One of the product formed when `K_2Cr_2O_7` reacts with conc.`H_2SO_4` in cold is

Options

A. `CrO_3`
B. `Cr_2(SO_4)_3`
C. `Cr_2O_3`
D. `CrO_4Cl_2`

Answer

To solve the question regarding the reaction of potassium dichromate (K₂Cr₂O₇) with concentrated sulfuric acid (H₂SO₄) in cold conditions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are potassium dichromate (K₂Cr₂O₇) and concentrated sulfuric acid (H₂SO₄).

2. **Understand the Reaction Conditions**: The reaction is taking place in cold conditions, which can affect the products formed.

3. **Write the Reaction Equation**: When K₂Cr₂O₇ reacts with concentrated H₂SO₄, the products formed are chromium trioxide (CrO₃), potassium hydrogen sulfate (KHSO₄), and water (H₂O). The balanced chemical equation for this reaction is:
\[ K_2Cr_2O_7 + 4 H_2SO_4 \rightarrow 2 CrO_3 + 2 KHSO_4 + 2 H_2O \]

4. **Identify the Products**: From the balanced equation, we can see that one of the products formed is chromium trioxide (CrO₃).

5. **Evaluate the Options**: Based on the products formed, we can evaluate the options given in the question:
- Option 1: CrO₃ (Correct)
- Option 2: Cr₂SO₄ (Incorrect)
- Option 3: Cr₂O₃ (Incorrect)
- Option 4: CrO₄Cl₂ (Incorrect)

6. **Conclusion**: The product formed when K₂Cr₂O₇ reacts with concentrated H₂SO₄ in cold is CrO₃.

### Final Answer:
The correct answer is **CrO₃**.

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