What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid ?

Options

A. `Cr^(3+) and Cr_2O_7^(2-)` are formed
B. `Cr_2O_7^(2-) and H_2O` are formed
C. `CrO_4^(2-)` is reduced to +3 state of Cr
D. `CrO_4^(2-)` is oxidised to +7 state of Cr.

Explanation

To solve the problem of what happens when a solution of potassium chromate (K₂CrO₄) is treated with an excess of dilute nitric acid (HNO₃), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are potassium chromate (K₂CrO₄) and dilute nitric acid (HNO₃).

### Step 2: Write the Reaction
When potassium chromate is treated with dilute nitric acid, the reaction can be written as:
\[ K_2CrO_4 + HNO_3 \rightarrow K_2Cr_2O_7 + KNO_3 + H_2O \]

### Step 3: Determine the Products
From the reaction, we can see that the products formed are:
- Potassium dichromate (K₂Cr₂O₇)
- Potassium nitrate (KNO₃)
- Water (H₂O)

### Step 4: Analyze the Oxidation States
In potassium chromate (K₂CrO₄), the oxidation state of chromium (Cr) is +6. In potassium dichromate (K₂Cr₂O₇), the oxidation state of chromium remains +6. Therefore, there is no change in the oxidation state of chromium during this reaction.

### Step 5: Conclusion
The reaction does not produce chromium in the +3 oxidation state or any other oxidation state; it simply converts chromate ions (CrO₄²⁻) into dichromate ions (Cr₂O₇²⁻) without changing the oxidation state of chromium.

### Final Answer
When a solution of potassium chromate is treated with an excess of dilute nitric acid, potassium dichromate (K₂Cr₂O₇) and water (H₂O) are formed, and the oxidation state of chromium remains +6.