Question

Which of the following lanthanoid has the outer electronic configuration \(4f^7 6s^2\) in its ground state

• A. Neodymium
• B. Gadolinium
• C. Europium
• D. Promethium
• E. Samarium

Hint:

Gadolinium (Gd, Z=64) also has a half-filled f-orbital, but its configuration is \(4f^7 5d^1 6s^2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This question is about the electronic configuration of lanthanoids. Lanthanoid configurations are generally of the form \([\text{Xe}] 4f^n 5d^0 6s^2\) or \([\text{Xe}] 4f^n 5d^1 6s^2\). Similar to d-block elements, there are exceptions due to the enhanced stability of half-filled (f⁷) and completely filled (f¹⁴) f-subshells.
Step 2: Detailed Explanation
Let's find the element with the configuration ending in \(4f^7 6s^2\). This configuration is particularly stable because the f-subshell is exactly half-filled. We need to check the configurations of the given elements.

Europium (Eu): Atomic number Z = 63. Its electronic configuration is \([\text{Xe}] 4f^7 6s^2\). It adopts this configuration to benefit from the high stability of the half-filled 4f subshell.
Gadolinium (Gd): Atomic number Z = 64. Its electronic configuration is \([\text{Xe}] 4f^7 5d^1 6s^2\). After Europium fills the 4f subshell to the stable half-filled state, the next electron for Gadolinium enters the 5d orbital to maintain the stable 4f⁷ configuration before starting to pair electrons in the 4f subshell.
Neodymium (Nd): Z = 60. Configuration is \([\text{Xe}] 4f^4 6s^2\).
Promethium (Pm): Z = 61. Configuration is \([\text{Xe}] 4f^5 6s^2\).
Samarium (Sm): Z = 62. Configuration is \([\text{Xe}] 4f^6 6s^2\).
The lanthanoid with the outer electronic configuration \(4f^7 6s^2\) is Europium.
Step 3: Final Answer
Europium (Eu) has the ground state outer electronic configuration \(4f^7 6s^2\).