Question

Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature ? (Given Z for Nd=60, Yb=70, La=57, Ce=58)

• A. Nd
• B. Yb
• C. La
• D. Ce

Hint:

Stability in lanthanide oxidation states usually follows the rule of stable subshells: empty (\( f^0 \)), half-filled (\( f^7 \)), or fully-filled (\( f^{14} \)).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Diamagnetism occurs when all electrons in an atom or ion are paired. In lanthanides, stability and magnetic properties often relate to the filling of the 4f subshell.
Step 2: Detailed Explanation:
1. Yb (Z=70): Ground state electronic configuration is \( [Xe] \, 4f^{14} \, 6s^2 \).
When it loses two electrons to form \( \text{Yb}^{2+} \), the configuration becomes \( [Xe] \, 4f^{14} \).
The 4f subshell is completely filled (14 electrons), and there are no unpaired electrons. Therefore, it is diamagnetic.
2. Nd (Z=60): \( \text{Nd}^{2+} \) would be \( [Xe] \, 4f^4 \), which has 4 unpaired electrons (paramagnetic).
3. La (Z=57): Typically shows +3. \( \text{La}^{2+} \) would be \( [Xe] \, 5d^1 \) (paramagnetic).
4. Ce (Z=58): Typically shows +3 or +4. \( \text{Ce}^{4+} \) is diamagnetic (\( 4f^0 \)), but the question asks for the +2 state.
Step 3: Final Answer:
\( \text{Yb}^{2+} \) is the lanthanide ion that is both in a +2 state and diamagnetic.