Question

What are the Principal & Azimuthal quantum number values of the valence electrons in tripositive Lutetium?

• A. \( n = 4 \) \( l = 2 \)
• B. \( n = 5 \) \( l = 2 \)
• C. \( n = 5 \) \( l = 1 \)
• D. \( n = 4 \) \( l = 3 \)

Hint:

For transition metals, the valence electrons are typically found in the \( (n-1)f, nd, \) and \( ns \) orbitals. In the case of tripositive Lutetium, electrons are removed first from the \( 6s \) and then from the \( 5d \) orbitals, leaving the \( 4f \) electrons as the valence electrons.

Answer 1

The Correct Option is D

The electronic configuration of Lutetium (Lu) is: \[ \text{Lu} = [Xe] 4f^{14} 5d^1 6s^2 \] For the tripositive state (Lu\(^{3+}\)), 3 electrons are removed, which will affect the \( 6s^2 \) and \( 5d^1 \) orbitals. The valence electrons left are in the \( 4f \) subshell, and the 4f orbital corresponds to \( n = 4 \) and \( l = 3 \). Thus, the correct quantum numbers are \( n = 4 \) and \( l = 3 \).