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VKC Notes · Spectroscopy · IFoS P-II

13.10 ESR (EPR) Spectroscopy: g-Value, Hyperfine Splitting, Radical Anions, Metal Ions and Spin Labels

★★ 2021–2025: 5 Mains parts, 25 marks, all IFoS Paper II, mostly 5-mark parts: count the hyperfine lines and give their intensities (¹³CF₂H, pyrazine, benzene anion), find the resonance field, or define a spin label. Civil Services and Geo-Scientist do not have this syllabus line.

Syllabus line. IFoS: “Electron spin resonance: inorganic complexes and free radicals” (Principles of spectroscopy and applications in structure elucidation).

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13.10.1 Principle and the resonance condition ★★★

(13.10.1)
  • ESR (EPR) is the resonant absorption of microwaves by an unpaired electron in a magnetic field. Only paramagnetic species show it: free radicals, radical ions, triplets, most d- and f-block ions. A closed-shell molecule is ESR-silent, however many magnetic nuclei it has.SN
  • In a field the two spin states split linearly; the β state (ms = −½) lies lower, because the electron’s magnetic moment is opposite to its spin. Resonance when hν = gμBB0 (Fig. 13.10.1). X-band: ν ≈ 9.5 GHz, B ≈ 0.34 T; Q-band 35 GHz, about 1.2 T.SN
  • Spectrometer: fixed-frequency microwave source → sample in a resonant cavity between the poles of an electromagnet → detector. The field is swept, and field modulation with phase-sensitive detection records the first derivative of the absorption; the line centre is where the derivative crosses zero.SN
Ems  = +½ (α)ms  = −½ (β)ΔE = hν = gμB  B0 gap ΔE = gμB  B grows with Babsorptionfirst derivative (recorded)B0 : centre where dA/dB = 0B (swept), ν fixed ≈ 9.5 GHz
Fig. 13.10.1 Electron Zeeman levels, the resonance condition, and the absorption and first-derivative signals on the same field axis (line shape computed).
Draw it in 60 s: two straight lines fanning out from zero field (α up, β down); a double arrow hν at B0; below, a bell curve and under it an S-shaped derivative crossing zero at B0.

Solved PYQ IFoS 2021 Q3(c)(i) 5m

H-atom ESR at 9.5 GHz with g = 2.0026: find the resonance field (μB = 9.274 × 10⁻²⁴ J T⁻¹).

  1. B0 = hν/gμB = (6.626 × 10⁻³⁴ × 9.5 × 10⁹)/(2.0026 × 9.274 × 10⁻²⁴) = 0.339 T (339 mT).SN
  2. This is the centre: the proton (I = ½) splits it into a doublet, one line on each side.

13.10.2 The g-value ★★

(13.10.2)
  • g plays the part of the chemical shift in NMR. For a free electron ge = 2.0023. In a molecule the field can drive small orbital currents through excited states; their local field shifts g by an amount ∝ (spin–orbit coupling constant ξ)/(excitation energy ΔE).SN
  • Organic radicals: g ≈ 2.003 (light atoms, small ξ). Inorganic radicals 1.9–2.1. d-Metal complexes: anywhere from about 0 to 6, because ligand-field gaps are small and ξ is large. g is anisotropic: in solids or frozen solutions an axial complex shows g∥ and g⊥; rapid tumbling in solution averages them.SN
  • Worked: methyl radical centred at 329.40 mT at 9.2330 GHz: g = (6.626 × 10⁻³⁴ × 9.2330 × 10⁹)/(9.274 × 10⁻²⁴ × 0.32940) = 2.0027; at 34.000 GHz (Q-band) it comes into resonance at 1.213 T.

