Question

What is the number of unpaired electrons in Lutetium (Lu) in the +3 oxidation state?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

The number of unpaired electrons can be determined by examining the electron configuration in the oxidation state of the element. If all electrons are paired, the element has no unpaired electrons.

Answer 1

The Correct Option is C

Lutetium (Lu) has an atomic number of 71, and its electron configuration in the ground state is: \[ \text{Lu}: [Xe] 4f^{14} 5d^1 6s^2 \]
Step 1: Electron Configuration in the +3 Oxidation State In the +3 oxidation state, Lutetium loses three electrons. The electrons are removed first from the 6s orbital, and then from the 5d orbital. Therefore, the electron configuration for \( \text{Lu}^{3+} \) is: \[ \text{Lu}^{3+}: [Xe] 4f^{14} \]
Step 2: Determine the Number of Unpaired Electrons In the \( 4f^{14} \) configuration, all 14 electrons are paired in the f-orbital because the f-orbitals can hold up to 14 electrons in 7 orbitals (with 2 electrons per orbital). Therefore, there are no unpaired electrons in the \( 4f^{14} \) configuration of \( \text{Lu}^{3+} \).
Step 3: Conclusion The number of unpaired electrons in \( \text{Lu}^{3+} \) is \( \boxed{0} \).