Question

What is the outer electronic configuration of Gd (Gadolinium, atomic number 64)?

• A. [Xe] \( 4f^7 \, 5d^1 \, 6s^2 \)
• B. [Xe] \( 4f^8 \, 6s^2 \)
• C. [Xe] \( 4f^6 \, 5d^2 \, 6s^2 \)
• D. [Xe] \( 4f^7 \, 5d^0 \, 6s^2 \)

Hint:

For lanthanides, prioritize placing electrons to stabilize the \( 4f \) subshell. Gadolinium prefers \( 4f^7 \) over \( 4f^8 \) due to half-filled shell stability.

Answer 1

The Correct Option is A

Step 1: Understand the atomic number
Gadolinium (Gd) has an atomic number of \( Z = 64 \). This means it has 64 electrons in the neutral state
Step 2: Start with the nearest noble gas
The noble gas preceding Gd is Xenon (Xe), which has 54 electrons. So, we begin with: \[ \text{[Xe]} = \text{1s}^2 \, \text{2s}^2 \, \text{2p}^6 \, \text{3s}^2 \, \text{3p}^6 \, \text{4s}^2 \, \text{3d}^{10} \, \text{4p}^6 \, \text{5s}^2 \, \text{4d}^{10} \, \text{5p}^6 \] Step 3: Add remaining electrons after [Xe]
We need to place \( 64 - 54 = 10 \) more electrons. These go into:

• \( 6s \) orbital → 2 electrons → \( 6s^2 \)
• \( 4f \) orbital → 7 electrons → \( 4f^7 \)
• \( 5d \) orbital → 1 electron → \( 5d^1 \)

Step 4: Write the complete configuration
\[ \boxed{\text{[Xe]} \, 4f^7 \, 5d^1 \, 6s^2} \] This is the most stable configuration for Gd because the \( 4f^7 \) is a half-filled subshell, which adds extra stability.