Question

Identify the pair of diamagnetic lanthanoid ions by the following?

• A. \(\text{Ce}^{4+}\) and \(\text{Yb}^{2+}\)
• B. \(\text{Ce}^{4+}\) and \(\text{Eu}^{2+}\)
• C. \(\text{Gd}^{3+}\) and \(\text{Eu}^{3+}\)
• D. \(\text{Pm}^{3+}\) and \(\text{Sm}^{3+}\)

Hint:

Diamagnetic ions have no unpaired electrons, so write the 4f configuration of each ion.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
An ion is diamagnetic when it has no unpaired electrons.
For lanthanoid ions this means the 4f subshell is empty (\(4f^{0}\)) or completely filled (\(4f^{14}\)).

Step 2: Write the configurations
\(\text{Ce}^{4+}\) (Z = 58) has \([\text{Xe}]\,4f^{0}\), so it has no unpaired electron.
\(\text{Yb}^{2+}\) (Z = 70) has \([\text{Xe}]\,4f^{14}\), which is completely filled.

Step 3: Check the other ions
\(\text{Eu}^{2+}\) is \(4f^{7}\), \(\text{Gd}^{3+}\) is \(4f^{7}\), \(\text{Eu}^{3+}\) is \(4f^{6}\), \(\text{Pm}^{3+}\) is \(4f^{4}\) and \(\text{Sm}^{3+}\) is \(4f^{5}\).
All of these have unpaired electrons, so they are paramagnetic.

Final Answer:
Ce4+ and Yb2+ are both diamagnetic. \[ \boxed{\text{(A) }\text{Ce}^{4+}\text{ and }\text{Yb}^{2+}} \]