Common oxidation state of lanthanoids is

The correct answer is Option (2) → +3

The lanthanoids (Ce to Lu) have the general configuration \( \mathrm{[Xe]\,4f^{1-14}\,5d^{0-1}\,6s^{2}} \). Losing the two 6s electrons and one more electron (the 5d electron, or one 4f electron) gives the \( \ce{Ln^3+} \) ion. The energy needed for this is well repaid by lattice or hydration energy, so +3 is the principal oxidation state of all the lanthanoids.

A few lanthanoids also show +2 (Eu, Yb) or +4 (Ce), where this gives an empty, half-filled or filled 4f subshell, but these ions change readily to +3.

So the common oxidation state of the lanthanoids is +3.

NCERT (2026-27 edition):
Oxidation States