A solution of FeCl3, when treated with K4[Fe(CN)6], gives a pruss… — Solution | JEE Chemistry

MCQEasyJEE 2023Salt Analysis (Cations & Anions)

A solution of $FeCl_3$ , when treated with $K_4[Fe(CN)_6]$ , gives a prussian blue precipitate due to the formation of:

Answer

Correct answer:D

Standard Method

Given: A solution of $FeCl_3$ is treated with $K_4[Fe(CN)_6]$ .

Find: The compound responsible for the prussian blue precipitate.

Ferric chloride gives $Fe^{3+}$ ions in solution, while $K_4[Fe(CN)_6]$ gives $[Fe(CN)_6]^{4-}$ ions.

The prussian blue precipitate is the insoluble complex formed by their reaction:

4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow

Therefore, the compound formed is $Fe_4[Fe(CN)_6]_3$ .

The correct option is D. The solution labels option C, but the worked reaction clearly gives $Fe_4[Fe(CN)_6]_3$ , which matches option D.

Confusing ferric ion $Fe^{3+}$ with ferrous ion $Fe^{2+}$ is incorrect because Prussian blue in this test is formed from $Fe^{3+}$ reacting with $[Fe(CN)_6]^{4-}$ . First identify the oxidation state of iron from $FeCl_3$ .
Choosing the option from the listed label in the solution without checking the chemical formula is wrong because the solution text itself forms $Fe_4[Fe(CN)_6]_3$ . Match the derived product with the option content, not only with the printed letter.
Ignoring charge balance in the complex formation can lead to incorrect stoichiometry such as $Fe_3[Fe(CN)_6]_2$ . Balance charges between $4Fe^{3+}$ and $3[Fe(CN)_6]^{4-}$ before selecting the product.

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