Question

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

• 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.

Answer 1

The Correct Option is B

To understand the given question about Cerium (IV), let's consider the properties and electronic configuration of Cerium. Cerium (IV) refers to the Ce⁴⁺ ion. Cerium, with an atomic number of 58, belongs to the lanthanide series. In its +4 oxidation state, Cerium has lost four electrons. This results in the electronic configuration akin to that of the noble gas Xenon (Xe):

• Cerium (Ce) Configuration: [Xe] 4f¹ 5d¹ 6s²
• Ce⁴⁺ Configuration: [Xe]

This noble gas configuration is highly stable due to its full shell. Now, let’s analyze the given options based on this stability and behavior of oxidizing/reducing agents:

• Option 1: "It will not prefer to undergo redox reactions." - While noble gas configurations are stable, Ce⁴⁺ is known for participating in redox reactions.
• Option 2: "It will prefer to gain electron and act as an oxidizing agent." - Ce⁴⁺ can easily gain an electron to form the Ce³⁺ ion, which is more stable due to partially filled 4f orbitals. Thus, it acts as an oxidizing agent, as it accepts electrons.
• Option 3: "It will prefer to give away an electron and behave as reducing agent." - This is incorrect for Ce⁴⁺ as it already has a stable noble gas configuration and will not easily lose an electron.
• Option 4: "It acts as both, oxidizing and reducing agent." - Although redox reactions are possible, Ce⁴⁺ primarily acts as an oxidizing agent; it does not frequently show dual behavior.

Therefore, the correct statement about Cerium (IV) is that "It will prefer to gain electron and act as an oxidizing agent".