Hyperfine Splitting in EPR

Introduction

If an unpaired electron sat in complete isolation, its EPR spectrum would be a single line. In real molecules the electron spends time near nuclei that have magnetic moments, such as ¹H, ¹⁴N, ¹³C or ⁶³Cu. Each nucleus produces a small local field that adds to or subtracts from the applied field, so the single line splits into a pattern of lines. This hyperfine structure is the richest source of structural information in EPR: it tells us which nuclei carry spin density and how many of each kind there are.

Core explanation

Origin of the splitting. A nucleus with spin quantum number I has 2I + 1 orientations, labelled m I = I, I − 1, …, −I. Each orientation shifts the local field felt by the electron by a slightly different amount. In a field-swept spectrum, resonance then occurs at 2I + 1 field positions:

B = B₀ − a·m I

where B₀ is the centre field and a is the hyperfine coupling constant . The EPR selection rule is Δmₛ = ±1 with Δm I = 0, so each nuclear orientation gives exactly one line. Because the nuclear orientations are almost equally populated at normal temperatures, the lines have equal intensity.

One nucleus. A single ¹H (I = ½) gives two lines; a single ¹⁴N (I = 1) gives three equal lines; ⁶³Cu or ⁶⁵Cu (I = 3/2) gives four; ⁵⁵Mn (I = 5/2) gives six; ⁵¹V (I = 7/2) gives eight.

Several equivalent nuclei. n equivalent nuclei of spin I give 2nI + 1 lines. For spin-½ nuclei this becomes n + 1 lines with binomial intensities from Pascal's triangle. The methyl radical, •CH₃, has three equivalent protons and shows four lines in the ratio 1:3:3:1, spaced by about 2.3 mT. Two equivalent ¹⁴N nuclei give five lines with intensities 1:2:3:2:1, obtained by overlapping three equal triplets.

Non-equivalent sets. If there are several sets of equivalent nuclei, the patterns multiply. The maximum number of lines is the product of (2nᵢIᵢ + 1) over all sets, although some lines may overlap by accident. The largest coupling usually defines the widest splitting, and each of its lines is then split further by smaller couplings.

Isotropic and anisotropic parts. The hyperfine interaction has two contributions. The Fermi contact term depends on unpaired spin density exactly at the nucleus, which only s orbitals possess; it is isotropic and survives rapid tumbling. The dipolar term is the through-space interaction between electron and nuclear magnetic moments; it depends on orientation and averages to zero in fast-tumbling solution. Solution spectra therefore reveal isotropic couplings, while frozen or solid samples also show anisotropic couplings.

Size of couplings. The hydrogen atom, with its electron entirely in a 1s orbital, has a = 50.7 mT. In organic π radicals, protons lie in the nodal plane of the p orbital, so their couplings are much smaller, typically 0.1–3 mT, and arise indirectly through spin polarisation of the C–H bond. Couplings are quoted in mT, gauss (1 mT = 10 G) or MHz; for g ≈ 2.0023, 1 mT corresponds to about 28.0 MHz.

Step-by-step reasoning

To predict a hyperfine pattern:

1. List every magnetic nucleus that could carry spin density, with its spin I and natural abundance. 2. Group the nuclei into sets of symmetry-equivalent atoms. 3. For each set, find 2nI + 1 lines and their relative intensities. 4. Start with the largest coupling and split each line by the next set. 5. Check the total number of lines against the product rule and note likely overlaps.

Visual explanation

Draw a stick "tree" diagram. The single central line branches into four equal-spaced sticks for three equivalent protons, with the inner sticks three times taller. If a nitrogen is also present, draw each of those sticks splitting into three equal branches beneath. The final row of sticks is the predicted spectrum.

Real-world analogy

Imagine a singer whose voice is picked up by several microphones slightly out of tune with one another. Each microphone shifts the pitch by a fixed amount up or down, and the listener hears several copies of one note. The number of copies reveals how many microphones there are, and the spacing reveals how strongly each one is coupled.

Real-world example

Nitroxide spin labels, such as derivatives of TEMPO, show a characteristic three-line spectrum from their ¹⁴N nucleus, with a N around 1.5–1.7 mT. Because this coupling and the line shapes are sensitive to polarity and motion, nitroxides attached to proteins or membranes report on their local environment.

Why?

Why do n equivalent spin-½ nuclei give binomial intensities? Each nucleus can be up or down with nearly equal probability. The central lines correspond to combinations with total m I near zero, which can be formed in many ways, while the outer lines need all spins aligned, which can happen in only one way.

Common misconception

"The number of lines equals the number of magnetic nuclei." For equivalent nuclei the count is 2nI + 1, not n; for non-equivalent sets it is the product of the individual multiplicities. Three equivalent protons give four lines, not three.

Worked example

Question: Predict the hyperfine pattern of the ethyl radical, CH₃CH₂•, given a(α-H, 2 H) ≈ 2.24 mT and a(β-H, 3 H) ≈ 2.69 mT.

Reasoning: The three β protons give a 1:3:3:1 quartet with 2.69 mT spacing. Each quartet line is split by the two α protons into a 1:2:1 triplet with 2.24 mT spacing. Total lines = 4 × 3 = 12.

Answer: Twelve lines: a quartet of triplets, with intensities in each triplet 1:2:1 scaled by 1, 3, 3 and 1; some lines lie close together because the two couplings are similar.

Quick check

1. How many lines, and in what intensity ratio, does a radical with two equivalent ¹⁴N nuclei show? Answer: Five lines (2 × 2 × 1 + 1) with intensities 1:2:3:2:1.

Exam focus

Learn the 2nI + 1 rule, the binomial pattern for spin-½ nuclei and the product rule for non-equivalent sets. Know that the Fermi contact term is isotropic and needs s-orbital spin density, while dipolar coupling averages to zero in fluid solution. Show unit conversions between mT and MHz clearly.

Advanced insight

When a coupling becomes large compared with the electron Zeeman energy, as for some metal ions, lines are no longer exactly equally spaced. Second-order (Breit–Rabi) corrections shift outer lines, so the centre of the pattern is not exactly at the field corresponding to g. Techniques such as ENDOR measure couplings directly as nuclear frequencies, resolving small couplings hidden within broad EPR lines.

Summary

Hyperfine splitting arises when an unpaired electron couples to magnetic nuclei. A nucleus of spin I gives 2I + 1 equal lines; n equivalent nuclei give 2nI + 1 lines, binomial for spin ½; non-equivalent sets multiply. The isotropic Fermi contact term reports s spin density, while the anisotropic dipolar term appears only in solids or frozen solutions.

Practice questions

1. How many lines does a radical show if it couples to one ⁵⁵Mn nucleus (I = 5/2)? Answer: 2 × 5/2 + 1 = 6 equal lines. 2. Predict the pattern for a radical with four equivalent protons. Answer: Five lines in the ratio 1:4:6:4:1. 3. A radical couples to one ¹⁴N and two equivalent protons, and the proton coupling differs from the nitrogen coupling. What is the maximum number of lines? Answer: 3 × 3 = 9 lines: a triplet of 1:2:1 triplets or vice versa. 4. Convert a proton coupling of 0.50 mT into MHz for a radical with g ≈ 2.0023. Answer: 0.50 × 28.0 ≈ 14 MHz. 5. Why does the dipolar hyperfine contribution not appear in fast-tumbling solution spectra? Answer: It depends on orientation as (3cos²θ − 1), which averages to zero when the molecule tumbles rapidly through all orientations.