JEE Chemistry 2024 Question with Solution

Given: Two statements about $\text{Ce}^{4+}$ in lanthanides.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

For Statement I: Cerium has atomic number $58$ and its electronic configuration is

$$[\text{Xe}]4f^15d^16s^2$$

On forming $\text{Ce}^{4+}$, four electrons are removed, giving

$$[\text{Xe}]$$

This is a noble gas configuration, so the formation of $\text{Ce}^{4+}$ is favored. Therefore, Statement I is true.

For Statement II: $\text{Ce}^{4+}$ readily reverts to the more common $\text{Ce}^{3+}$ state and therefore behaves as a strong oxidizing agent. Hence, Statement II is also true.

So, both statements are true. The solution also explains that although one header marks option A, the written conclusion states both Statement I and Statement II are true, which matches option B in the given options.

Therefore, the correct option is B.

Given: Statements about stability and oxidizing nature of $\text{Ce}^{4+}$.

Find: The correct option.

Cerium in elemental form is written as

$$[\text{Xe}]4f^15d^16s^2$$

When it forms $\text{Ce}^{4+}$, it loses four electrons and becomes

$$[\text{Xe}]$$

Since this is a noble gas configuration, Statement I is true.

Also, $\text{Ce}^{4+}$ has a strong tendency to gain an electron and convert to $\text{Ce}^{3+}$. That is why the $\text{Ce}^{4+}/\text{Ce}^{3+}$ couple behaves as a strong oxidizing agent. Hence, Statement II is true.

Therefore, both Statement I and Statement II are true, so the correct option is B.