Question

Precipitate X is

• A. Fe₄[Fe(CN)₆]₃
• B. Fe₄[Fe(CN)₆]
• C. K₂Fe[Fe(CN)₆]
• D. KFe[Fe(CN)₆]
• A. \([Fe(NO)_2(SO_4)_2]^{2-}\)
• B. \([Fe(NO)_2(H_2O)_4]^{3+}\)
• C. \([Fe(NO)_4(SO_4)_2]\)
• D. \([Fe(NO)(H_2O)_5]^{2+}\)

Answer 1

The Correct Option is C

Step 1: Reaction of K₃[Fe(CN)₆] with FeSO₄:
When K₃[Fe(CN)₆] reacts with freshly prepared FeSO₄ solution, a dark blue precipitate is formed. This precipitate is called Turnbull’s blue. This reaction occurs when Fe²⁺ ions from FeSO₄ interact with [Fe(CN)₆]^3- ions, resulting in the formation of Turnbull's blue.
The reaction is as follows:
\( K_3[Fe(CN)_6] + FeSO_4 \rightarrow \text{Turnbull’s blue (dark blue precipitate)} \).

Step 2: Reaction of K₄[Fe(CN)₆] with FeSO₄ in the absence of air:
When K₄[Fe(CN)₆] reacts with FeSO₄ solution in the complete absence of air, a white precipitate is initially formed. This precipitate is denoted as X. However, when exposed to air, this white precipitate turns blue, indicating the formation of a complex that contains iron in both +2 and +3 oxidation states. This suggests that X is a compound formed by the interaction of Fe²⁺ and [Fe(CN)₆]^4-.
The precipitate X formed is \( K_2Fe[Fe(CN)_6] \), a white precipitate that turns blue upon exposure to air due to the oxidation of Fe²⁺ to Fe³⁺.

Step 3: Reaction of FeSO₄ with NaNO₃ and H₂SO₄:
When FeSO₄ solution is mixed with NaNO₃, followed by slow addition of concentrated H₂SO₄ through the side of the test tube, a brown ring forms at the junction of the two liquids. This brown ring is due to the formation of the complex ion [Fe(H₂O)₅NO]²⁺, which is a characteristic test for the presence of Fe²⁺ ions in solution.
The reaction is as follows:
\( FeSO_4 + NaNO_3 + H_2SO_4 \rightarrow \text{brown ring} \) (due to the formation of [Fe(H₂O)₅NO]²⁺).

Conclusion:
The precipitate X formed in the absence of air is identified as \( K_2Fe[Fe(CN)_6] \), which is a white precipitate that turns blue upon exposure to air due to the oxidation of Fe²⁺ to Fe³⁺.

Final Answer:
Precipitate X is \( K_2Fe[Fe(CN)_6] \).

Answer 2

Step 1: Reaction of K₃[Fe(CN)₆] with FeSO₄:
When K₃[Fe(CN)₆] reacts with freshly prepared FeSO₄ solution, a dark blue precipitate called Turnbull’s blue is formed. This reaction occurs when Fe²⁺ ions from FeSO₄ react with [Fe(CN)₆]^3- ions. The resulting precipitate is Turnbull’s blue.
The reaction can be represented as:
\( K_3[Fe(CN)_6] + FeSO_4 \rightarrow \text{Turnbull's blue (dark blue precipitate)} \).

Step 2: Reaction of K₄[Fe(CN)₆] with FeSO₄ in the absence of air:
When K₄[Fe(CN)₆] reacts with FeSO₄ solution in the absence of air, a white precipitate X is formed. This precipitate is identified as \( K_2Fe[Fe(CN)_6] \), which remains white in the absence of air but turns blue when exposed to air due to oxidation of Fe²⁺ to Fe³⁺.
The reaction can be written as:
\( K_4[Fe(CN)_6] + FeSO_4 \rightarrow K_2Fe[Fe(CN)_6] \text{ (white precipitate)} \),
which turns blue upon exposure to air.

Step 3: Brown ring test:
When FeSO₄ solution is mixed with NaNO₃, followed by the slow addition of concentrated H₂SO₄ through the side of the test tube, a brown ring forms at the interface of the two liquids.
This brown ring is due to the formation of the complex ion \([Fe(NO)(H_2O)_5]^{2+}\), which is characteristic of Fe²⁺ ions in the presence of nitric acid and sulfuric acid.
The reaction responsible for the brown ring formation is as follows:
\( FeSO_4 + NaNO_3 + H_2SO_4 \rightarrow [Fe(NO)(H_2O)_5]^{2+} \) (brown ring at the interface).

Final Answer:
The brown ring is due to the formation of \([Fe(NO)(H_2O)_5]^{2+}\), which is the characteristic complex ion formed in the brown ring test.