Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]
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Approach Solution - 1
Precipitates with \( K_4[{Fe(CN)}_6] \)
Step 1: Identify the Cations that Give Precipitates
The cations that react with \( K_4[{Fe(CN)}_6] \) to form precipitates are:
- \( Cu^{2+} \): Forms a blue precipitate of \( Cu_2[{Fe(CN)}_6] \).
- \( Fe^{3+} \): Forms a blue precipitate of \( Fe_4[{Fe(CN)}_6]_3 \).
- \( Zn^{2+} \): Forms a white precipitate of \( Zn_2[{Fe(CN)}_6] \).
Step 2: Identify the Cations that Do Not Give Precipitates
The following cations do not form characteristic precipitates with \( K_4[{Fe(CN)}_6] \):
- \( Ba^{2+} \)
- \( Ca^{2+} \)
- \( NH_4^+ \)
- \( Mg^{2+} \)
Conclusion
The cations that form characteristic precipitates with \( K_4[{Fe(CN)}_6] \) are:
- \( Cu^{2+} \)
- \( Fe^{3+} \)
- \( Zn^{2+} \)
This gives us a total of 3 cations. Therefore, the correct answer is \( \boxed{(3)} \).
Approach Solution -2
Analysis of Cations with \( K_4[Fe(CN)_6] \)
We are given the following cations and their reaction with \( K_4[Fe(CN)_6] \). Let's determine which cations will form characteristic precipitates:
- Cu\(^{2+}\) (Copper(II) ion): Copper(II) ions react with \( K_4[Fe(CN)_6] \) to form a characteristic yellow precipitate of copper(II) ferrocyanide, \( Cu_2[Fe(CN)_6] \). Therefore, Cu\(^{2+}\) will give a precipitate.
- Fe\(^{3+}\) (Iron(III) ion): Iron(III) ions react with \( K_4[Fe(CN)_6] \) to form a characteristic blue precipitate of Prussian blue, \( Fe_4[Fe(CN)_6]_3 \). Therefore, Fe\(^{3+}\) will give a precipitate.
- Ba\(^{2+}\) (Barium ion): Barium ions do not react with \( K_4[Fe(CN)_6] \) to form a precipitate. Thus, Ba\(^{2+}\) will not give a precipitate.
- Ca\(^{2+}\) (Calcium ion): Calcium ions do not form a precipitate with \( K_4[Fe(CN)_6] \). Therefore, Ca\(^{2+}\) will not give a precipitate.
- NH\(_4^+\) (Ammonium ion): Ammonium ions do not react with \( K_4[Fe(CN)_6] \) to form a precipitate. Therefore, NH\(_4^+\) will not give a precipitate.
- Mg\(^{2+}\) (Magnesium ion): Magnesium ions do not form a characteristic precipitate with \( K_4[Fe(CN)_6] \). Therefore, Mg\(^{2+}\) will not give a precipitate.
- Zn\(^{2+}\) (Zinc ion): Zinc ions react with \( K_4[Fe(CN)_6] \) to form a characteristic white precipitate of zinc ferrocyanide, \( Zn_2[Fe(CN)_6] \). Thus, Zn\(^{2+}\) will give a precipitate.
From the analysis above, the cations that will give characteristic precipitates with \( K_4[Fe(CN)_6] \) are Cu\(^{2+}\), Fe\(^{3+}\), and Zn\(^{2+}\). Therefore, the number of cations that will give a characteristic precipitate is \( \boxed{3} \).
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