Number of ions produced when `[Cu(NH_3)_4]Cl_2` is dissolved in water is/are

Options

A. (a) `3`
B. (b) `2`
C. (c) `1`
D. (d) `5`

Explanation

To determine the number of ions produced when the complex `[Cu(NH₃)₄]Cl₂` is dissolved in water, we can follow these steps:

### Step 1: Identify the complex and its components
The given complex is `[Cu(NH₃)₄]Cl₂`. Here, `Cu` is the central metal ion, and `NH₃` (ammonia) is a ligand that coordinates to the copper ion. The `Cl₂` indicates that there are two chloride ions associated with this complex.

### Step 2: Determine the charge of the complex
In this complex, the ammonia ligands are neutral, and the copper ion typically has a +2 oxidation state when coordinated with four ammonia molecules. Therefore, the overall charge of the complex ion `[Cu(NH₃)₄]` is +2.

### Step 3: Write the dissociation in water
When `[Cu(NH₃)₄]Cl₂` is dissolved in water, it dissociates into its constituent ions. The dissociation can be represented as follows:

\[ [Cu(NH₃)₄]^{2+} + 2Cl^{-} \]

### Step 4: Count the total number of ions produced
From the dissociation:
- We have 1 complex ion `[Cu(NH₃)₄]^{2+}`.
- We have 2 chloride ions `Cl^{-}`.

Adding these together gives us:
- Total ions = 1 (from the complex) + 2 (from chloride) = 3 ions.

### Conclusion
When `[Cu(NH₃)₄]Cl₂` is dissolved in water, it produces a total of **3 ions**.