On treatment of 100 mL of 0.1 M solution of `COCl_(3).6H_(2)O` with excess of `AgNO_(3), 1.2 xx 10^(22)` ions are precipitated. The complex is

Options

A. `[Co(H_(2)O)_(4) Cl_(2)]Cl. 2H_(2)O`
B. `[Co(H_(2)O)_(3)]. 3H_(2) O`
C. `[Co(H_(2)O)_(6)] Cl_(3)`
D. `[Co(H_(2)O)_(5) Cl]Cl_(2). H_(2)O`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of `CoCl3.6H2O`
Given:
- Volume of the solution = 100 mL = 0.1 L
- Molarity of the solution = 0.1 M

Using the formula for moles:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Number of moles} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]

### Step 2: Calculate the number of moles of `AgCl` precipitated
Given:
- Number of ions precipitated = \(1.2 \times 10^{22}\) ions
- Avogadro's number = \(6 \times 10^{23}\) ions/mol

Using the formula to find moles from the number of ions:
\[ \text{Number of moles} = \frac{\text{Number of ions}}{\text{Avogadro's number}} \]
\[ \text{Number of moles of } AgCl = \frac{1.2 \times 10^{22}}{6 \times 10^{23}} = 0.02 \, \text{moles} \]

### Step 3: Relate the moles of `AgCl` to the moles of `Cl^-` ions
Since each mole of `AgCl` corresponds to one mole of `Cl^-` ions, the moles of `Cl^-` ions that precipitated is also 0.02 moles.

### Step 4: Determine the number of ionizable `Cl^-` ions in `CoCl3.6H2O`
From the formula of the complex, we have:
\[ \text{CoCl}_3.6\text{H}_2\text{O} \rightarrow \text{Co}^{3+} + 3 \text{Cl}^- + 6 \text{H}_2\text{O} \]
Let \(y\) be the number of `Cl^-` ions that are ionizable. The total number of moles of `CoCl3.6H2O` is 0.01, and since the ionizable `Cl^-` ions are released, we can set up the equation:
\[ y \times 0.01 = 0.02 \]
Solving for \(y\):
\[ y = \frac{0.02}{0.01} = 2 \]

### Step 5: Determine the structure of the complex
Since \(y = 2\), we have 2 `Cl^-` ions outside the coordination sphere. The remaining `Cl^-` ions (1) and water molecules (5) will be inside the coordination sphere:
\[ \text{Complex} = \text{CoCl}_2 \cdot 5\text{H}_2\text{O} \]

### Final Structure of the Complex
The final structure of the complex can be written as:
\[ \text{[CoCl}_2(\text{H}_2\text{O})_5] \]

### Conclusion
Thus, the complex is \(\text{CoCl}_2 \cdot 5\text{H}_2\text{O}\).

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