Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ______.

Answer

To determine the oxidation number of chromium (Cr) in the product formed when dichromate ion (\(Cr_2O_7^{2-}\)) is treated with a base, we can follow these steps:

### Step 1: Identify the reactant and the product
The reactant is the dichromate ion, which has the formula \(Cr_2O_7^{2-}\). When it reacts with a base, it converts to chromate ion, which has the formula \(CrO_4^{2-}\).

### Step 2: Write the balanced reaction
In a basic medium, the reaction can be represented as:
\[ Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O \]
This shows that dichromate ion reacts with hydroxide ions to form chromate ions and water.

### Step 3: Determine the oxidation state of chromium in the product
To find the oxidation number of chromium in \(CrO_4^{2-}\), we can set up the following equation based on the known oxidation states of oxygen and the overall charge of the ion.

Let the oxidation number of chromium be \(X\). In \(CrO_4^{2-}\):
- The oxidation number of oxygen (O) is \(-2\).
- There are 4 oxygen atoms, contributing a total of \(4 \times (-2) = -8\).
- The overall charge of the chromate ion is \(-2\).

Setting up the equation:
\[ X + 4(-2) = -2 \]
\[ X - 8 = -2 \]
\[ X = -2 + 8 \]
\[ X = +6 \]

### Conclusion
The oxidation number of chromium in the chromate ion (\(CrO_4^{2-}\)) is \(+6\).

### Final Answer
The oxidation number of Cr in the product formed is **6**.

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