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

Options

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

Explanation

To solve the question regarding Cerium (IV) and its noble gas configuration, we will analyze the properties of Cerium (IV) and determine the correct statement about it.

### Step-by-Step Solution:

1. **Understanding Cerium (IV) Configuration**:
- Cerium (Ce) has an atomic number of 58. Its electron configuration is [Xe] 6s² 4f¹, which means it has two electrons in the 6s subshell and one in the 4f subshell.
- When Cerium is in the +4 oxidation state (Ce⁴⁺), it loses four electrons, resulting in the configuration [Xe], which is the noble gas configuration.

2. **Identifying Oxidizing and Reducing Properties**:
- A species with a noble gas configuration is generally stable and less likely to participate in redox reactions.
- However, Cerium (IV) is known to act as a strong oxidizing agent. This means it can accept electrons from other substances, causing those substances to be oxidized while it itself is reduced.

3. **Evaluating the Redox Behavior**:
- The standard reduction potential (E°) for the half-reaction of Ce⁴⁺ to Ce³⁺ is +1.74 V. A positive E° value indicates that the reaction is favorable, meaning Ce⁴⁺ has a strong tendency to gain electrons and be reduced to Ce³⁺.
- Therefore, while Ce⁴⁺ can act as an oxidizing agent (causing oxidation of other substances), it also has a strong tendency to undergo reduction itself.

4. **Conclusion**:
- The correct statement regarding Cerium (IV) is that it acts as a strong oxidizing agent due to its ability to accept electrons and undergo reduction.

### Final Answer:
Cerium (IV) acts as a strong oxidizing agent due to its noble gas configuration and high positive standard reduction potential.

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