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Question #44145
The Question
Consider the central atoms in the following species: (I) \ce{CrO5} (II) \ce{H2SO5} (III) \ce{XeF6} (IV) \ce{ClO2}
Which of the following statements correctly describes the ability of the central atom in these species to undergo disproportionation (for III, specifically into other binary fluorides) under standard conditions, and identifies the underlying structural reasons?
Options
- A.
The central atoms in (I) and (II) cannot disproportionate as they are already in their maximum +6 oxidation states due to peroxy linkages; (III) disproportionates into \ce{XeF4} and \ce{XeF8} because Xe can expand its octet to +8.
- B.
The central atoms in (I), (II), and (III) cannot disproportionate; (I) and (II) are at their maximum +6 oxidation state due to peroxy linkages, and (III) cannot form the sterically and thermodynamically unstable \ce{XeF8}. The central atom in (IV) disproportionates in an alkaline medium.
- C.
The central atoms in (I) and (II) undergo disproportionation because their formal oxidation states (+10 and +8) are highly unstable; (IV) cannot disproportionate as Cl is in a stable +4 state.
- D.
The central atoms in (II) and (IV) disproportionate to yield S(+8)/S(+4) and Cl(+5)/Cl(+3) respectively, while (I) and (III) cannot disproportionate due to the absence of higher accessible oxidation states for Cr and Xe.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
Disproportionation Constraints|Stability of High O.S. Fluorides/Oxides|Peroxy Linkage Identification
Why This Question Is Tricky
Common Mistakes
This question evaluates the student's ability to differentiate between algebraically calculated formal oxidation states and actual structural oxidation states.
A common pedagogical trap is blindly applying the algebraic sum rule, leading to impossible oxidation states (e.g., +10 for Cr in \ce{CrO5} or +8 for S in \ce{H2SO5}).
Recognizing the peroxy linkages reveals the central atoms are already at their maximum group oxidation states (+6), making disproportionation impossible.
Furthermore, the question tests the 'Stability of High O.S.
Fluorides' exception: while Xe can reach a +8 oxidation state with oxygen, \ce{XeF8} is sterically forbidden.
This prevents the disproportionation of \ce{XeF6} into binary fluorides, acting as a secondary trap for students who only consider electronic configurations without steric constraints.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Redox Reactions - Class 11 Part 2 |
| Concept | Disproportionation Constraints |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |