In context of the lanthanoids, which of the following statements is not correct?

Lanthanoids (atomic numbers 57-71) have their differentiating electron entering the $$4f$$ subshell. Their chemistry shows several characteristic trends which help us analyse each statement.

Option A: All the members exhibit $$+3$$ oxidation state.
Every lanthanoid loses two $$6s$$ electrons and one $$5d/4f$$ electron readily, giving the stable configuration $$[Xe]\,4f^{\,n}$$ and oxidation state $$+3$$. Hence this statement is correct.

Option B: Because of similar properties the separation of lanthanoids is not easy.
All $$Ln^{3+}$$ ions have almost the same ionic size (differences are only a few picometres) and display nearly identical chemistry (hard-acid nature, preference for oxygen donors, colourless complexes, etc.). Therefore their mutual separation requires special methods such as ion-exchange or solvent extraction. Statement is correct.

Option C: Availability of $$4f$$ electrons results in the formation of compounds in $$+4$$ state for all the members of the series.
Only a few lanthanoids (notably Ce, Pr, Nd, Tb, and occasionally Dy) can reach the $$+4$$ state because an extra electron can be removed to attain stable empty, half-filled or exactly filled $$4f$$ configurations (e.g. $$Ce^{4+}:\,4f^{\,0}$$, $$Tb^{4+}:\,4f^{\,7}$$). The rest of the elements cannot stabilise the $$+4$$ state under ordinary conditions. Hence the statement is incorrect.

Option D: There is a gradual decrease in the radii of the members with increasing atomic number in the series.
This is the well-known lanthanoid contraction caused by poor shielding of the increasing nuclear charge by $$4f$$ electrons. The ionic radius of $$Ln^{3+}$$ ions systematically falls from $$La^{3+}$$ to $$Lu^{3+}$$, so the statement is correct.

Thus, the only incorrect statement is:

Option C which is: Availability of $$4f$$ electrons results in the formation of compounds in $$+4$$ state for all the members of the series.