The Question is from neet-chemistry classification-of-elements-and-periodicity-in-prop spdf-block, and it was created by the author THIRUPATHI REDDY

Question

Which of the following lanthanides exhibits the most stable +4 oxidation state?

THIRUPATHI REDDY · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Electronic Configuration and Stability of Oxidation States in Lanthanides
	**Related Topics**:  f-block elements, lanthanide contraction, oxidation states
	**Prerequisites**: Understanding of electronic configurations, stability of half-filled and fully-filled orbitals.
## SOLUTION PATHWAY
**Given Information**:
	Options: Cerium (Ce), Praseodymium (Pr), Neodymium (Nd), Promethium (Pm)
	Required: Lanthanide with most stable +4 oxidation state.
**Method & Steps**:
	1. Analyze the electronic configurations of the given lanthanides.
		Reasoning: The stability of a +4 oxidation state is related to the electronic configuration and the ease of losing four electrons.
		Formula/Rule:  Consider the electronic configuration of the neutral atom and the resulting configuration after losing four electrons.  A stable configuration (like half-filled or fully-filled subshells) will favor the +4 state.
	2. Determine the electronic configurations and assess stability after losing four electrons.
		Working:
			Ce: [Xe] 4f${^1}$5d${^1}$6s${^2}$  → Ce${^{4+}}$: [Xe] (Stable, empty 4f subshell)
			Pr: [Xe] 4f${^3}$6s${^2}$ → Pr${^{4+}}$: [Xe] 4f${^1}$ (Less stable than Ce${^{4+}}$)
			Nd: [Xe] 4f${^4}$6s${^2}$ → Nd${^{4+}}$: [Xe] 4f${^2}$ (Less stable than Ce${^{4+}}$)
			Pm: [Xe] 4f${^5}$6s${^2}$ → Pm${^{4+}}$: [Xe] 4f${^3}$ (Less stable than Ce${^{4+}}$)
		Key Point:  Cerium's +4 oxidation state leads to an empty 4f subshell, which is a particularly stable configuration.
	3. Identify the lanthanide with the most stable +4 oxidation state.
		Result: Cerium (Ce)
		Verification: The empty 4f subshell in Ce${^{4+}}$ explains its relatively high stability compared to the other options.
## OPTION ANALYSIS
A ✓: Cerium (Ce):  The +4 oxidation state of cerium is relatively stable due to the attainment of an empty 4f subshell after losing four electrons. 
B :x: Praseodymium (Pr): The +4 oxidation state is less stable than Ce${^{4+}}$ because the resulting 4f${^1}$ configuration is not as stable.
C :x: Neodymium (Nd): Similar to Pr, the +4 state is less stable due to its 4f${^2}$ configuration.
D :x: Promethium (Pm):  The +4 state is also less stable because of the resulting 4f${^3}$ configuration.
## LEARNING AIDS
	**Common Mistakes**:
		1. Ignoring the electronic configurations:  Focusing solely on the number of electrons lost without considering the resulting stability.
			How to avoid: Carefully write out the electronic configurations of each ion to assess the stability of the resulting electronic structure.
		2. Misunderstanding the stability of subshells:  Not recognizing the exceptional stability of empty, half-filled, or fully-filled subshells.
			How to avoid: Review the rules of electronic configuration and the relative stability of different subshell occupancy.
	**Quick Tips**:
		1. Remember that empty f subshells are particularly stable in lanthanides.
		2.  Make a table comparing the electronic configurations of the neutral atoms and their +4 ions to easily compare their stability.

THIRUPATHI REDDY · Answer

The correct Option is "Cerium (Ce)"

THIRUPATHI REDDY · Answer

Praseodymium (Pr)

THIRUPATHI REDDY · Answer

Neodymium (Nd)

THIRUPATHI REDDY · Answer

Promethium (Pm)

NEET Chemistry Classification of Elements and Periodicity in Properties S,P,D,F-Block MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company