Answer:

Introduction:
The magnetic moment of a lanthanide (3+) ion can be calculated by considering the total angular momentum of the unpaired electrons. The magnetic moment closest to the spin-only value for ND(3), Dy(3), and Lu(3) ions can be calculated using the formula μ = √n(n+2)BM, where n is the number of unpaired electrons and BM is the Bohr magneton.

Calculations:
- ND(3) ion: Nd(3+) has 6 unpaired electrons, giving a magnetic moment μ = √6(6+2)BM = 6.24BM. The spin-only value is 6.82BM, so the magnetic moment of Nd(3+) is relatively close to the spin-only value.
- Dy(3) ion: Dy(3+) has 5 unpaired electrons, giving a magnetic moment μ = √5(5+2)BM = 5.92BM. The spin-only value is 10.35BM, so the magnetic moment of Dy(3+) is further from the spin-only value than Nd(3+).
- Lu(3) ion: Lu(3+) has no unpaired electrons and therefore has no magnetic moment.

Explanation:
The magnetic moment of a lanthanide (3+) ion is affected by several factors, including the number of unpaired electrons and the electronic configuration of the ion. The closer the magnetic moment is to the spin-only value, the higher the degree of spin-only behavior exhibited by the ion. In the case of ND(3), the magnetic moment is relatively close to the spin-only value, indicating a high degree of spin-only behavior. In contrast, Dy(3) has a magnetic moment that is further from the spin-only value, indicating a lower degree of spin-only behavior. Lu(3) has no unpaired electrons and therefore has no magnetic moment, indicating that it exhibits no spin-only behavior.