Key concept

The number of oxidation states shown by f-block elements depends on how easily the f-electrons can participate in bonding.

Compare lanthanoids and actinoids

• In lanthanoids, the electrons enter the 4f orbitals.
• In actinoids, the electrons enter the 5f orbitals.

Why lanthanoids show fewer oxidation states

The 4f orbitals are highly buried inside the atom and are shielded by outer electrons. Because of this, 4f electrons do not participate easily in bonding. Hence, lanthanoids usually show fewer oxidation states, with $+3$ being the most common.

Why actinoids show more oxidation states

The 5f orbitals are less deeply buried and extend farther from the nucleus than the 4f orbitals. Therefore, 5f, 6d, and 7s electrons can all take part in bonding more readily. As a result, actinoids exhibit a larger variety of oxidation states.

Option-wise check

• A: "the 5f orbitals are more buried than the 4f orbitals"
  This is false. If 5f were more buried, they would be less available for bonding.

• B: "there is a similarity between 4f and 5f orbitals in their angular part of the wave function"
  This does not explain the greater number of oxidation states.

• C: "the actinoids are more reactive than the lanthanoids"
  This is not the fundamental reason for multiple oxidation states.

• D: "the 5f orbitals extend further from the nucleus than the 4f orbitals"
  This correctly explains why actinoids show more oxidation states.

Final answer

Therefore, the correct option is: $\boxed{\text{D}}$