The total multiplicity of the R-S term 5I4 is :
65
First decode the term symbol \(^{5}I_{4}\), which is written in the form \(^{2S+1}L_J\).
Spin. The superscript gives \(2S + 1 = 5\), so \(S = 2\).
Orbital angular momentum. The letter I corresponds to \(L = 6\), following the sequence S = 0, P = 1, D = 2, F = 3, G = 4, H = 5, I = 6.
Total angular momentum. The subscript gives \(J = 4\).
Now count the states. The total multiplicity of the term is the number of microstates it contains, which is the product of the spin and orbital degeneracies:
\((2S+1)(2L+1) = 5 \times 13 = 65\).
The reasoning is that \(M_S\) can take \(2S+1 = 5\) values (from +2 to -2) and \(M_L\) can independently take \(2L+1 = 13\) values (from +6 to -6), and every combination is a distinct microstate.
The remaining options correspond to the individual factors and to a related but different quantity. 5 alone is the spin multiplicity, 13 alone the orbital degeneracy, and 9 is \(2J + 1\) — the degeneracy of the particular \(J = 4\) level, which is what would be asked for if the question concerned splitting in a magnetic field.
Hence the total multiplicity of 5I4 is 65.
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