When one mole of `CoCl_(3).5NH_(3)` was treated with excess of silver nitrate solution, 2 mol of `AgCl` was precipitated. The formula of the compound is:

Options

A. `[Co(NH_(3))_(5)Cl_(2)]Cl`
B. `[Co(NH_(3))_(5)Cl]Cl_(2)`
C. `[Co(NH_(3))_(4)Cl_(2)](NH_(3))Cl`
D. `[Co(NH_(3))_(3)Cl_(3)](NH_(3))_(2)`

Answer

To determine the formula of the compound when one mole of CoCl₃·5NH₃ is treated with excess silver nitrate solution, resulting in the precipitation of 2 moles of AgCl, we can follow these steps:

### Step 1: Understand the Reaction
When CoCl₃·5NH₃ is treated with silver nitrate (AgNO₃), the chloride ions (Cl⁻) from CoCl₃ react with silver ions (Ag⁺) to form silver chloride (AgCl), which precipitates out of the solution.

### Step 2: Identify the Chloride Ions
In the compound CoCl₃·5NH₃, there are 3 chloride ions (Cl⁻) present. When excess AgNO₃ is added, these chloride ions will react with silver ions to form AgCl.

### Step 3: Calculate the Amount of AgCl Precipitated
According to the problem, 2 moles of AgCl are precipitated. This means that only 2 out of the 3 chloride ions are involved in the precipitation reaction.

### Step 4: Determine the Coordination of Ammonia
Since CoCl₃·5NH₃ contains 5 ammonia (NH₃) ligands, it suggests that the cobalt ion (Co) is in a coordination complex. The presence of 5 ammonia ligands indicates that the cobalt is likely in a +3 oxidation state, and the complex can be represented as [Co(NH₃)₅Cl₂]Cl.

### Step 5: Write the Final Formula
The final formula of the compound can be deduced as [Co(NH₃)₅Cl₂]Cl, where 2 Cl⁻ ions are coordinated to the cobalt and 1 Cl⁻ is present as a counterion.

### Conclusion
The formula of the compound is [Co(NH₃)₅Cl₂]Cl.

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