13.10.3 Hyperfine splitting: number of lines and intensities ★★★

(13.10.3)
  • Each magnetic nucleus adds a local field amI (hyperfine coupling constant a, in mT). One nucleus of spin I splits a line into 2I + 1 equal lines; n equivalent nuclei give 2nI + 1 lines; inequivalent sets multiply. Intensities: Pascal’s triangle for I = ½ (1:1, 1:2:1, 1:3:3:1 …); for I = 1 the modified triangle (1:1:1; 1:2:3:2:1; 1:3:6:7:6:3:1).SN
  • Spins to remember: ¹H, ¹³C, ¹⁹F, ³¹P: ½; ²H (D), ¹⁴N: 1; ⁶³Cu: 3/2; ⁵⁵Mn: 5/2; ⁵¹V: 7/2; ¹²C, ¹⁶O, ³²S: 0 (no splitting). Same counting rule as NMR multiplets, with electron–nucleus instead of nucleus–nucleus coupling: 13.3 NMR.SN
  • •CH₃ (aH = 2.3 mT): 4 lines 1:3:3:1. •CD₃: 7 lines 1:3:6:7:6:3:1, packed closer because aD = aH × (gN(D)/gN(H)) = 0.154 aH (Fig. 13.10.2).SN
•CH3: 3 equivalent H (I = ½) → 4 lines1 : 3 : 3 : 1•CD3: 3 equivalent D (I = 1) → 7 lines1 : 3 : 6 : 7 : 6 : 3 : 1aH  = 2.3 mTaD  = 0.154 aH  ≈ 0.35 mT (same field scale)
Fig. 13.10.2 Stick spectra of •CH3 and •CD3 on the same field scale (computed).
Draw it in 60 s: a horizontal base line with four sticks 1:3:3:1 spaced a; below, seven short close sticks 1:3:6:7:6:3:1 with a/6.5 spacing.

Solved PYQ IFoS 2021 Q1(h) 5m

How many ESR lines can the radical ¹³CF₂H show?

  1. ¹³C (I = ½) → 2; two equivalent ¹⁹F (I = ½) → 2 × 2 × ½ + 1 = 3 (1:2:1); ¹H (I = ½) → 2.
  2. Lines = 2 × 3 × 2 = 12, provided the couplings are all different and resolved (Fig. 13.10.3). With ¹²C instead, 6.SN
one line at g13C (I = ½): doublet2 19F: each → 1:2:11H: each → doublet2 × 3 × 2 = 12 lines
Fig. 13.10.3 Splitting tree for ¹³CF2H (coupling sizes drawn to show the order only).
Draw it in 60 s: one line; split into two (13C); each into a 1:2:1 triplet (2F); each of the six into a doublet (H); count the bottom row: 12.

13.10.4 Aromatic radical anions and the McConnell equation ★★★

(13.10.4)

Solved PYQ IFoS 2023 Q8(c) 5m

Benzene with an alkali metal in THF: product, technique, total spin, number of lines and intensities.

  1. Na or K gives one electron to benzene’s LUMO (e2u): the benzene radical anion C₆H₆•⁻ (ion pair with M⁺). One unpaired electron: S = ½; study by ESR.SN
  2. The electron is spread evenly over the six C (ρ = ⅙ each), so six equivalent H: 2 × 6 × ½ + 1 = 7 lines, 1:6:15:20:15:6:1, spacing 0.375 mT (Fig. 13.10.4).SN
  • McConnell: aH = 0.375 mT at ρ = ⅙ gives Q = 6 × 0.375 = 2.25 mT. Naphthalene•⁻: α-H (1,4,5,8) a = 0.490 mT → ρ = 0.22; β-H (2,3,6,7) a = 0.183 mT → ρ = 0.08; 5 × 5 = 25 lines. ESR thus maps the orbital that holds the unpaired electron (the Hückel LUMO).SN
C6H6•−: 7 lines 1:6:15:20:15:6:1, a = 0.375 mTC10H8•−: 5 × 5 = 25 lines, aα  0.490, aβ  0.183 mT−1.0−0.50+0.5+1.0B − B0  / mT
Fig. 13.10.4 Computed stick spectra of the benzene and naphthalene radical anions (book coupling constants, same mT scale).
Draw it in 60 s: top: seven sticks rising to the centre 1:6:15:20:15:6:1; bottom: a five-group pattern, each group itself a small quintet.

Solved PYQ IFoS 2022 Q7(c) 5m

Predict the ESR peaks of pyrazine and their intensity ratios.

  1. Neutral pyrazine is closed-shell: no ESR signal. Its radical anion (from Na or K) has one unpaired electron.
  2. Two equivalent ¹⁴N (I = 1): 2 × 2 × 1 + 1 = 5 lines, 1:2:3:2:1. Each split by four equivalent H: 5 lines, 1:4:6:4:1. Total 5 × 5 = 25 lines.SN
•−C6H6•− (ρ = ⅙) 0.220.220.220.220.080.080.080.08C10H8•−: ρα 0.22, ρβ 0.08 •−pyrazine•− •TEMPO
Fig. 13.10.5 Radical anions met in PYQs, with naphthalene spin densities from McConnell, and the TEMPO spin label.
Draw it in 60 s: a hexagon with •− inside; naphthalene with 0.22 at the four corner (α) carbons and 0.08 at the four β; pyrazine with N at 1,4; a six-ring with four methyls flanking N–O•.

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