# 4.2.2.1.6: Magnetic Moments of Transition Metals

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Magnetic moments are often used in conjunction with electronic spectra to gain information about the valence and stereochemistry of the central metal atom in coordination complexes. A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.

For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3d orbitals are degenerate.

A simple crystal field theory approach to the bonding in these ions assumes that when they form octahedral complexes, the energy of the d orbitals are no longer degenerate but are split such that two orbitals, the dx2-y2 and the dz2 (eg subset) are at higher energy than the dxy, dxz, dyz orbitals (the t2g subset).

For octahedral ions with between 4 and 7 d electrons, this gives rise to 2 possible arrangements called either high spin/weak field or low spin/strong field respectively. The energy gap is dependent on the position of the coordinated ligands in the SPECTROCHEMICAL SERIES.

Note

A good starting point is to assume that all Co(III), d6 complexes are octahedral and LOW spin, i.e. t2g6.

In tetrahedral complexes, the energy levels of the orbitals are again split, such that the energy of two orbitals, the $$d_{x^2-y^2}$$ and the $$d_{z^2}$$ (e subset) are now at lower energy (more favored) than the remaining three $$d_{xy}$$, $$d_{xz}$$, $$d_{yz}$$ (the $$t_2$$ subset) which are destabilized.

Tetrahedral complexes are ALL high spin since the difference between the 2 subsets of energies of the orbitals is much smaller than is found in octahedral complexes.

The usual relationship quoted between them is:

$\Delta_{tet} \approx \dfrac{4}{9} \Delta_{oct}$

Square planar complexes are less common than tetrahedral and d8 e.g. Ni(II), Pd(II), Pt(II), etc, have a strong propensity to form square planar complexes. As with octahedral complexes, the energy gap between the $$d_{xy}$$ and $$d_{x^2-y^2}$$ is $$\Delta_{oct}$$ and these are considered strong field / low spin hence they are all diamagnetic, μ=0 Bohr Magneton (B.M.)

The formula used to calculate the spin-only magnetic moment can be written in two forms; the first based on the number of unpaired electrons, n, and the second based on the electron spin quantum number, $$S$$. Since for each unpaired electron, $$n=1$$ and $$S=1/2$$ then the two formulae are related and the answer obtained must be identical.

$\mu_{so}= \sqrt{n(n+2)}$

$\mu_{so}= \sqrt{4S(S+1)}$

Comparison of calculated spin-only magnetic moments with experimental data for some octahedral complexes
Metal Config μso / B.M. μobs / B.M.
Ti(III) d1 (t2g1) √3 = 1.73 1.6-1.7
V(III) d2 (t2g2) √8 = 2.83 2.7-2.9
Cr(III) d3 (t2g3) √15 = 3.88 3.7-3.9
Cr(II) d4 high spin (t2g3 eg1) √24 = 4.90 4.7-4.9
Cr(II) d4 low spin (t2g4) √8 = 2.83 3.2-3.3
Mn(II)/ Fe(III) d5 high spin (t2g3 eg2) √35 = 5.92 5.6-6.1
Mn(II)/ Fe(III) d5 low spin (t2g5) √3 = 1.73 1.8-2.1
Fe(II) d6 high spin (t2g4 eg2) √24 = 4.90 5.1-5.7
Co(III) d6 low spin (t2g6) 0 0
Co(II) d7 high spin (t2g5 eg2) √15 = 3.88 4.3-5.2
Co(II) d7 low spin (t2g6 eg1) √3 = 1.73 1.8
Ni(II) d8 (t2g6 eg2) √8 = 2.83 2.9-3.3
Cu(II) d9 (t2g6 eg3) √3 = 1.73 1.7-2.2
Comparison of calculated spin-only magnetic moments with experimental data for some tetrahedral complexes
Ion Config μso / B.M. μobs / B.M.
Cr(V) d1 (e1) √3 = 1.73 1.7-1.8
Cr(IV) / Mn(V) d2 (e2) √8 = 2.83 2.6 - 2.8
Fe(V) d3 (e2 t21) √15 = 3.88 3.6-3.7
- d4 (e2 t22) √24 = 4.90 -
Mn(II) d5 (e2 t23) √35 = 5.92 5.9-6.2
Fe(II) d6 (e3 t23) √24 = 4.90 5.3-5.5
Co(II) d7 (e4 t23) √15 = 3.88 4.2-4.8
Ni(II) d8 (e4 t24) √8 = 2.83 3.7-4.0
Cu(II) d9 (e4 t25) √3 = 1.73 -