The number of ions given by `[Pt(NH_(3))_(6)]Cl_(4)` in aqueous solution will be

Options

A. two
B. three
C. five
D. eleven .

Explanation

To determine the number of ions produced by the coordination compound \([Pt(NH_3)_6]Cl_4\) in aqueous solution, we can follow these steps:

### Step 1: Identify the Coordination Compound
The given compound is \([Pt(NH_3)_6]Cl_4\). This indicates that platinum (Pt) is surrounded by six ammonia (NH₃) ligands and is associated with four chloride (Cl⁻) ions.

### Step 2: Understand the Ionization in Water
When \([Pt(NH_3)_6]Cl_4\) is dissolved in water, it dissociates into its constituent ions. The coordination complex \([Pt(NH_3)_6]^{4+}\) will remain intact as a cation, while the chloride ions will dissociate as anions.

### Step 3: Write the Dissociation Equation
The dissociation of \([Pt(NH_3)_6]Cl_4\) in water can be represented as follows:
\[ [Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^{-} \]

### Step 4: Count the Ions
From the dissociation equation, we can see that:
- There is **1 cation**: \([Pt(NH_3)_6]^{4+}\)
- There are **4 anions**: \(4Cl^{-}\)

### Step 5: Total Number of Ions
To find the total number of ions produced in the solution, we add the number of cations and anions:
\[ \text{Total number of ions} = 1 \text{ (cation)} + 4 \text{ (anions)} = 5 \]

### Conclusion
Thus, the total number of ions given by \([Pt(NH_3)_6]Cl_4\) in aqueous solution is **5**.