In acidic medium, `H_(2)O_(2)` changes `Cr_(2)O_(7)^(2-)` ion to `CrO_(5)`. The oxidation states of Cr in `Cr_(2)O_(7)^(2-)` and `CrO_(5)` are respectively :

Options

A. `+7,+5`
B. `+6,+6`
C. `+6,+10`
D. `+6,+5`

Answer

To determine the oxidation states of chromium in the compounds \( \text{Cr}_2\text{O}_7^{2-} \) and \( \text{CrO}_5 \), we will follow these steps:

### Step 1: Determine the oxidation state of Cr in \( \text{Cr}_2\text{O}_7^{2-} \)

1. **Identify the formula**: The compound is \( \text{Cr}_2\text{O}_7^{2-} \).
2. **Assign oxidation states**: Let the oxidation state of chromium be \( x \). The oxidation state of oxygen is typically \(-2\).
3. **Set up the equation**:
\[ 2x + 7(-2) = -2 \]
This accounts for the two chromium atoms and seven oxygen atoms contributing to the overall charge of \(-2\).
4. **Solve the equation**:
\[ 2x - 14 = -2 \\ 2x = 12 \\ x = +6 \]
Therefore, the oxidation state of chromium in \( \text{Cr}_2\text{O}_7^{2-} \) is \( +6 \).

### Step 2: Determine the oxidation state of Cr in \( \text{CrO}_5 \)

1. **Identify the formula**: The compound is \( \text{CrO}_5 \).
2. **Assign oxidation states**: Let the oxidation state of chromium be \( y \). In \( \text{CrO}_5 \), four of the oxygen atoms are in a peroxide linkage, which means they have an oxidation state of \(-1\), while one oxygen is in the normal \(-2\) state.
3. **Set up the equation**:
\[ y + 4(-1) + (-2) = 0 \]
This accounts for the chromium and the five oxygen atoms contributing to an overall neutral charge.
4. **Solve the equation**:
\[ y - 4 - 2 = 0 \\ y - 6 = 0 \\ y = +6 \]
Therefore, the oxidation state of chromium in \( \text{CrO}_5 \) is also \( +6 \).

### Conclusion

The oxidation states of chromium in \( \text{Cr}_2\text{O}_7^{2-} \) and \( \text{CrO}_5 \) are both \( +6 \).

### Final Answer
The oxidation states of Cr in \( \text{Cr}_2\text{O}_7^{2-} \) and \( \text{CrO}_5 \) are respectively: \( +6 \) and \( +6 \).

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