# Is Co[Hg(SCN)4] diamagnetic?

I was reading a book and it was mentioned that $\ce{Co[Hg(SCN)4]}$ is diamagnetic. I know that it is made up of $\ce{Hg^2+}$ and $\ce{Co^2+}$. $\ce{Co^2+}$ has 7 electrons which means that it should always have an unpaired electron but then why is this compound diamagnetic? Am I missing something?

There was definitely a typo and the book meant to write $\ce{Ca}$ instead of $\ce{Co}$. In the magnetic properties of a coordination complex, both the counter-ions and the coordination sphere metal ion are contributors. Since $\ce{Hg^2+}$ is diamagnetic and the overall compoud is given to be diamagnetic, the counter-ion has to be diamagnetic as well. Hence, it cannot be $\ce{Co^2+}$, which happens to have unpaired electrons.
What the book might have meant to say is that calcium is the counter-ion here. When checking diamagnetism or paramagnetism, you have to instead check the metal ion inside the coordination sphere, which in this case is $\ce{Hg^2+}$. And note that $\ce{Hg^2+}$ will have a $\ce{5d^10}$ electronic configuration.