JEE Mains · Chemistry · STD 12 - 4. d and f- block elements
The electronic configuration for Neodymium is: [Atomic Number for Neodymium \(60\)]
- A \([\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2\)
- B \([\mathrm{Xe}] 5 \mathrm{f}^4 7 \mathrm{~s}^2\)
- C \([\mathrm{Xe}] 4 \mathrm{f}^6 6 \mathrm{~s}^2\)
- D \([\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2\)
Answer & Solution
Correct Answer
(A) \([\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2\)
Step-by-step Solution
Detailed explanation
Electronic configuration of \(\mathrm{Nd}(\mathrm{Z}=60)\) is; \([\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2\)
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