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I have a little question that are confusing me at bit. I have to argue, from the Pauli Exclusion Principle, that $M_{s} = 2$ is the maximum $M_{s}$ quantum number for the $nd^{6}$ configuration.

Now, from what I can find, $M_{s}$ is given by:

$M_{s} = \sum\nolimits_{i} m_{s_{i}}$

And with 6 valence electrons, that can have $m_{s} = \pm \frac{1}{2}$, I can't seem to see how I end up with a maximum of 2 ?

So anyone who could give me a little hint ?

Thanks in advance.

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up vote 1 down vote accepted

The $d$ orbitals are a set of five orbitals, each of which can hold two electrons. So if you distribute six electrons between the five $d$ orbitals you will get one pair of electrons and four unpaired electrons. The maximum spin possible is if the four unpaired electrons all have the same spin, in which case the total spin is $4 \times 1/2 = 2$.

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