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Question #45028

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The Question

Consider the following three reaction sequences involving metal oxides:

\textbf{Sequence 1:} $1.0 \text{ mol}$ of red lead (\ce{Pb3O4}) is treated with excess dilute \ce{HNO3} at $298 \text{ K}$ . The insoluble dark brown solid residue $X$ is filtered and heated with excess concentrated \ce{H2SO4}, evolving $V_1$ liters of \ce{O2(g)} at STP.

\textbf{Sequence 2:} $1.0 \text{ mol}$ of potassium superoxide (\ce{KO2}) is reacted with excess cold dilute \ce{H2SO4} at $273 \text{ K}$ , evolving $V_2$ liters of \ce{O2(g)} at STP.

\textbf{Sequence 3:} $1.0 \text{ mol}$ of barium peroxide (\ce{BaO2}) is reacted with excess cold dilute \ce{H2SO4} at $273 \text{ K}$ , evolving $V_3$ liters of \ce{O2(g)} at STP.

Assume ideal gas behavior ( $1 \text{ mol} = 22.4 \text{ L}$ at STP).

Which of the following correctly identifies $V_1, V_2, V_3$ and the true oxidation states of Pb in \ce{Pb3O4}?

Options

A.
$V_1 = 11.2 \text{ L}$ , $V_2 = 11.2 \text{ L}$ , $V_3 = 0.0 \text{ L}$ , Pb states: $+2, +4$
B.
$V_1 = 0.0 \text{ L}$ , $V_2 = 11.2 \text{ L}$ , $V_3 = 11.2 \text{ L}$ , Pb state: $+8/3$
C.
$V_1 = 11.2 \text{ L}$ , $V_2 = 22.4 \text{ L}$ , $V_3 = 0.0 \text{ L}$ , Pb states: $+2, +4$
D.
$V_1 = 0.0 \text{ L}$ , $V_2 = 0.0 \text{ L}$ , $V_3 = 11.2 \text{ L}$ , Pb state: $+8/3$

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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

Metal Activity Series Logic|Peroxy Linkage Identification|Fractional Oxidation State Misconception

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to differentiate between dioxides, peroxides, and superoxides, and their respective reactions with acids.

The primary trap is the 'Fractional Oxidation State Misconception' where students might assume \ce{Pb3O4} has a true $+8/3$ state.

Furthermore, students often falsely identify \ce{PbO2} as a peroxide (expecting it to yield \ce{H2O2} and $V_1 = 0$ ) or misbalance the disproportionation of the superoxide \ce{KO2}.

Recognizing that \ce{BaO2} is a true peroxide that yields only \ce{H2O2} ( $V_3 = 0$ ) solidifies the conceptual distinction.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Metal Activity Series Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free