Number of electron involved in the reduction of ` Cr_(2)O_(7)^(2-)` ion in acidic solution of `Cr^(3+)` is

Options

A. 4
B. 6
C. 3
D. 4

Explanation

To determine the number of electrons involved in the reduction of the dichromate ion \((Cr_2O_7^{2-})\) to chromium ions \((Cr^{3+})\) in an acidic solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation states of chromium in the reactants and products:**
- In the dichromate ion \((Cr_2O_7^{2-})\), we need to calculate the oxidation state of chromium.
- The overall charge of the dichromate ion is \(-2\). The oxidation state of oxygen is \(-2\), and there are 7 oxygen atoms, contributing a total of \(-14\).
- Let the oxidation state of chromium be \(x\). The equation can be set up as:
\[ 2x + 7(-2) = -2 \]
Simplifying this gives:
\[ 2x - 14 = -2 \implies 2x = 12 \implies x = +6 \]
- Therefore, the oxidation state of chromium in \((Cr_2O_7^{2-})\) is \(+6\).

2. **Determine the oxidation state of chromium in the product:**
- In the product \((Cr^{3+})\), the oxidation state of chromium is \(+3\).

3. **Calculate the change in oxidation state:**
- The change in oxidation state for one chromium atom is:
\[ +6 \text{ (in } Cr_2O_7^{2-}\text{) to } +3 \text{ (in } Cr^{3+}\text{)} = 6 - 3 = 3 \]

4. **Account for the number of chromium atoms:**
- Since there are 2 chromium atoms in the dichromate ion, the total change in oxidation state for both chromium atoms is:
\[ 3 \text{ (change per chromium)} \times 2 \text{ (chromium atoms)} = 6 \]

5. **Determine the number of electrons involved:**
- Each decrease in oxidation state corresponds to the gain of one electron. Therefore, the total number of electrons involved in the reduction of \((Cr_2O_7^{2-})\) to \((Cr^{3+})\) is \(6\).

### Final Answer:
The number of electrons involved in the reduction of \((Cr_2O_7^{2-})\) to \((Cr^{3+})\) is **6**.