Explanation:
The maximum number of different oxidation states in a compound refers to the highest number of possible valence states that an element can exhibit in its compounds. The element with the highest number of possible oxidation states is Americium (Am), which is represented by the symbol 'D' in the given options.

Reasoning:
To determine the element with the maximum number of different oxidation states, we can analyze the electronic configuration and valence electrons of each element mentioned in the options.

1. Eu (Europium):
Europium has an atomic number of 63 and belongs to the lanthanide series. Its electronic configuration is [Xe] 4f7 6s2. Europium commonly exhibits an oxidation state of +3 in its compounds. However, it can also exhibit an oxidation state of +2 in some compounds.

2. La (Lanthanum):
Lanthanum has an atomic number of 57 and is the first element in the lanthanide series. Its electronic configuration is [Xe] 5d1 6s2. Lanthanum commonly exhibits an oxidation state of +3 in its compounds. However, it can also exhibit an oxidation state of +2 in some compounds.

3. Ge (Germanium):
Germanium has an atomic number of 32 and belongs to Group 14 of the periodic table. Its electronic configuration is [Ar] 3d10 4s2 4p2. Germanium commonly exhibits an oxidation state of +4 in its compounds. However, it can also exhibit oxidation states of +2 and -4 in some compounds.

4. Am (Americium):
Americium has an atomic number of 95 and belongs to the actinide series. Its electronic configuration is [Rn] 5f7 7s2. Americium can exhibit various oxidation states, including +2, +3, +4, +5, and +6 in its compounds, giving it the maximum number of different oxidation states among the options.

Conclusion:
Among the given options, Americium (Am) exhibits the maximum number of different oxidation states in its compounds. This is due to its electronic configuration and the availability of different valence states for its valence electrons.