By using the magnetic susceptibility, calculate the number of unpaired elections in the complex
Let's start with "what is the species?"

While you write it $\ce{HgCo(NCS)4}$ it's probably better written as $\ce{Hg^{+2} [Co(NCS)4]^{2-}}$.

You can take the two metal ions separately as far as electronic structure.

$\ce{Hg^{+2}: d^{10}}$ So we don't expect any unpaired electrons.

$\ce{[Co(NCS)4]^{2-}}$ Well, $\ce{Co}$ must be $\ce{Co^{+2}}$, so it's $\ce{d^7}$ on paper. Next you'd need to consider the shape of the complex and possibly where the ligand is in the spectrochemical series and the likely orbital occupations.


   Here, $\ce{Co^{+2}}$ is tetrahedral, so it doesn't matter where NCS falls on the series. You expect a high spin complex with 3 unpaired spins.


Note: (NCS) can be a tricky ligand, since it can coordinate with either N or S atoms, but the crystal structure is clearly N-coordination.

I mention this, because while the Hg ion is, on paper, separated from the Co complex, in the crystal structure, there's clearly some Hg-S interactions.

So let's get to your question about the calculation. Yes, if you need an integer, you'd round up. But since you have many, many more than one molecule, your answer implies there are multiple magnetic species.


   This is clear from the electronic structure. Since the calculation above gives three unpaired spins in an isolated $\ce{Co^{+2}}$ complex, there must be some magnetic interactions between multiple species in the solid state. Indeed, this is observed at different temperatures. I'm hardly an expert in magnetic coupling here, but it certainly implies Hg must be involved as well.