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Question #44125

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The Question

A $100 \text{ mL}$ of $0.2 \text{ M}$ $MnSO_4$ aqueous solution is completely oxidized to a deeply purple-colored species by exactly $x$ moles of $NaBiO_3(s)$ in dilute $HNO_3$ at $298 \text{ K}$ . The resulting purple species is then completely reduced by exactly $y$ moles of Mohr's salt, $\ce{FeSO4.(NH4)2SO4.6H2O}$ , in an acidic medium.

Determine the values of $x$ and $y$ , and identify the primary quantum mechanical phenomenon responsible for the exceptional oxidizing power of $NaBiO_3$ compared to its lighter group 15 congeners.

Options

A.
$x = 0.05$ , $y = 0.10$ , Poor shielding by $4f$ and $5d$ electrons
B.
$x = 0.10$ , $y = 0.10$ , Poor shielding by $3d$ and $4d$ electrons
C.
$x = 0.05$ , $y = 0.10$ , High hydration enthalpy of $Bi^{3+}$
D.
$x = 0.02$ , $y = 0.04$ , Poor shielding by $4f$ and $5d$ electrons

Check Your Answer

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

Oxidation State Extremes|Variable Oxidation State Stability|d-orbital contraction

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to integrate quantitative redox stoichiometry with advanced periodic trends.

The 'Hidden Variable' is the $n$ -factor of $NaBiO_3$ , which relies on knowing that Bi(V) exclusively reduces to Bi(III) due to the Inert Pair Effect.

Students often mechanically assign an incorrect $n$ -factor (e.g., assuming reduction to Bi(0)) or fail to recognize that the poor shielding of both 4f (Lanthanoid contraction) and 5d orbitals is the root quantum mechanical cause, falling for distractors involving 3d/4d orbitals or hydration enthalpy.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Oxidation State Extremes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free