5
$\begingroup$

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?

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ 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. $\endgroup$
    – Martin - マーチン ♦
    Nov 21 '18 at 14:59
  • 1
    $\begingroup$ 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. $\endgroup$
    – orthocresol ♦
    Nov 21 '18 at 18:03
  • 1
    $\begingroup$ 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. $\endgroup$
    – orthocresol ♦
    Nov 21 '18 at 18:06
  • 1
    $\begingroup$ 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. $\endgroup$
    – orthocresol ♦
    Nov 21 '18 at 18:18
  • 1
    $\begingroup$ 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). $\endgroup$
    – orthocresol ♦
    Nov 21 '18 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.