Question

Which lanthanoid from following exhibits \(f^{14}\) configuration in +3 state?

• A. Tm
• B. Lu
• C. Ce
• D. Dy

Hint:

Lanthanoid 4f runs from f0 (La) to f14 (Lu). In +3 state, Lu3+ has 4f14.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The lanthanoids fill the 4f subshell across the series from cerium (Z = 58) to lutetium (Z = 71). Lanthanum has none and lutetium completes the subshell. In the +3 state, the ions lose the 6s and 5d or 4f electrons so that only the 4f configuration remains.

Step 2: Key Formula or Approach:
Ln\(^{3+}\) has the configuration \([\text{Xe}]4f^n\), where \(n = Z - 57\).

Step 3: Detailed Explanation:
Lutetium (Z = 71): \(n = 71 - 57 = 14\), so \(\text{Lu}^{3+} = [\text{Xe}]4f^{14}\).
Thulium (Z = 69): \(n = 12\), \(4f^{12}\).
Dysprosium (Z = 66): \(n = 9\), \(4f^{9}\).
Cerium (Z = 58): \(n = 1\), \(4f^{1}\).
Only lutetium reaches the full \(4f^{14}\) set.

Final Answer:
Lu in the +3 state has the \(f^{14}\) configuration, option (B). \[ \boxed{\text{Lu (B)}} \]