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Question #105432

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The Question

A $100 \text{ mL}$ aqueous solution of $0.05 \text{ M } \ce{K2Cr2O7}$ is acidified with dilute $\ce{H2SO4}$ and treated with excess $\ce{H2O2}$ in the presence of diethyl ether. The chromium species is quantitatively extracted into the ether layer as a deep blue complex, $\mathbf{X}$ (assume no unreacted $\ce{H2O2}$ transfers to the ether layer).

The ether layer is separated and divided into two equal halves, Part I and Part II. Part I is immediately treated with excess acidified $\ce{KI}$ solution. The liberated iodine requires $V \text{ mL}$ of $0.5 \text{ M } \ce{Na2S2O3}$ for complete titration. Part II is shaken with dilute aqueous acid and allowed to stand until $\mathbf{X}$ completely decomposes into a green aqueous solution and a colorless gas $\mathbf{Y}$ .

Assuming ideal gas behavior, calculate $V$ , the volume of gas $\mathbf{Y}$ evolved at $298 \text{ K}$ and $1 \text{ bar}$ ( $R = 0.08314 \text{ L bar K}^{-1} \text{ mol}^{-1}$ ), and identify the number of peroxy linkages in $\mathbf{X}$ .

Options

A.
$V = 70 \text{ mL}$ , $V_{\mathbf{Y}} = 216.8 \text{ mL}$ , Peroxy linkages = 2
B.
$V = 30 \text{ mL}$ , $V_{\mathbf{Y}} = 216.8 \text{ mL}$ , Peroxy linkages = 2
C.
$V = 70 \text{ mL}$ , $V_{\mathbf{Y}} = 196.0 \text{ mL}$ , Peroxy linkages = 0
D.
$V = 30 \text{ mL}$ , $V_{\mathbf{Y}} = 92.9 \text{ mL}$ , Peroxy linkages = 1

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Peroxy Linkage Identification|Stability of High O.S. Fluorides/Oxides|Redox Reactions - Class 11 Part 2

Why This Question Is Tricky

Examiner Levers

Approximation Bait

Options are extremely close; tests precision vs. early rounding.

Difficulty Boost : +3

Common Mistakes

This question tests the structural reality versus the formal oxidation state of high-oxidation state species like $\ce{CrO5}$ .

A common trap is calculating the n-factor of $\ce{CrO5}$ in iodometry by only considering the reduction of $\ce{Cr^{+6}}$ to $\ce{Cr^{3+}}$ (n-factor = 3), ignoring that the peroxy ligands ( $\ce{O2^{2-}}$ ) are also potent oxidizing agents that reduce to $\ce{O^{2-}}$ , giving a true n-factor of 7.

Furthermore, the decomposition stoichiometry requires balancing both mass and charge to reveal that 4 moles of $\ce{CrO5}$ yield 7 moles of $\ce{O2}$ , trapping students who attempt to balance the equation using incorrect oxidation states or incomplete redox half-reactions.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Peroxy Linkage Identification
Difficulty	5
Cognitive Load	6
Target Solve Time	150 sec
Access	Free