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NEET CHEMISTRYThe d-and f-Block ElementsMedium

Question

Identify the pair of Lanthanoides with one strong oxidant and one strong reductant:

A

Ce(IV), Tb(IV)

B

Yb(II), Eu(II)

C

Eu(IV), Lu(III)

D

Ce(IV), Eu(II)

Step-by-Step Solution

In the lanthanoids, the most common and stable oxidation state is +3. Ions that exhibit +2 and +4 states tend to revert to the +3 state to achieve stability. The formation of CeIVCe^{IV} is favoured by its stable noble gas configuration, but it acts as a strong oxidant (oxidising agent) because it readily accepts an electron to revert to the common +3 state. Conversely, Eu2+Eu^{2+} is formed by losing its two s electrons, leaving a highly stable half-filled f7f^7 configuration. However, it acts as a strong reducing agent (reductant) as it loses an electron to change back to the common +3 state . Thus, the pair with one strong oxidant and one strong reductant is Ce(IV) and Eu(II).

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 Elementsidentifylanthanoidesstrongoxidantstrong

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