Boiling `CuCl_(2)` with Cu in conc. HCl gives:

Options

A. `CuCl`
B. `CuCl_(2)`
C. `H[CuCl_(2)]`
D. `Cu_(2)Cl`

Answer

To solve the question regarding the reaction of boiling CuCl₂ with Cu in concentrated HCl, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Copper(II) chloride (CuCl₂)
- Copper (Cu)
- Concentrated hydrochloric acid (HCl)

### Step 2: Understand the Reaction Conditions
When CuCl₂ is boiled with Cu in concentrated HCl, the presence of concentrated HCl indicates that we may be forming a complex ion. The reaction typically involves the reduction of Cu²⁺ ions from CuCl₂ by Cu.

### Step 3: Write the Reaction
The reaction can be represented as follows:
\[ \text{CuCl}_2 + \text{Cu} + 2 \text{HCl} \rightarrow \text{HCuCl}_2 \]

### Step 4: Identify the Product
The product formed in this reaction is known as hydrogen dichloride cuprate, which can be represented as HCuCl₂. This complex indicates that copper is in a +1 oxidation state in the product.

### Step 5: Choose the Correct Option
From the options provided:
1. CuCl
2. Cu₂Cl₂
3. HCuCl₂
4. Cu₂Cl

The correct answer is **option 3: HCuCl₂**.

### Final Answer
Boiling CuCl₂ with Cu in concentrated HCl gives **HCuCl₂**.

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