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

Question

Mateo Aguilar · Accepted Answer

I got #3# unpaired electrons.

Written as follows is the spin-only magnetic moment:

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

where

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#

To obtain, solve the quadratic formula:

#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#:

#ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul( color(white)(uarrdarr)) " " ul( color(white)(uarr darr))#

#d^7#:

#ul(uarr darr) " " ul(uarr darr) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))#

Kai Bowman · Answer

undefined The number of unpaired electrons in a transition metal complex with a spin-only magnetic moment of √(15) is 3.