MCQ
The electronic configuration for Neodymium is: [Atomic Number for Neodymium $60$]
  • $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$
  • 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

Correct option: A.
$[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$
a
Electronic configuration of $\mathrm{Nd}(\mathrm{Z}=60)$ is; $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$

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