Question

The total number of ions produced from the complex \( [Cr(NH_3)_6]Cl_3 \) in aqueous solution will be?

• A. 3
• B. 6
• C. 9
• D. 12

Hint:

When a coordination complex dissociates in water, the total number of ions is equal to the sum of the ions produced from the complex and its counterions (such as \( Cl^- \) or other anions).

Answer 1

The Correct Option is B

We are asked to find the total number of ions produced when the complex \( [Cr(NH_3)_6]Cl_3 \) dissociates in aqueous solution. 
The given complex is \( [Cr(NH_3)_6]Cl_3 \). This complex consists of a central metal ion \( Cr^{3+} \) coordinated with six ammonia molecules (denoted as \( NH_3 \)) and three chloride ions \( Cl^- \) as counterions. 
Step 1: Dissociation in water 
When the complex \( [Cr(NH_3)_6]Cl_3 \) is placed in water, it dissociates into its components. The dissociation reaction is as follows: \[ [Cr(NH_3)_6]Cl_3 \rightarrow [Cr(NH_3)_6]^{3+} + 3Cl^- \] Here: 
- \( [Cr(NH_3)_6]^{3+} \) is the complex ion. 
- The 3 \( Cl^- \) ions are the counterions. 
Step 2: Counting the ions produced 
The dissociation produces: 
- 1 ion of \( [Cr(NH_3)_6]^{3+} \) 
- 3 ions of \( Cl^- \) Thus, the total number of ions produced is: \[ 1 \, \text{(from the complex ion)} + 3 \, \text{(from the chloride ions)} = 6 \, \text{ions} \] Conclusion: 
The total number of ions produced from the dissociation of \( [Cr(NH_3)_6]Cl_3 \) is 6.