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

Question

Given below are two statements :
Statement I: The boiling point of hydrides of Group 16 elements follow the order
H₂O > H₂Te > H₂Se > H₂S.
Statement II: On the basis of molecular mass, H₂O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H₂O, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below :

Answer 1

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".

Answer 2

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".

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

The Correct Option is B

Solution and Explanation

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