Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being

The number of oxidation states that an element can exhibit depends on how many of its outer-lying electrons (those in $$s,\,p,\,d$$ or $$f$$ subshells) can participate in bonding. For such participation, the energy separation between the subshells must be small; then removal or sharing of electrons from more than one subshell becomes feasible.

Lanthanoids involve the $$4f,\,5d$$ and $$6s$$ subshells. After the first one or two electrons enter $$5d$$, the $$4f$$ orbitals get rapidly stabilised (shielded by the filled $$5s$$ and $$5p$$ subshells) and lie appreciably lower than the $$5d$$ orbitals. Hence the energy gap between $$4f$$ and $$5d$$ is comparatively large; electrons of the $$4f$$ subshell remain core-like and do not take part readily in bonding. Consequently, the lanthanoids show only a few oxidation states, most commonly $$+3$$ (and occasionally $$+2, +4$$).

Actinoids involve the $$5f,\,6d$$ and $$7s$$ subshells. Because the shielding by the intervening $$6s$$ and $$6p$$ electrons is poorer, the $$5f$$ orbitals are not stabilised to the same extent as $$4f$$. The energies of $$5f$$ and $$6d$$ orbitals remain very close to each other through much of the series. This small energy separation enables the actinoid atoms/ions to lose or share varying numbers of electrons from both $$5f$$ and $$6d$$ subshells in addition to $$7s$$ electrons, giving rise to a wide range of oxidation states (from $$+2$$ up to $$+7$$ in different members).

Therefore, the chief reason why actinoids exhibit many more oxidation states than lanthanoids is the smaller energy difference between their $$5f$$ and $$6d$$ orbitals compared with the $$4f$$-$$5d$$ energy gap in lanthanoids.

Option B is correct: lesser energy difference between $$5f$$ and $$6d$$ than between $$4f$$ and $$5d$$ orbitals.