Question

Choose the correct sets with respective observations: 
(1) \( \text{CuSO}_4 \) (acidified with acetic acid) + \( K_2\text{Fe(CN)}_6 \) (neutralized with NaOH) → Blue precipitate 
(2) \( 2\text{CuSO}_4 \) + \( K_2\text{Fe(CN)}_6 \) → Blue precipitate 
(3) \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) → \( \frac{1}{2}K_4\text{Fe(CN)}_6 \) 
(4) \( 37\text{Cl}_2 \) + \( 2\text{KFe(CN)}_6 \) → 6KC1 
In the light of the above options, choose the correct set:

• A. \( \text{CuSO}_4 \) (acidified with acetic acid) + \( K_2\text{Fe(CN)}_6 \) (neutralized with NaOH) → Blue precipitate
• B. \( 2\text{CuSO}_4 \) + \( K_2\text{Fe(CN)}_6 \) → Blue precipitate
• C. \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) → \( \frac{1}{2}K_4\text{Fe(CN)}_6 \)
• D. \( 37\text{Cl}_2 \) + \( 2\text{KFe(CN)}_6 \) → 6KC1

Hint:

Calculate the empirical formula

Answer 1

The Correct Option is C

To solve the given problem, we need to analyze each chemical reaction and identify which set of reactions is correct. Here are the options:
(1) \( \text{CuSO}_4 \) (acidified with acetic acid) + \( K_2\text{Fe(CN)}_6 \) (neutralized with NaOH) \(\rightarrow\) Blue precipitate
(2) \( 2\text{CuSO}_4 \) + \( K_2\text{Fe(CN)}_6 \) \(\rightarrow\) Blue precipitate
(3) \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) \(\rightarrow\) \( \frac{1}{2}K_4\text{Fe(CN)}_6 \)
(4) \( 37\text{Cl}_2 \) + \( 2\text{KFe(CN)}_6 \) \(\rightarrow\) 6KC1
Consideration of known inorganic chemistry reactions indicates that (3) \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) \(\rightarrow\) \( \frac{1}{2}K_4\text{Fe(CN)}_6 \) is the valid reaction. 

Answer 2

The problem requires us to identify the correct chemical reaction and its associated observation from the given list. This involves knowledge of qualitative inorganic analysis, specifically the precipitation reactions of transition metal ions with complex cyanide anions.

Concept Used:

Precipitation reactions occur when two soluble ionic compounds are mixed, and the resulting combination of ions forms an insoluble product called a precipitate. The color of the precipitate is often a key characteristic used for identifying specific ions.

Two key reactions relevant to the options are:

1. Test for Ferric Ion (\( \text{Fe}^{3+} \)): When a solution containing ferric ions (e.g., from \( \text{FeCl}_3 \)) is treated with a solution of potassium hexacyanoferrate(II) (also known as potassium ferrocyanide, \( \text{K}_4[\text{Fe(CN)}_6] \)), an intensely colored deep blue precipitate called Prussian blue is formed.
2. Test for Copper(II) Ion (\( \text{Cu}^{2+} \)): When a solution containing copper(II) ions (e.g., from \( \text{CuSO}_4 \)) is treated with potassium hexacyanoferrate(II) (\( \text{K}_4[\text{Fe(CN)}_6] \)), a reddish-brown precipitate of copper(II) hexacyanoferrate(II) is formed.

Step-by-Step Solution:

We will analyze each option to determine its correctness based on established chemical principles.

Step 1: Analyze Option (1) and (2)

Both options describe the reaction of copper(II) sulfate (\( \text{CuSO}_4 \)) with a potassium hexacyanoferrate compound, and both state that a "Blue precipitate" is formed. As noted in the concept section, the reaction between \( \text{Cu}^{2+} \) ions and hexacyanoferrate(II) ions (\( [\text{Fe(CN)}_6]^{4-} \)) produces a reddish-brown precipitate. Therefore, the observation of a blue precipitate is incorrect for this reaction.

(Note: The formula \( \text{K}_2\text{Fe(CN)}_6 \) is chemically incorrect and likely a typo for \( \text{K}_4\text{Fe(CN)}_6 \)).

Step 2: Analyze Option (4)

The equation \( 37\text{Cl}_2 + 2\text{KFe(CN)}_6 \rightarrow 6\text{KCl} \) is chemically nonsensical. The starting material \( \text{KFe(CN)}_6 \) is not a standard compound, and the equation is severely unbalanced in terms of atoms and charge. This option can be dismissed as incorrect.

Step 3: Analyze Option (3)

This option presents the reaction between ferric chloride (\( \text{FeCl}_3 \)) and potassium hexacyanoferrate(II) (\( \text{K}_4\text{Fe(CN)}_6 \)). As described in the concept section, these reactants produce the classic Prussian blue precipitate (\( \text{Fe}_4[\text{Fe(CN)}_6]_3 \)). The stoichiometric coefficients \( 4\text{FeCl}_3 \) and \( 3\text{K}_4\text{Fe(CN)}_6 \) are correct for the balanced equation. Although the product side of the equation in the option (\( \frac{1}{2}K_4\text{Fe(CN)}_6 \)) is a clear typographical error, the set of reactants shown is the correct and well-known combination for producing a blue precipitate.

Conclusion:

Comparing the options, options (1) and (2) are incorrect because the reaction of copper(II) ions with ferrocyanide gives a reddish-brown precipitate, not blue. Option (4) is a chemically invalid equation. Option (3) correctly identifies the reactants required for the formation of the Prussian blue precipitate. Despite the error in writing the product, this option represents the only chemically correct system among the choices that leads to a blue precipitate.

Therefore, the correct set is represented by the reactants in option (3): \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \).