# Total magnetic moment of atom

Whenever I read about coordination compounds in my textbooks, I always find a discussion about spin-only magnetic moment which is given by $\sqrt{n(n+2)}\cdot\mu_\mathrm{B}$, where $n$ is the number of unpaired electrons and $\mu_\mathrm{B}$ the Bohr magneton.

I'd like to know how to calculate the total magnetic moment of a given atom (spin only + orbital). For example let's take $\ce{Ni^2+}$. The spin only magnetic moment comes out to be $\sqrt{8}$ BM. Now how do I add the orbital magnetic moment? I've seen a bunch of expressions for orbital magnetic moment in terms on quantum numbers $n$ and $l$, but I am unable to understand which one to add.

## 1 Answer

Firstly I recommend you to understand the concept of "Magnetic Susceptibility". My source for this is Miessler,Tarr Inorganic Chemistry Book 2nd edition.

Consider Ni2+, its electronic structure is 4s03d8. This is a d8 ion. Then it has 8 electrons in d orbitals. d orbitals have 5 degenerate level of orbitals.

Also you should know gyromagnetic ratio which is the ratio of magnetic dipole moment to angular momentum of the system. (We got a system where quantum mechanics may be obeyed).

So, considering orbitals and electrons that fill these orbitals we got two magnetic moment here. Spin magnetic moment(sum of it is S) and angular magnetic moment(sum of it is L). I don't know properly write all calculations with codes. So I attached an image. 