If `K_(3)[Fe(CN)_(6)]` gets ionized completely in a solution, number of particles in the solution from 1 molecule solute is …………………..

Answer

To solve the question regarding the ionization of \( K_3[Fe(CN)_6] \) in a solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Coordination Compound**:
The compound in question is \( K_3[Fe(CN)_6] \). Here, \( [Fe(CN)_6]^{3-} \) is the coordination sphere, and \( K^+ \) are the counter ions.

2. **Determine the Ionization**:
When \( K_3[Fe(CN)_6] \) is dissolved in water, it dissociates into its constituent ions. The dissociation can be represented as:
\[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \]

3. **Count the Number of Particles**:
From the dissociation equation, we can see that:
- 3 potassium ions (\( K^+ \))
- 1 complex ion (\( [Fe(CN)_6]^{3-} \))

Therefore, the total number of particles produced from 1 molecule of \( K_3[Fe(CN)_6] \) is:
\[ 3 + 1 = 4 \text{ particles} \]

### Final Answer:
The number of particles in the solution from 1 molecule of \( K_3[Fe(CN)_6] \) is **4**.

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