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