How many ions are produced from `[Co(NH_(3))_(6)]Cl_(3)` in solution ?

Options

A. `6`
B. `4`
C. `3`
D. `2`

Explanation

To determine how many ions are produced from the coordination compound \([Co(NH_3)_6]Cl_3\) in solution, we can follow these steps:

### Step 1: Identify the Coordination Compound
The given compound is \([Co(NH_3)_6]Cl_3\). This indicates that cobalt (Co) is coordinated with six ammonia (NH₃) ligands.

### Step 2: Determine the Charge on the Complex Ion
In this compound, the ammonia (NH₃) is a neutral ligand, which means it does not contribute any charge. Therefore, the overall charge of the complex ion \([Co(NH_3)_6]^{3+}\) is +3. This is because cobalt typically has a +3 oxidation state in this complex.

### Step 3: Identify the Counter Ions
The compound contains three chloride ions (Cl⁻) as counter ions. Since the formula indicates that there are three chloride ions, we can deduce that they will dissociate in solution.

### Step 4: Write the Dissociation Equation
When \([Co(NH_3)_6]Cl_3\) dissolves in water, it dissociates into its constituent ions:
\[ [Co(NH_3)_6]^{3+} + 3Cl^{-} \]

### Step 5: Count the Total Number of Ions
From the dissociation:
- There is **1 cation**: \([Co(NH_3)_6]^{3+}\)
- There are **3 anions**: \(3Cl^{-}\)

Adding these together gives:
\[ 1 + 3 = 4 \]

### Conclusion
Thus, the total number of ions produced from \([Co(NH_3)_6]Cl_3\) in solution is **4**.