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:

C > A > B > D > E

The correct answer is option 3. C > A > B > D > E.

The elements provided are all part of the lanthanide series, which includes elements from Lanthanum (La) to Lutetium (Lu). These elements have atomic numbers ranging from 57 to 71. As you move across the lanthanide series from La to Lu:

Effective Nuclear Charge: The number of protons in the nucleus increases, which increases the effective nuclear charge (the net positive charge experienced by electrons).

Shielding Effect: The 4f orbitals, which are gradually filled as you move across the series, have poor shielding effectiveness. This means the added electrons do not effectively shield the increasing nuclear charge.

Result: The increasing nuclear charge pulls the electrons closer to the nucleus, leading to a decrease in the size of the ions across the series. This gradual decrease in ionic radii across the series is known as the **lanthanide contraction**.

Analyzing the Given Ions:

C. \(La^{3+}\) (Lanthanum, Z = 57):

As the first element in the lanthanide series, Lanthanum has the largest ionic radius among the given ions because it has the fewest protons pulling on the electrons.

A. \(Ce^{3+}\) (Cerium, Z = 58):

Cerium is next in the series, with a slightly smaller ionic radius than Lanthanum due to the increased nuclear charge.

B. \(Pr^{3+}\) (Praseodymium, Z = 59):

Praseodymium follows Cerium, with a further decrease in ionic radius.

D. \(Nd^{3+}\) (Neodymium, Z = 60):

Neodymium comes next, with an even smaller ionic radius than Praseodymium.

E. \(Pm^{3+}\) (Promethium, Z = 61):

Promethium has the smallest ionic radius among the given ions because it has the highest atomic number, meaning the most significant nuclear charge pulling the electrons closer to the nucleus.

Decreasing Order of Ionic Radii:

Given the trend of lanthanide contraction, the decreasing order of the ionic radii for the given ions is:

\(\text{C }(La^{3+}) > \text{A }(Ce^{3+}) > \text{B }(Pr^{3+}) > \text{D }(Nd^{3+}) > \text{E }(Pm^{3+})\)

Thus, the correct answer is (3) C > A > B > D > E