Sodium nitroprusside is used to test `S^(2-)` ion How many `CN^(Θ)` ion acts as ligand in the compound .

Explanation

To solve the question regarding the number of cyanide ions (CN⁻) acting as ligands in sodium nitroprusside, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**:
Sodium nitroprusside is represented as Na₂[Fe(CN)₅(NO)].

2. **Understand the Structure**:
In the compound, the iron (Fe) is coordinated with five cyanide (CN⁻) ions and one nitrosyl (NO) ligand.

3. **Determine the Ligands**:
The ligands in this complex are CN⁻ ions and the NO group. We are specifically interested in the cyanide ligands.

4. **Count the CN⁻ Ligands**:
From the formula Na₂[Fe(CN)₅(NO)], we can see that there are five cyanide ions coordinated to the iron center.

5. **Conclusion**:
Therefore, the number of cyanide ions (CN⁻) acting as ligands in sodium nitroprusside is **5**.

### Final Answer:
The number of CN⁻ ions acting as ligands in sodium nitroprusside is **5**.

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