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Question #44660

Difficulty: 8/10 | 🟢 Free

The Question

A synthetic sample of chromium is prepared by mixing the isotopes \ce{^{50}Cr} and \ce{^{53}Cr} in a $1:2$ molar ratio. This isotopic mixture is used to synthesize the blue-colored, ether-soluble complex Chromium pentoxide (\ce{CrO5}). A $1.32\text{ g}$ sample of this complex is added directly to an acidified $0.5\text{ M}$ aqueous solution of \ce{FeSO4}, where it is completely reduced.

Identify the true oxidation state of Chromium in the \ce{CrO5} complex and the exact volume of the \ce{FeSO4} solution required for complete reduction.

Options

A.
$+6$ and $140\text{ mL}$
B.
$+6$ and $60\text{ mL}$
C.
$+10$ and $140\text{ mL}$
D.
$+10$ and $60\text{ mL}$

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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

Isotopic Average Atomic Mass|Equivalent Concept in Redox|Peroxy Linkage Identification

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of isotopic mass calculations, structural chemistry, and redox stoichiometry.

The primary trap lies in the dual nature of the oxidizing agent.

Students often memorize that the n-factor is determined solely by the change in the oxidation state of the central metal atom.

If a student correctly identifies the oxidation state of Cr as +6 but calculates the n-factor based only on the Cr(+6) to Cr(+3) transition, they will incorrectly use an n-factor of 3 (yielding 60 mL).

Conversely, calculating the oxidation state algebraically without structural knowledge yields a 'fake' +10 state; while this accidentally gives the correct n-factor of 7 (10 - 3 = 7), it demonstrates a fundamental misunderstanding of the molecule's peroxy linkages, which is penalized by the multiple-choice options.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Some Basic Concepts of Chemistry - Class 11 Part 1
Concept	Isotopic Average Atomic Mass
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free