Question

In \( \text{Fe}_x[\text{Fe}_y(\text{CN})_6]_3 \), x, y respectively, are

• A. 3, 2
• B. 4, 1
• C. 2, 3
• D. 1, 4

Hint:

- Prussian Blue: Formula \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \). Outer Fe is Fe(III), inner Fe in complex is Fe(II). Comparing to \( \text{Fe}_x[\text{Fe}_y(\text{CN})_6]_3 \): \(x=4\), and \(y=1\) (as there's one Fe atom in the \( [\text{Fe(CN)}_6] \) unit). - Turnbull's Blue: Historically \( \text{Fe}_3[\text{Fe(CN)}_6]_2 \). Outer Fe is Fe(II), inner Fe is Fe(III). This formula does not fit the \( [\dots]_3 \) pattern. It is now known that Prussian Blue and Turnbull's Blue have the same structure. - The notation \( \text{Fe}_y(\text{CN})_6 \) almost certainly means y=1, referring to one Fe atom in the hexacyanoferrate complex. - If the question implies \(y\) is an oxidation state, the format is very unusual. - Given standard coordination chemistry, option (2) \(x=4, y=1\) for Prussian blue is the most chemically sensible interpretation if \(y\) is a stoichiometric index for Fe within the complex unit.

Answer 1

The Correct Option is A

Analysis of Iron Complex

The compound is represented as \( \text{Fe}_x[\text{Fe}_y(\text{CN})_6]_3 \).
This structure could refer to compounds such as Prussian Blue or Turnbull's Blue.

Prussian Blue typically has the formula \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \), with outer Fe in the +3 oxidation state and inner Fe in the +2 oxidation state.

Interpretation of \( x \) and \( y \):

- \( x \) is interpreted as the number of outer Fe atoms in the complex.

- \( y \) refers to the oxidation state of the inner Fe atom.

There is some confusion about whether \( y \) represents the oxidation state or a coefficient for Fe atoms in the complex.

Case 1: Inner Fe is Fe(II):

If inner Fe is Fe(II), the complex becomes \( [\text{Fe(II)}(\text{CN})_6]^{4-} \).

The formula would then be \( \text{Fe}_x[\text{Fe(II)}(\text{CN})_6]_3 \), where outer Fe is in the +3 oxidation state to balance the charge.

Charge Balance:

If \( \text{Fe(II)} \) forms \( [\text{Fe(II)}(\text{CN})_6]^{4-} \), the overall charge of the complex needs to balance, so outer Fe would need to be Fe(III), leading to a formula of:

\[ \text{Fe}_4[\text{Fe(II)}(\text{CN})_6]_3 \]

This is the formula for Prussian Blue, where \( x = 4 \) and the inner Fe is Fe(II).

Case 2: Inner Fe is Fe(III):

If inner Fe is Fe(III), the complex becomes \( [\text{Fe(III)}(\text{CN})_6]^{3-} \).

The formula would then be \( \text{Fe}_x[\text{Fe(III)}(\text{CN})_6]_3 \), where the outer Fe would have to be Fe(II) to balance the charge.

Formula for Prussian Blue and Turnbull's Blue:

The formula for Turnbull's Blue is similar to Prussian Blue, but the oxidation states of the iron atoms differ.

For Prussian Blue, the formula is:

\[ \text{Fe}_4[\text{Fe(CN)}_6]_3 \quad \text{with outer Fe as Fe(III) and inner Fe as Fe(II)}.
\]

For Turnbull's Blue, the formula is:

\[ \text{Fe}_3[\text{Fe(CN)}_6]_2 \quad \text{with outer Fe as Fe(II) and inner Fe as Fe(III)}.
\]

Conclusion:

The formula \( \text{Fe}_x[\text{Fe}_y(\text{CN})_6]_3 \) most likely refers to Prussian Blue, with \( x = 4 \) and \( y = 1 \) for the oxidation state of inner Fe being Fe(II) and outer Fe being Fe(III).