The 'spin only' magnetic moment $\mu$ (in Bohr magnetons) of a transition metal ion with $n$ unpaired d-electrons is given by:
- $\mu = n$
- $\mu = \sqrt{n}$
- $\mu = \sqrt{n(n+2)}$
- $\mu = n(n+1)$
Show answer and solution
Correct answer: C. $\mu = \sqrt{n(n+2)}$
The spin-only formula is $\mu = \sqrt{n(n + 2)}\,\text{BM}$, where $n$ is the number of unpaired electrons.
It comes from the spin angular momentum quantum number $S = n/2$ and the relation $\mu = 2\sqrt{S(S+1)}\,\text{BM} = \sqrt{n(n+2)}$.
The actual magnetic moment can differ slightly due to orbital contribution
this is small for first-row transition ions in octahedral fields.