Q19 / 155 · WBJEE 2026 · Chemistry
Which one of the following cations gives a chocolate brown precipitate upon addition of aqueous solution of $K_4[Fe(CN)_6]$?
- A$Fe^{3+}$
- Correct answerB$Cu^{2+}$
- C$Zn^{2+}$
- D$Ca^{2+}$
Solution
Reaction of $Cu^{2+}$ with potassium ferrocyanide, $K_4[Fe(CN)_6]$, yields a chocolate brown precipitate of cupric ferrocyanide.
$2Cu^{2+} + K_4[Fe(CN)_6] \rightarrow Cu_2[Fe(CN)_6] \downarrow + 4K^+$
The other ions give different coloured precipitates with $K_4[Fe(CN)_6]$:
$Fe^{3+}$ gives a Prussian blue precipitate of $Fe_4[Fe(CN)_6]_3$.
$Zn^{2+}$ gives a white precipitate of $Zn_2[Fe(CN)_6]$.
$Ca^{2+}$ gives a white precipitate of $CaK_2[Fe(CN)_6]$.
Answer: $Cu^{2+}$
$2Cu^{2+} + K_4[Fe(CN)_6] \rightarrow Cu_2[Fe(CN)_6] \downarrow + 4K^+$
The other ions give different coloured precipitates with $K_4[Fe(CN)_6]$:
$Fe^{3+}$ gives a Prussian blue precipitate of $Fe_4[Fe(CN)_6]_3$.
$Zn^{2+}$ gives a white precipitate of $Zn_2[Fe(CN)_6]$.
$Ca^{2+}$ gives a white precipitate of $CaK_2[Fe(CN)_6]$.
Answer: $Cu^{2+}$
Hence, the correct answer is option: 'B'.