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Chemistry · d and f Block Elements

JEE Main 2024 — 27 January, Shift 1 — Question 72

The electronic configuration for Neodymium is: [Atomic Number for Neodymium 60]

  1. Option A:

    [Xe]4f46 s2[\mathrm{Xe}] 4 \mathrm{f}^{4} 6 \mathrm{~s}^{2}

    Correct
  2. Option B:

    [Xe]5f47 s2[\mathrm{Xe}] 5 \mathrm{f}^{4} 7 \mathrm{~s}^{2}

  3. Option C:

    [Xe]4f66 s2[\mathrm{Xe}] 4 \mathrm{f}^{6} 6 \mathrm{~s}^{2}

  4. Option D:

    [Xe]4f15 d16 s2[\mathrm{Xe}] 4 \mathrm{f}^{1} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}

Answer: A

Step-by-step solution

Electronic configuration of Nd(Z=60)\mathrm{Nd}(Z=60) is; [Xe]4f46 s2[\mathrm{Xe}] 4 \mathrm{f}^{4} 6 \mathrm{~s}^{2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 1
Subject
Chemistry
Chapter
d and f Block Elements
Topic
Inner transition elements (f-Block elements)

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