The number of electrons involve in the equation `CrO_4^(2-)rarrCr_2O_7^(2-)` Then value of x is equal to

Options

A. 0
B. 1
C. 3
D. 5

Answer

To find the number of electrons involved in the reaction \( \text{CrO}_4^{2-} \rightarrow \text{Cr}_2\text{O}_7^{2-} \), we need to analyze the oxidation states of chromium in both species.

### Step 1: Determine the oxidation state of chromium in \( \text{CrO}_4^{2-} \)

1. In \( \text{CrO}_4^{2-} \), let the oxidation state of chromium be \( x \).
2. The oxidation state of oxygen is \(-2\).
3. The total charge of the ion is \(-2\).

Using the formula for the charge:
\[ x + 4(-2) = -2 \]
\[ x - 8 = -2 \]
\[ x = +6 \]

### Step 2: Determine the oxidation state of chromium in \( \text{Cr}_2\text{O}_7^{2-} \)

1. In \( \text{Cr}_2\text{O}_7^{2-} \), let the oxidation state of chromium be \( y \).
2. There are 2 chromium atoms, so the equation becomes:
\[ 2y + 7(-2) = -2 \]
\[ 2y - 14 = -2 \]
\[ 2y = 12 \]
\[ y = +6 \]

### Step 3: Compare the oxidation states

- The oxidation state of chromium in both \( \text{CrO}_4^{2-} \) and \( \text{Cr}_2\text{O}_7^{2-} \) is \( +6 \).
- Since there is no change in the oxidation state of chromium during the conversion from \( \text{CrO}_4^{2-} \) to \( \text{Cr}_2\text{O}_7^{2-} \), there are no electrons transferred.

### Conclusion

The number of electrons involved in the reaction is \( 0 \).

Thus, the value of \( x \) is equal to \( 0 \).

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