A certain Mn2+ (I = 5/2) high spin octahedral complex displays an EPR (ESR) spectrum of five sextets (5×6=30 lines). Ignoring the spin-spin interaction between metal and ligands, this spectrum can be interpreted as
Non degenerate Kramer doublets in zero field
High-spin Mn2+ is d5 with S = 5/2, so it has six MS levels: ±5/2, ±3/2 and ±1/2. Because the ion has an odd number of electrons, Kramers’ theorem keeps each ±MS pair degenerate in zero field, but a small distortion from perfect octahedral symmetry (zero-field splitting) separates the three Kramers doublets from one another.
In a magnetic field the allowed \(\Delta M_S = \pm 1\) transitions between the six Zeeman levels are then no longer coincident, giving five fine-structure lines. Each line is split by the 55Mn nucleus into \(2I+1 = 2(5/2)+1 = 6\) hyperfine components, giving five sextets, 30 lines in total.
Only the ±½ levels would give a single sextet. Integral spin states 0, ±1, ±2 belong to an even-electron (S = 2) ion, not to d5 Mn2+. Four spin states would give three transitions, not five.
Non degenerate Kramer doublets in zero field option is correct.
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