Discussion
For a rare earth ion the ground state energy level is characterized by the term value \(^{4}I_{9/2}\). The number of 4f unpaired electrons in this ion is :
Answer: B
The ion has 3 unpaired 4f electrons.
In the term \(^{4}I_{9/2}\), the multiplicity 2S + 1 = 4 gives S = 3/2, which means three unpaired electrons each contributing 1/2. The letter I means L = 6, and J = L - S = 9/2, as Hund's third rule requires for a shell less than half full.
This is the ground term of \(Nd^{3+}\), which has the configuration \(4f^{3}\). Two unpaired electrons would give a triplet, four a quintet and five a sextet.
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