The 'f' orbitals are half and completely filled, respectively in lanthanide ions [Given : Atomic no. Eu, 63 : Sm, 62 : Tm, 69; Tb, 65; Yb, 70; Dy, 66]

We need to find lanthanide ions where the f orbitals are half-filled and completely filled, respectively.

The 4f subshell can hold a maximum of 14 electrons. It is half-filled at 7 electrons and completely filled at 14 electrons.

Determine the electron configuration of each ion in Option C.

Tb (Z = 65): Ground state configuration is $$[Xe] 4f^9 6s^2$$.

For $$Tb^{4+}$$, we remove 4 electrons (2 from 6s, 1 from 5d if present, and remaining from 4f):

$$Tb^{4+}: [Xe] 4f^7$$

This gives 7 electrons in 4f, which is a half-filled configuration.

Yb (Z = 70): Ground state configuration is $$[Xe] 4f^{14} 6s^2$$.

For $$Yb^{2+}$$, we remove 2 electrons (both from 6s):

$$Yb^{2+}: [Xe] 4f^{14}$$

This gives 14 electrons in 4f, which is a completely filled configuration.

Verify other options are incorrect.

Option A: $$Eu^{2+}$$: Eu (Z = 63) has $$[Xe] 4f^7 6s^2$$. $$Eu^{2+}: [Xe] 4f^7$$ (half-filled). $$Tm^{3+}$$: Tm (Z = 69) has $$[Xe] 4f^{13} 6s^2$$. $$Tm^{3+}: [Xe] 4f^{12}$$ (neither half nor completely filled). So this pair does not satisfy the condition.

Option B: $$Sm^{2+}$$: Sm (Z = 62) has $$[Xe] 4f^6 6s^2$$. $$Sm^{2+}: [Xe] 4f^6$$ (not half-filled). This fails.

Option D: $$Dy^{3+}$$: Dy (Z = 66) has $$[Xe] 4f^{10} 6s^2$$. $$Dy^{3+}: [Xe] 4f^9$$ (not half-filled). This fails.

Therefore, Option C is correct: $$Tb^{4+}$$ has half-filled 4f orbitals ($$4f^7$$) and $$Yb^{2+}$$ has completely filled 4f orbitals ($$4f^{14}$$).