# What is the magnetic moment of potassium hexacyanochromate(II)?

I calculated the magnetic moment of $$\ce{K4[Cr(CN)6]}$$? in the following way:

$$\ce{Cr^{+2}}$$ has $$4$$ d electrons. And since $$\ce{CN-}$$ is a strong field ligand, the electrons will pair up in the $$\mathrm{t_{2g}}$$. The electronic configuration would be $$\mathrm{t_{2g}^4}\;\mathrm{e_g^0}$$. Or more specifically, $$\mathrm{d}_x^2\; \mathrm{d}_y^1\; \mathrm{d}_z^1\; \mathrm{d}_{z^2}^0\; \mathrm{d}_{x^2-y^2}^0$$.
The magnetic moment will be a contribution of both orbital and spin magnetic moments. $$\mu = \root\of {n(n+2)+L(L+1)},$$ where $$n$$ will be $$2$$ as the unpaired electrons are $$2$$. $$L$$ will be $$2\times 2 + 1\times 1 +0\times 1 = 5$$ corresponding to $$\Sigma m_l$$.

Solving I get $$\mu = 3.74$$. But the experimental value is $$3.2$$.

Why is there this deviation?

• I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. – Martin - マーチン Nov 21 '18 at 14:59
• For transition metal complexes, to a first approximation you should not include any orbital angular momentum, so the magnetic moment is described by the spin-only formula: $\mu = \sqrt{n(n+1)}\mu_\mathrm B = 2.83 \mu_\mathrm B$. In general, this is a very rough approximation (it holds well in some cases), and bearing that in mind, $2.83$ is pretty close to the experimental value of $3.2$ you quote. To go beyond the spin-only formula, however, is more difficult and involved than one might think. – orthocresol Nov 21 '18 at 18:03
• Note also that the "real" d-orbitals $\mathrm d_{xy}$, $\mathrm d_{yz}$, $\mathrm d_{zx}$, $\mathrm d_{x^2-y^2}$ do not possess well-defined values of $m_l$ (they are linear combinations of $m_l = \pm 1$ and $\pm 2$), so your summation of $m_l$ values is inappropriate anyway. – orthocresol Nov 21 '18 at 18:06
• Atomic orbitals are solutions to the Schrodinger equation (eigenfunctions of the Hamiltonian), but the choice of solutions is not unique, as any linear combination $a\chi_1 + b\chi_2$ of degenerate eigenfunctions $\chi_1, \chi_2$ is still an eigenfunction itself. So, when you have degenerate orbitals (like the five d-orbitals), you can add and subtract them as you wish to make a "new" set of d-orbitals. The five d-orbitals that possess specific values of $m_l$ are not the same as the five d-orbitals that are usually drawn ($\mathrm d_{xz}, \cdots$) - they are linear combinations of each other. – orthocresol Nov 21 '18 at 18:18
• Back on the topic of the magnetic moment - you are correct that the ground state (which has a $T$ term symbol) possesses orbital angular momentum. I don't want to try to explain how this leads to deviations in the magnetic moment from the spin-only formula - it is really quite involved, and although I learnt about it I don't feel comfortable writing an answer - so maybe someone else will answer, but in the meantime, consider looking for a book on physical inorganic chemistry (it's too specialised for the general inorganic texts). – orthocresol Nov 21 '18 at 18:34