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Question #44669

Difficulty: 8/10 | 🟢 Free

The Question

A deep blue solid $X$ is precipitated when an aqueous solution of iron(III) chloride is treated with potassium hexacyanoferrate(II) at 298 K. Let $n_{inner}$ and $n_{outer}$ be the number of unpaired electrons present on a single iron atom in the inner coordination sphere and outer ionization sphere of $X$ , respectively. What is the value of $|n_{outer} - n_{inner}|$ and the average oxidation state of iron in the anhydrous empirical formula of $X$ ?

Options

A.
5 and $+\frac{18}{7}$
B.
3 and $+\frac{12}{5}$
C.
1 and $+\frac{18}{7}$
D.
5 and $+\frac{8}{3}$

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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

Empirical vs Molecular Formula|Complex Ion Oxidation States|Fractional Oxidation State Misconception

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to distinguish between the stoichiometric average oxidation state and the actual localized oxidation states in a mixed-valence complex.

A common trap is calculating a bulk average (like +8/3 for Fe3O4) or confusing the reaction with the formation of Turnbull's Blue (historically written as Fe3[Fe(CN)6]2), which would incorrectly yield an average oxidation state of +12/5 and a spin difference of 3.

Furthermore, it rigorously tests Crystal Field Theory (CFT) by requiring the evaluation of the spin state of the inner Fe(II) (low spin due to C-coordinated strong field CN-) versus the outer Fe(III) (high spin due to N-coordinated weak field).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Some Basic Concepts of Chemistry - Class 11 Part 1
Concept	Empirical vs Molecular Formula
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free