$ ext{How many ions are produced from the complex } [Co(NH_3)_6]Cl_2 ext{ in solution ?}$

Q: $\text{How many ions are produced from the complex } [Co(NH_3)_6]Cl_2 \text{ in solution ?}$

To determine how many ions are produced from the complex $[Co(NH_3)_6]Cl_2$ in solution, we need to consider the dissociation of this complex in water. 1. **Dissociation of the Complex:** - The complex $[Co(NH_3)_6]Cl_2$ consists of a coordination complex $[Co(NH_3)_6]^{3+}$ and two chloride ions $Cl^-$. - When this complex dissolves in water, it dissociates into its constituent ions. 2. **Ions Produced:** - The dissociation can be represented as: \[ [Co(NH_3)_6]Cl_2 \rightarrow [Co(NH_3)_6]^{3+} + 2Cl^- \] - From this equation, we can see that one molecule of $[Co(NH_3)_6]Cl_2$ produces one $[Co(NH_3)_6]^{3+}$ ion and two $Cl^-$ ions. 3. **Total Number of Ions:** - Therefore, the total number of ions produced is: - 1 ion of $[Co(NH_3)_6]^{3+}$ - 2 ions of $Cl^-$ - Total = 1 + 2 = 3 ions Thus, the correct answer is Option 3, which states that 3 ions are produced. **Explanation of Incorrect Options:** - **Option 1 (9 ions):** This option is incorrect because it suggests a much higher number of ions than what is actually produced. The complex only dissociates into 3 ions, not 9. - **Option 2 (8 ions):** This option is incorrect because it overestimates the number of ions. The dissociation of $[Co(NH_3)_6]Cl_2$ results in only 3 ions, not 8. - **Option 4 (2 ions):** This option is incorrect because it underestimates the number of ions. The complex dissociates into 3 ions, not just 2. In conclusion, the correct answer is Option 3, as the complex $[Co(NH_3)_6]Cl_2$ dissociates into 3 ions in solution.

Question

$	ext{How many ions are produced from the complex } [Co(NH_3)_6]Cl_2 	ext{ in solution ?}$

Accepted Answer

To determine how many ions are produced from the complex $[Co(NH_3)_6]Cl_2$ in solution, we need to consider the dissociation of this complex in water.

1. **Dissociation of the Complex:**
   - The complex $[Co(NH_3)_6]Cl_2$ consists of a coordination complex $[Co(NH_3)_6]^{3+}$ and two chloride ions $Cl^-$.
   - When this complex dissolves in water, it dissociates into its constituent ions.

2. **Ions Produced:**
   - The dissociation can be represented as:
     $$
     [Co(NH_3)_6]Cl_2 \rightarrow [Co(NH_3)_6]^{3+} + 2Cl^-
     $$
   - From this equation, we can see that one molecule of $[Co(NH_3)_6]Cl_2$ produces one $[Co(NH_3)_6]^{3+}$ ion and two $Cl^-$ ions.

3. **Total Number of Ions:**
   - Therefore, the total number of ions produced is:
     - 1 ion of $[Co(NH_3)_6]^{3+}$
     - 2 ions of $Cl^-$
   - Total = 1 + 2 = 3 ions

Thus, the correct answer is Option 3, which states that 3 ions are produced.

**Explanation of Incorrect Options:**

- **Option 1 (9 ions):** This option is incorrect because it suggests a much higher number of ions than what is actually produced. The complex only dissociates into 3 ions, not 9.

- **Option 2 (8 ions):** This option is incorrect because it overestimates the number of ions. The dissociation of $[Co(NH_3)_6]Cl_2$ results in only 3 ions, not 8.

- **Option 4 (2 ions):** This option is incorrect because it underestimates the number of ions. The complex dissociates into 3 ions, not just 2.

In conclusion, the correct answer is Option 3, as the complex $[Co(NH_3)_6]Cl_2$ dissociates into 3 ions in solution.