An isomer of the complex `Co(en)_(2)(H_(2)O)IC l_(2)`, on reaction with concentrated `H_(2)SO_(4)`, it suffers loss in weight and on reaction with `AgNO_(3)` solution gives a yellow precipitated which is isoluble in `NH_(3)` solution.
Q. If one mole of original is treated with excess `Pb(NO_(3))_(2)` solution, then the number of moles of white precipitate formed will be

Options

A. `2.0`
B. `1.0`
C. `0.0`
D. `3.0`

Answer

To solve the problem step by step, we will analyze the given coordination compound and its reactions.

### Step 1: Identify the coordination compound
The coordination compound given is `Co(en)₂(H₂O)ICl₂`. Here, `en` stands for ethylenediamine, which is a bidentate ligand. The compound contains cobalt (Co), ethylenediamine, one water molecule (H₂O), one iodide ion (I⁻), and two chloride ions (Cl⁻).

**Hint:** Identify the ligands and their coordination numbers in the complex.

### Step 2: Understand the reaction with concentrated H₂SO₄
When the complex reacts with concentrated sulfuric acid (H₂SO₄), it acts as a dehydrating agent. This means it will remove the water molecule from the coordination sphere. The loss of weight indicates that the water is removed.

**Hint:** Concentrated sulfuric acid is a dehydrating agent that removes water from the coordination sphere.

### Step 3: Analyze the reaction with AgNO₃
The complex, after losing water, is likely to be `Co(en)₂ICl₂`. When this complex is treated with silver nitrate (AgNO₃), a yellow precipitate forms. The yellow precipitate indicates the presence of iodide ions (I⁻), which react with Ag⁺ to form silver iodide (AgI), a yellow precipitate.

**Hint:** The formation of a yellow precipitate suggests the presence of iodide ions in the solution.

### Step 4: Determine the structure of the complex after reactions
After the reactions, the remaining components in the coordination sphere are `Co(en)₂` and the chloride ions (Cl⁻) are still coordinated. The iodide ion (I⁻) is now outside the coordination sphere after the reaction with AgNO₃.

**Hint:** Identify which ligands remain in the coordination sphere after the reactions.

### Step 5: Reaction with Pb(NO₃)₂
Now, if we treat one mole of the original compound `Co(en)₂(H₂O)ICl₂` with excess lead(II) nitrate (Pb(NO₃)₂), we need to consider the chloride ions. However, since the chloride ions are coordinated within the complex, they do not dissociate to form lead(II) chloride (PbCl₂), which is a white precipitate.

**Hint:** Only the anions that are outside the coordination sphere can react to form precipitates.

### Step 6: Conclusion
Since the chloride ions are part of the coordination sphere and do not dissociate, no white precipitate will form when treated with lead(II) nitrate. Therefore, the number of moles of white precipitate formed will be zero.

**Final Answer:** 0 moles of white precipitate will be formed.