The present paper is concerned with the calculation of the paramagnetic susceptibility of highly hydrated crystals of the iron group elements Ni, Cr and Co. On the assumption that the metallic ion is subject to a crystalline electric field, predominantly cubic but also with a smaller rhombic term, the Hamiltonian function in a magnetic field $H$ is given by $D(x^4+y^4+z^4)+A{x}^2+B{y}^2-(A+B)\mathrm{}{z}^2+\lambda \left(L·S\right)+\beta H·\mathrm{}(L+2S)\mathrm{}$ the numerical value of $\lambda$ being known from the work of Laporte but the other constants yet to be determined. It actually proves possible to formulate and solve approximately the resulting secular equations and so obtain the first and second order Zeeman effects and hence the susceptibility. For all three ions $L=3\mathrm{}$, so that the orbital problem is the same for all. This problem is exactly soluble, the energy levels consisting of two triplets and a singlet, the singlet not lying between the triplets. The effect of the introduction of the spin and its coupling to the orbit then leads to a determinant of order 21 for Ni and of order 28 for Cr and Co. That for Ni factors into one of order 10 and one of order 11, while those for Cr and Co factor into two determinants, identical except for the sign of the coefficient of $H$. On the assumption of a cubic field of the same sign and of approximately the same magnitude for all three ions the orbit-spin, together with the rhombic field, is able to remove the degeneracy of the lowest level in Ni and Cr only in a high approximation, while with Co the degeneracy is removed in first approximation. This difference accounts for the isotropy of Ni and Cr compared with the anisotropy of Co. In order to obtain agreement with experiment it is necessary to assume that in Ni the singlet of the orbital problem lies lowest. It then follows from the work of Van Vleck that the singlet also lies lowest for Cr but that for Co the singlet lies highest. When the singlet lies lowest, the square of the magneton number is given by the "spin only" value $4S(S+1\mathrm{})\mathrm{}$, together with a small orbital contribution of order $\frac{\lambda }{D}$, whose sign can be either positive or negative. Actually it is positive for Ni and negative for Cr. In order to fit the results on the principal susceptibilities of Ni, it is necessary to take $D=1260\mathrm{}$ ${\mathrm{cm}}^{-1\mathrm{}}$, $A=176\mathrm{}$ ${\mathrm{cm}}^{-1\mathrm{}}$, $B=352\mathrm{}$ ${\mathrm{cm}}^{-1\mathrm{}}$, the magnitude of $\lambda$ being -335 ${\mathrm{cm}}^{-1\mathrm{}}$. For Ni and Cr the theory requires that for the mean susceptibility $\chi =Q+\frac{P}{T}$, where $P$ and $Q$ are constants, $Q$ being uniquely determined when $P$ is fixed. Choosing $P$ so that $\chi T$ passes through the experimental point at 170°K we find that good agreement is obtained over the whole temperature range. For Cr $\lambda =87\mathrm{}$ ${\mathrm{cm}}^{-1\mathrm{}}$ and we find $D=3730\mathrm{}$ ${\mathrm{cm}}^{-1\mathrm{}}$, but we cannot determine $A$ or $B$ since there are no data on the principal susceptibilities.

Computational difficulties prevent the accurate solution of the Co problem. The situation is complicated by the experimental data not being complete. It proves necessary to consider a sextet which is soluble only numerically in the general case but perturbation theory can be applied when either the orbit-spin is large compared with the rhombic field or vice-versa. We obtain fair agreement with experiment and our calculations indicate that good agreement would be obtained in an intermediate case.

• Received 13 October 1932

DOI:https://doi.org/10.1103/PhysRev.42.666