Diamagnetic materials are those that have all their electrons paired up in their atomic or molecular orbitals, resulting in no net magnetic moment. On the other hand, paramagnetic materials have unpaired electrons and exhibit a net magnetic moment.

In the given options, we need to identify the lanthanoid anion that is diamagnetic.

Let's analyze each option one by one:

a) Sm2: Samarium (Sm) has an atomic number of 62. In its neutral state, it has the electron configuration [Xe] 4f6 6s2. Sm2 would mean that two electrons are removed from the 6s orbital. The resulting electron configuration would be [Xe] 4f6. Since there are unpaired electrons present, Sm2 is paramagnetic and not diamagnetic.

b) Eu2: Europium (Eu) has an atomic number of 63. In its neutral state, it has the electron configuration [Xe] 4f7 6s2. Eu2 would mean that two electrons are removed from the 6s orbital. The resulting electron configuration would be [Xe] 4f7. Since there are unpaired electrons present, Eu2 is paramagnetic and not diamagnetic.

c) Yb2: Ytterbium (Yb) has an atomic number of 70. In its neutral state, it has the electron configuration [Xe] 4f14 6s2. Yb2 would mean that two electrons are removed from the 6s orbital. The resulting electron configuration would be [Xe] 4f14. Since all the orbitals are completely filled and there are no unpaired electrons, Yb2 is diamagnetic.

d) Ce2: Cerium (Ce) has an atomic number of 58. In its neutral state, it has the electron configuration [Xe] 4f1 5d1 6s2. Ce2 would mean that two electrons are removed from the 6s orbital. The resulting electron configuration would be [Xe] 4f1 5d1. Since there is an unpaired electron present in the 4f orbital, Ce2 is paramagnetic and not diamagnetic.

Therefore, the correct answer is option C, Yb2, as it is the only lanthanoid anion that has a completely filled f-orbital and no unpaired electrons, making it diamagnetic.