Hyperfine lines expected in the EPR spectrum of tetranuclear rhodium (I = \(\tfrac{1}{2}\)) complex :
5
The number of hyperfine lines produced by a set of equivalent nuclei is given by
\(\text{number of lines} = 2nI + 1\)
where \(n\) is the number of equivalent nuclei and \(I\) the nuclear spin of each.
Substitute the values. The complex is tetranuclear, so \(n = 4\), and rhodium has \(I = \tfrac{1}{2}\):
\(2 \times 4 \times \tfrac{1}{2} + 1 = 4 + 1 = 5\) lines.
For spin-\(\tfrac{1}{2}\) nuclei this reduces to the familiar \(n + 1\) rule, the same rule that governs multiplicity in NMR.
The intensities follow the binomial coefficients, here 1 : 4 : 6 : 4 : 1. The reason is combinatorial: the five lines correspond to total nuclear spin projections of +2, +1, 0, -1, -2, and there is only one way to have all four rhodium spins aligned but six ways to have two up and two down. The central line is therefore the most intense.
Two conditions underlie the simple formula, and both are stated or implied in the question. The four rhodium nuclei must be equivalent — if they were inequivalent, each would split the signal independently and up to \(2^{4} = 16\) lines could appear. And the unpaired electron must interact with all four, which is what makes the observation informative: seeing five equally spaced lines is direct evidence that the unpaired electron is delocalised over the whole tetranuclear cluster rather than localised on one metal centre.
Hence 5 hyperfine lines are expected.
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