# By using the magnetic susceptibility, calculate the number of unpaired elections in the complex

I'm having trouble making sense of this answer. I believe I am doing all the steps right, but my main concern is that I am not arriving at a whole number. If my answer is correct, should I just round up or down?

## Question

By using the known values of $\chi$ for $\ce{HgCo(NCS)4}$ calculate the number of unpaired electrons in each of these complexes, and interpret the results in terms of the electronic structures of the complexes.

## My Attempt

$$\ce{HgCo(NCS)4}\ 's \ \chi = 16.44 x10^{-6}\ \mathrm{CGS (g^{-1}\cdot cm^3) \ at \ 20°C}$$

$$\mathrm{\chi_M =16.44 \cdot 10^{-6}g^{-1}\cdot cm^3 * \frac{491.8528 g}{1 mol} =8.09\cdot 10^{-3} mol^{-1}cm^3}$$ $$\chi ' _M= \chi_M (\text{metal core electrons}) + \chi_M (\text{ligands}) + \chi_M (\text{other}) - \chi_M$$

Diamagnetic corrections include:

\begin{align} \ce{Hg^{2+}}&=40.0\cdot 10^{-6}\\ \ce{Co^{2+}}&=12.8\cdot 10^{-6} \\ \ce{NCS^-} &=26.2\cdot 10^{-6}\end{align} Source

$$\chi ' _M= (40\cdot 10^{-6})+(12.8\cdot 10^{-6})+(26.2\cdot 10^{-6})(3)-(16.44\cdot 10^{-6}) =-0.02418$$

then using the equation

$$\mathrm{\chi '_MT=\frac{1}{8}n(n+2)}$$ Assuming 298K and setting to zero $$\mathrm{n^2 +2n-57.6722=0}$$

quadradic formula and i get $$\mathrm{n=6.65}$$

I figure with all of that arithmetic it would be difficult to arrive at a whole number. So, my intuition is to round up to 7.

Is this approach correct? Which way should I round if so?

And I don't know how I would determine the electronic structure because there's two metal ions in the complex?

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.

• Okay so I follow most of that, but if there's 3 expected unpaired spins, why would my calculation show 7? – John Snow May 8 '15 at 19:08
• Well, that certainly implies that it's not just Co d7 involved in the material, doesn't it. :-) – Geoff Hutchison May 9 '15 at 15:28
• There are multiple academic articles on this compound, depending on the level of detail you want. The article linked in the last section gives some useful information into the temperature-dependent behavior and many relevant citations. – Geoff Hutchison May 9 '15 at 15:30