Question:medium

Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]

Show Hint

In tests with \( {K}_4[{Fe(CN)}_6] \), only certain cations like Cu²⁺, Fe³⁺, and Zn²⁺ give characteristic precipitates. Cations such as Ba²⁺, Ca²⁺, NH₄⁺, and Mg²⁺ typically do not react in this way.
Updated On: Jan 22, 2026
Show Solution

Solution and Explanation

Precipitates with \( K_4[{Fe(CN)}_6] \)

Step 1: Identify Precipitating Cations

Cations forming precipitates with \( K_4[{Fe(CN)}_6] \) are:

  • \( Cu^{2+} \): Forms \( Cu_2[{Fe(CN)}_6] \) (blue).
  • \( Fe^{3+} \): Forms \( Fe_4[{Fe(CN)}_6]_3 \) (blue).
  • \( Zn^{2+} \): Forms \( Zn_2[{Fe(CN)}_6] \) (white).

Step 2: Identify Non-Precipitating Cations

Cations that do not form characteristic precipitates with \( K_4[{Fe(CN)}_6] \) are:

  • \( Ba^{2+} \)
  • \( Ca^{2+} \)
  • \( NH_4^+ \)
  • \( Mg^{2+} \)

Conclusion

The cations yielding characteristic precipitates with \( K_4[{Fe(CN)}_6] \) are \( Cu^{2+} \), \( Fe^{3+} \), and \( Zn^{2+} \).

Total precipitating cations: 3. Thus, the answer is \( \boxed{(3)} \).

Was this answer helpful?
0