Discussion
Mn3+ ion has four 3d electrons. The spectroscopic term factor of this ion is :
Answer: D
The ground term is \({}^{5}D_{0}\).
Apply Hund's rules to \(3d^{4}\):
- Maximum S: four parallel spins, \(S = 2\), multiplicity \(2S + 1 = 5\).
- Maximum L for that S: \(m_{l} = 2, 1, 0, -1\), so \(L = 2\), a D term.
- The shell is less than half full, so \(J = |L - S| = 0\).
The term is \({}^{5}D_{0}\). \({}^{1}D_{2}\) has the wrong spin, and the terms with half-integer J require an odd number of electrons.
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