# 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.