Why is the spin magnetic moment for nickel(II) given as 1.7 Bohr magnetons in some references?

I read some theoretical articles concerning $\ce{Ni-O}$ system, where $\ce{Ni}$ sits at the center of the octahedron formed by $\ce{O}$. The calculated magnetic moment on the $\ce{Ni}$ sites is about $1.7$. And these articles said that this is the typical magnetic moment of $\mathrm d^8\ \ce{Ni^2+}$. I think according to $\mu=\sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons, $\mathrm d^8\ \ce{Ni^2+}$ should have 2 unpaired electrons and $\mu=2.8$.

• Can you link the article? My guess is that one has to think about something like the effective number of unpaired electrons or something weird like that. – jheindel Nov 26 '16 at 2:05