Question

Correct order of ionic radius of the following \(Ln^{3+}\) ions is

• A. \(Ho > Lu > Ce > Sm\)
• B. \(Lu > Ce > Ho > Sm\)
• C. \(Ce > Sm > Ho > Lu\)
• D. \(Lu > Ho > Sm > Ce\)

Hint:

For \(Ln^{3+}\) ions, size decreases across the lanthanoid series due to lanthanoid contraction.

Answer 1

The Correct Option is C

Concept:
In lanthanoids, ionic radius decreases from left to right in the series. This decrease is due to lanthanoid contraction. For \(Ln^{3+}\) ions, as atomic number increases, effective nuclear charge increases and ionic radius decreases.

Step 1: Arrange lanthanoids in increasing atomic number.
The order in the lanthanoid series is: \[ Ce,\ Sm,\ Ho,\ Lu \] Here \(Ce\) comes earlier and \(Lu\) comes later.

Step 2: Apply lanthanoid contraction.
Due to lanthanoid contraction: \[ \text{ionic radius decreases from }Ce^{3+}\text{ to }Lu^{3+} \] So: \[ Ce^{3+} > Sm^{3+} > Ho^{3+} > Lu^{3+} \]

Step 3: Check the options.
Option (A) is incorrect because \(Lu^{3+}\) cannot be larger than \(Ce^{3+}\).
Option (B) is incorrect because it places \(Lu^{3+}\) largest.
Option (C) is correct.
Option (D) is reverse of the correct trend. Hence, the correct answer is: \[ \boxed{(C)\ Ce > Sm > Ho > Lu} \]