A complex compound of chromium chromium contains five `NH_(3)` molecules, one nitro group and two chloride ions for one `Cr^(3+)` cation. One molecule of this compound produces three ions in aq. Solution, on reacting with excess of `AgNO_(3)` solution, two moles of `AgCl` get precipitated
Q. The formula of the complex compound is:

Options

A. `[CrCl(NO_2)(NH_(3))_(4)]NH_(3)*Cl`
B. `[CrCl(NH_(3))_(5)]Cl*NO_(2)`
C. `[Cr(NO_(2))(NH_(3))_(5)]Cl_(2)`
D. `[Cr(NH_(3))_(5)]NO_(2)*Cl_(2)`

Answer

To determine the formula of the complex compound of chromium based on the provided information, we will follow these steps:

### Step 1: Identify the components of the complex
The problem states that the complex contains:
- 1 chromium ion (Cr³⁺)
- 5 ammonia molecules (NH₃)
- 1 nitro group (NO₂)
- 2 chloride ions (Cl⁻)

### Step 2: Write the preliminary formula
From the components identified, we can start writing the formula for the complex. The ammonia and nitro group will be part of the coordination sphere, while the chloride ions will be outside the coordination sphere.

The preliminary formula can be written as:
\[ \text{[Cr(NH}_3\text{)}_5\text{(NO}_2\text{)]Cl}_2 \]

### Step 3: Analyze the ionization in aqueous solution
The problem states that one molecule of this compound produces three ions in aqueous solution. This means that when the compound dissolves, it dissociates into three ions.

The dissociation can be represented as:
\[ \text{[Cr(NH}_3\text{)}_5\text{(NO}_2\text{)]}^{+} + 2 \text{Cl}^{-} \rightarrow \text{3 ions} \]

### Step 4: Confirm the precipitation with AgNO₃
When this complex reacts with excess AgNO₃, it produces two moles of AgCl. This indicates that the two chloride ions are indeed outside the coordination sphere, confirming that they are counter ions.

### Step 5: Finalize the formula
Putting all the information together, the final formula for the complex is:
\[ \text{[Cr(NH}_3\text{)}_5\text{(NO}_2\text{)]Cl}_2 \]

### Conclusion
Thus, the formula of the complex compound is:
\[ \text{[Cr(NH}_3\text{)}_5\text{(NO}_2\text{)]Cl}_2 \]

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