In coordination chemistry there are a variety of methods applied to find out the structure of complexes.One method involves treating the complex with known reagents and from the nature of reaction, the formula of the complex can be predicted.An isomer of the six coordination number complex `Co(en)_2(H_2O)Cl_2Br`,on reaction with concentrated `H_2SO_4` (dehydrating agent) it suffers loss in weight and on reaction with `AgNO_3` solution it gives a white precipitate which is soluble in `NH_3`(aq).
If one mole of original complex is treated with excess `Pb(NO_3)_2` solution, then the number of moles of white precipitate (of `PbCl_2`) formed will be :

Options

A. 0.5
B. `1.0`
C. `0.0`
D. `3.0`

Explanation

To solve the problem, we need to analyze the coordination complex `Co(en)2(H2O)Cl2Br` and the reactions it undergoes.

### Step-by-Step Solution:

1. **Identify the Coordination Complex**:
The coordination complex is `Co(en)2(H2O)Cl2Br`. Here, `Co` is the central metal ion, `en` (ethylenediamine) is a bidentate ligand, and there are two chloride ions (Cl-) and one bromide ion (Br-) along with one water molecule (H2O).

2. **Determine the Isomer**:
The problem states that the complex is an isomer and upon reaction with concentrated `H2SO4`, it loses weight. This suggests that the water molecule is not part of the coordination sphere but is a free ligand. Therefore, the structure can be represented as:
\[ [Co(en)_2(Cl)_2(Br)] + H_2O \]
This indicates that the two Cl- ions are part of the coordination sphere, while the water molecule is outside.

3. **Reaction with `AgNO3`**:
When the complex reacts with `AgNO3`, it produces a white precipitate of `AgCl`. The formation of this precipitate indicates that Cl- ions are available in the free state. Since the complex has two Cl- ions, we can conclude that one of them is free and reacts with `AgNO3` to form AgCl, which is soluble in `NH3`.

4. **Reaction with `Pb(NO3)2`**:
Now, when the complex is treated with excess `Pb(NO3)2`, we need to determine how many moles of `PbCl2` will be formed. The reaction can be summarized as:
\[ Pb(NO_3)_2 + 2Cl^- \rightarrow PbCl_2 \downarrow + 2NO_3^- \]
Since we have established that there is only **one free Cl- ion** from the complex, we can deduce how many moles of `PbCl2` will form.

5. **Calculate Moles of `PbCl2`**:
From the reaction stoichiometry, 2 moles of Cl- are required to produce 1 mole of `PbCl2`. Therefore, if we have only 1 mole of Cl-, it will yield:
\[ \text{Moles of } PbCl_2 = \frac{1 \text{ mole of } Cl^-}{2} = 0.5 \text{ moles of } PbCl_2 \]

### Final Answer:
The number of moles of white precipitate (of `PbCl2`) formed will be **0.5 moles**.

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