Number of ions present in `K_4[Fe(CN)_6]`

Options

A. (a) `2`
B. (b) `10`
C. (c) `3`
D. (d) `5`

Explanation

To determine the number of ions present in the compound \( K_4[Fe(CN)_6] \), we will break down the dissociation of the compound step by step.

### Step 1: Identify the components of the compound
The compound \( K_4[Fe(CN)_6] \) consists of:
- 4 potassium ions (\( K^+ \))
- 1 complex ion \( [Fe(CN)_6]^{4-} \)

### Step 2: Understand the dissociation of the compound
When \( K_4[Fe(CN)_6] \) dissociates in solution, it separates into its constituent ions:
- The 4 potassium ions will dissociate as \( 4 K^+ \).
- The complex ion \( [Fe(CN)_6]^{4-} \) remains intact and does not dissociate further into smaller ions.

### Step 3: Count the total number of ions
Now, we will count the total number of ions produced:
- From \( K_4 \), we get 4 potassium ions: \( 4 K^+ \)
- From \( [Fe(CN)_6]^{4-} \), we get 1 complex ion.

Thus, the total number of ions is:
\[ 4 \text{ (from K)} + 1 \text{ (from the complex ion)} = 5 \text{ ions} \]

### Conclusion
Therefore, the total number of ions present in \( K_4[Fe(CN)_6] \) is **5 ions**.

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