Medium MCQ +4 / -1 PYQ · JEE Mains 2022

Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?

  1. A It will not prefer to undergo redox reactions.
  2. B It will prefer to gain electron and act as an oxidizing agent. Correct answer
  3. C It will prefer to give away an electron and behave as reducing agent.
  4. D It acts as both, oxidizing and reducing agent.

Solution

Cerium exists in two different oxidation state $+3$, $+4$<br/><br/> $$ \begin{array}{ll} \mathrm{Ce}^{+4}+\mathrm{e}^{-} \rightarrow \mathrm{Ce}^{3+} & \mathrm{E}^{0}=+1.61 \mathrm{~V} \\ \mathrm{Ce}^{+3}+3 \mathrm{e}^{-} \rightarrow \mathrm{Ce} & \mathrm{E}^{0}=-2.336 \mathrm{~V} \end{array} $$<br/><br/> It shows $\mathrm{Ce}^{+4}$ acts as a strong oxidising agent & accepts electron.

About this question

Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals

This question is part of PrepWiser's free JEE Main question bank. 155 more solved questions on d and f Block Elements are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →