Magnetic Moments of Transition Metals

Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion 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.

See an interactive JAVA applet for examples.

Note: For CHEM1902 (C10K), we 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 for CHEM1902 we will assume that the only ions forming square planar complexes are d8 e.g. Ni(II), Pd(II), Pt(II), etc. As with octahedral complexes, the energy gap between the dxy and dx2-y2 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 clearly 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
Ion 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 tetahedral 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 -

Contributors

• The Department of Chemistry, University of the West Indies)