Question

What is the number of unpaired electrons in f orbitals of lutetium in its +3 oxidation state?

• A. 7
• B. 5
• C. 4
• D. 0 (zero)

Hint:

$Lu^{3+}$ and $Yb^{2+}$ have completely filled f-orbitals ($f^{14}$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need the electronic configuration of \( \text{Lu}^{3+} \).

Step 3: Detailed Explanation:
Lutetium (\( Z = 71 \)) is the last lanthanide.
Neutral Lu: \( [Xe] 4f^{14} 5d^1 6s^2 \).
\( \text{Lu}^{3+} \): Loses two 6s electrons and one 5d electron.
Configuration of \( \text{Lu}^{3+} \): \( [Xe] 4f^{14} \).
The 4f orbital is completely filled with 14 electrons (all paired).
Number of unpaired electrons \( = 0 \).

Step 4: Final Answer:
The number of unpaired electrons in \( \text{Lu}^{3+} \) is 0.