How many unpaired electrons are in a transition metal complex that has a spin-only magnetic moment of #sqrt(15)#?

I got #3# unpaired electrons.

The spin-only magnetic moment is written as:

#\mathbf(mu_S = 2.00023sqrt(S(S + 1)))#

where:

• #g = 2.00023# is the gyromagnetic ratio.
• #S# is the total spin of all electrons in the atom. Paired electrons contribute #0# to #S# because #+1/2 + (-1/2) = 0#.

Hence, solving for #S# allows us to determine the number of unpaired electrons.

There would also be the contribution by the orbital magnetic moment, #mu_L#, but to find the unpaired electrons in the ion of a relatively light transition metal (i.e. first/second row transition metals), it's sufficiently accurate to use #mu_S#.

For example, for #"Fe"^(3+)#, a #d^5# metal, the calculated #mu_S = 2.00023sqrt(5/2(5/2 + 1)) = 5.92#, whereas the observed (including spin and orbital magnetic moments) #mu_(S+L) ~~ 5.9#, which is pretty excellent agreement.

What we have is #mu_S ~~ sqrt15# bohr magnetons. Therefore:

#sqrt15 = 2.00023sqrt(S(S + 1))#

#15 = 2.00023^2(S(S + 1))#

#15/2.00023^2 = S^2 + S#

#0 = S^2 + S - 15/2.00023^2#

Solve the quadratic formula to get:

#color(blue)(S ~~ 1.5 => 3/2)#

Therefore, the total spin #S# is #3/2#, the number of unpaired electrons is #\mathbf(3)#, and the atomic ion could be either a #d^3# or #d^7# configuration, as you would see here:

#d^3#:

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#d^7#:

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