$K_{4} [F e (C N)_{6}]$ can be used to identify $F e^{3 +}, Z n^{2 +}$ and $C d^{2 +}$ by forming different colors upon reaction.
If above statement is true, enter 1, else enter 0.

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Solution

The reagent is potassium ferrocyanide $(K_{4} F e (C N)_{6})$ .
(a) Potassium ferrocyanide reacts with Fe(III) ions to form a purssian blue precipitate of ferric ferrocyanide $(F e_{4} [F e (C N)_{6}]_{3})$ .
(b) Potassium ferrocyanide reacts with Zn(II) ions to form bluish white precipitate of $Z n_{2} [F e (C N)_{6}]$ .
(c) Potassium ferrocyanide reacts with Cd(II) ions to form bluish white precipitate of $C d_{2} [F e (C N)_{6}]$ .
(d) Potassium ferrocyanide reacts with Cu(II) ions to form reddish brown precipitate of $C u_{2} [F e (C N)_{6}]$ .
Hence, it can be used to identify different cations like ${F e}^{3 +}, Z n^{2 +}, C d^{2 +}$ etc.

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