, what is the value of J for less than half-filled vs more than half-filled magnetic ions?

The answer is Hund rule #3:

For atoms with less than half-filled shells, the level with the lowest value of J lies lowest in energy. When the shell is more than half full, the opposite rule holds (highest J lies lowest in energy).

So, for Cr$^{3+}$ the following applies:

  • S(Cr$^{3+}$)=3/2
  • L(Cr$^{3+}$)=3
  • J(Cr$^{3+}$)=L-S (since less than half filled, per Hund's rule #3)=3-3/2=3/2

Whereas for Co$^{2+}$ the following applies:

  • S(Co$^{2+}$)=3/2
  • L(Co$^{2+}$)=3
  • J(Co$^{2+}$)=L+S (since more than half filled, per Hund's rule #3)=3+3/2=9/2

To calculate the magnetic moment one would also need to calculate $g_J$. Higher values of $J$ result in smaller values of $g_J$, like in the case of lanthanides, but overall larger $J$ still result in higher magnetic moments. To visualize this general principle one can check the case of lanthanides: Lanthanide magnetism table, from https://www.radiochemistry.org/periodictable/la_series/L8.html

So, to explicitly answer the question, while these are just approximated models and the experiment is always preferrable, yes, just by looking at the electronic configurations and applying Hund's rules you can have a good deal of information about the relative magnetic moment of these two ions.

  • $\begingroup$ In the equation for the magnetic moment, don't you have a Lande g-factor in front of $\sqrt{J(J+1)}$ ? As the g factor contains a term with J(J+1) at the numerator and 2J(J+1) for the denominator, so for higher J values the g-factor should be lower (?). So without doing the full calculations, is it really possible to say that the higher J means higher magnetic moment? $\endgroup$
    – S R Maiti
    Commented Jun 21, 2021 at 19:15
  • $\begingroup$ As an example, here are the g·(sqrt(J·(J+1)) values for lanthanides. It is fairly clear that the less than half-filled have overall smaller values: radiochemistry.org/periodictable/la_series/L8.html $\endgroup$ Commented Jun 21, 2021 at 20:12
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    $\begingroup$ Is the Lande formula applicable to d-block as well? $\endgroup$
    – S R Maiti
    Commented Jun 21, 2021 at 20:14

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