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Question #44120

Difficulty: 8/10 | 🟢 Free

The Question

An acidified aqueous solution of \ceK2Cr2O7\ce{K2Cr2O7} reacts with \ceH2O2\ce{H2O2} in the presence of diethyl ether at 273 K273\text{ K} to yield a deep blue colored compound X\mathbf{X}, which is extracted into the organic layer. Consider two other compounds: Y\mathbf{Y} is sodium tetrathionate (\ceNa2S4O6\ce{Na2S4O6}) and Z\mathbf{Z} is tribromooctaoxide (\ceBr3O8\ce{Br3O8}). Let pp be the actual oxidation state of the central metal atom in X\mathbf{X}, qq be the number of peroxy linkages (\ce−O−O−\ce{-O-O-}) present in one molecule of X\mathbf{X}, rr be the magnitude of the difference between the highest and lowest formal oxidation states of the sulfur atoms in Y\mathbf{Y}, and ss be the average oxidation state of the bromine atoms in Z\mathbf{Z}. Evaluate the expression: Ω=(p×q)+r+3s\Omega = (p \times q) + r + 3s.

Options

  • A.

    21

  • B.

    28

  • C.

    33

  • D.

    41

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

Stock Notation Logic|Average vs. Actual Oxidation State|Peroxy Linkage Identification

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to distinguish between average and actual oxidation states derived from molecular structures.

The primary trap is calculating the oxidation state of Cr in CrO5 algebraically as +10, ignoring the two peroxy linkages that limit it to +6.

Secondary traps include assuming all sulfur atoms in Na2S4O6 share the average +2.5 state (which would incorrectly yield r=0) and failing to correctly average the mixed oxidation states in the Br3O8 trimer.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterRedox Reactions - Class 11 Part 2
ConceptStock Notation Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree