Q5(e). Ground-state term symbol of Ho³⁺

1. Ho (Z=67): [Xe]4f¹¹6s². The ion loses 6s² and one 4f electron: Ho³⁺ = [Xe]4f¹⁰.
2. Hund's first rule (maximum S): the seven 4f orbitals (mₗ=+3 to -3) take one electron each with parallel spin, and the remaining three pair up in mₗ=+3,+2,+1. Unpaired electrons =4: S=7×12-3×12=2, multiplicity 2S+1=5.
3. Hund's second rule (maximum L): the seven singly filled orbitals add to Σ mₗ=0; the three paired electrons add 3+2+1: M_L=6, so L=6, letter I (S P D F G H I for L=0 to 6).
4. Hund's third rule: the shell is more than half full, so the lowest level has the highest J: J=L+S=6+2=8.
5. Ground term: ⁵I₈ (read “quintet I eight”).
6. Check by holes: 4f¹⁰ has 4 holes, like 4f⁴ (Pm³⁺, ⁵I₄); the same L and S, but the hole rule turns J=|L-S|=4 into L+S=8.
7. For lanthanide ions the 4f electrons are shielded, so J remains a good quantum number. With g_J=1.25, μ_(eff)=g_J√(J(J+1))=10.6 B.M., close to the observed moment of Ho³⁺ salts.

Application. the term symbol decides magnetic and optical behaviour. Ho³⁺ and Dy³⁺ carry the largest magnetic moments of all ions (about 10.6 B.M.), and the sharp f–f lines of Ho³⁺ make holmium oxide filters the standard for checking the wavelength scale of spectrophotometers.