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JEE Main 202331 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
31 January
Shift
Morning Shift
Subject
Chemistry
Chapter
d and f Block Elements

JEE Main 2023 · 31 January · Morning Shift · Chemistry · d and f Block Elements

Nd ^ 2 + = ______

{Nd}^{2 +} = ______

Options

  1. A4f26s24{f}^{2}6{s}^{2}
  2. B4f44{f}^{4}
  3. C4f34{f}^{3}
  4. D4f46s24{f}^{4}6{s}^{2}

Correct answer

B. 4f44{f}^{4}

Step-by-step solution

The atomic number of Neodymium (Nd) is 60. Its electronic configuration is [Xe]4f46s2\left[Xe\right]4{f}^{4}6{s}^{2}.

Now, by removing two electrons from the outermost shell 6s, Nd2+{Nd}^{2 +} ion is formed. Hence, electronic configuration of Nd2+is[Xe]4f4{Nd}^{2 +} is \left[Xe\right]4{f}^{4}.

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