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Question #44886

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following parameters for Chromium and its compounds:

$P =$ Number of equivalent terminal \ce{Cr-O} bonds in the dichromate ion (\ce{Cr2O7^{2-}}). $Q =$ Highest oxidation state of Chromium in a characterized binary fluoride. $R =$ Number of $d\pi-p\pi$ bonds in a monomeric \ce{CrO3} molecule. $S =$ Oxidation state of Chromium in the deep blue complex formed when acidified \ce{K2Cr2O7} reacts with \ce{H2O2} in the presence of ether.

Calculate the value of $(P \times Q) - (R \times S)$ .

Options

A.
12
B.
18
C.
0
D.
6

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

Dichromate-Chromate Equilibrium|Stability of High O.S. Fluorides/Oxides|E° (M²⁺/M) Anomalies

Why This Question Is Tricky

Common Mistakes

This problem tests the limits of oxidation state stability and structural anomalies in transition metals.

The primary trap is assuming that because Cr is in Group 6, it forms a stable \ce{CrF6} ( $Q=6$ ), ignoring the steric and electronic limitations of fluorine compared to oxygen's $p\pi-d\pi$ stabilization.

A secondary trap is the classic miscalculation of the oxidation state in \ce{CrO5} as +10 ( $S=10$ ) by ignoring the peroxo linkages.

Distractors 18, 0, and 6 correspond exactly to falling for one or both of these traps.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	Dichromate-Chromate Equilibrium
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free