Arrange these ions into decreasing order of their ionic radii:
(A) \(Ce^{3+}\)
(B) \(Pr^{3+}\)
(C) \(La^{3+}\)
(D) \(Nd^{3+}\)
(E) \(Pm^{3+}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C > A > B > D > E
Across the lanthanoids the M3+ ions get steadily smaller with increasing atomic number (lanthanoid contraction). Each step adds one proton to the nucleus and one electron to the 4f subshell, and the 4f electrons shield one another poorly from the nuclear charge, so the effective nuclear charge rises and pulls the ion in.
So the larger the atomic number, the smaller the ion:
C \( \ce{La^3+} \) (Z = 57) — largest.
A \( \ce{Ce^3+} \) (Z = 58)
B \( \ce{Pr^3+} \) (Z = 59)
D \( \ce{Nd^3+} \) (Z = 60)
E \( \ce{Pm^3+} \) (Z = 61) — smallest.
NCERT (2026-27 edition):
Atomic and Ionic Sizes – “The overall decrease in atomic and ionic radii from lanthanum to lutetium (the lanthanoid contraction) is a unique feature in the chemistry of the lanthanoids. … The decrease in atomic radii (derived from the structures of metals) is not quite regular as it is regular in M3+ ions (Fig. 4.6). … There is fairly regular decrease in the sizes with increasing atomic number.”
