Question

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

Answer 1

I got `3` unpaired electrons.

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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`:

`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))`