Concept involved

The number of oxidation states shown by an element depends on how many electrons can participate in bonding.

• Lanthanoids: involve mainly $4f$, $5d$, and $6s$ electrons.
• Actinoids: involve mainly $5f$, $6d$, and $7s$ electrons.

Key difference between lanthanoids and actinoids

In lanthanoids, the energy gap between $4f$ and $5d$ orbitals is comparatively larger, so fewer electrons are available for variable bonding.

In actinoids, the energies of $5f$ and $6d$ orbitals are very close. Because of this small energy difference, electrons from both $5f$ and $6d$ orbitals can participate in bonding.

Hence, actinoids exhibit a larger number of oxidation states.

Evaluation of options

A: $4f$ orbitals more diffused than $5f$ orbitals
This is false. In fact, $5f$ orbitals are less effectively shielded and can participate more in bonding than $4f$ in early actinoids. So this is not the reason.

B: lesser energy difference between $5f$ and $6d$ than between $4f$ and $5d$ orbitals
This is correct. This directly explains why actinoids show more oxidation states.

C: more energy difference between $5f$ and $6d$ than between $4f$ and $5d$ orbitals
Opposite of the correct reason, so false.

D: more reactive nature of the actinoids than the lanthanoids
Reactivity alone is not the main reason for larger number of oxidation states.

Final answer

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