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NEET CHEMISTRYThe d-and f-Block ElementsMedium

Question

A diamagnetic lanthanoid ion among the following is : (At. numbers: Ce=58\text{Ce} = 58, Sm=62\text{Sm} = 62, Eu=63\text{Eu} = 63, Yb=70\text{Yb} = 70)

A

Ce2+\text{Ce}^{2+}

B

Eu2+\text{Eu}^{2+}

C

Sm2+\text{Sm}^{2+}

D

Yb2+\text{Yb}^{2+}

Step-by-Step Solution

A species is diamagnetic if it has no unpaired electrons. The electronic configurations of the given lanthanoid ions are as follows:

  • Ce\text{Ce} (Z=58Z=58): [Xe]4f15d16s2    Ce2+:[Xe]4f2[\text{Xe}] 4f^1 5d^1 6s^2 \implies \text{Ce}^{2+}: [\text{Xe}] 4f^2 (2 unpaired electrons, paramagnetic)
  • Eu\text{Eu} (Z=63Z=63): [Xe]4f76s2    Eu2+:[Xe]4f7[\text{Xe}] 4f^7 6s^2 \implies \text{Eu}^{2+}: [\text{Xe}] 4f^7 (7 unpaired electrons, highly paramagnetic)
  • Sm\text{Sm} (Z=62Z=62): [Xe]4f66s2    Sm2+:[Xe]4f6[\text{Xe}] 4f^6 6s^2 \implies \text{Sm}^{2+}: [\text{Xe}] 4f^6 (6 unpaired electrons, paramagnetic)
  • Yb\text{Yb} (Z=70Z=70): [Xe]4f146s2    Yb2+:[Xe]4f14[\text{Xe}] 4f^{14} 6s^2 \implies \text{Yb}^{2+}: [\text{Xe}] 4f^{14} (0 unpaired electrons, diamagnetic)

Since Yb2+\text{Yb}^{2+} has a completely filled ff-subshell and no unpaired electrons, it is diamagnetic.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from The d-and f-Block Elements. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThe d-and f-Block Elementsdiamagneticlanthanoidfollowingnumberstextce

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