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

##### 1 Answer

I got

The **spin-only magnetic moment** is written as:

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

#g = 2.00023# is thegyromagnetic ratio.#S# is thetotal spinof all electrons in the atom. Paired electrons contribute#0# to#S# because#+1/2 + (-1/2) = 0# .

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

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

For example, for *calculated* *observed* (including spin *and* orbital magnetic moments)

What we have is

#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**

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

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