In acidic medium, `H_(2)O_(2)` changes `Cr_(2)O_(7)^(2-)` to `CrO_(5)` which has two `(-O-O-)` bonds. Oxidation state of Cr in `CrO_(5)` is

Options

A. `+5`
B. `+3`
C. `+6`
D. `-10`

Explanation

To determine the oxidation state of chromium in CrO5, we can follow these steps:

### Step 1: Write the formula for CrO5
We start with the compound CrO5.

### Step 2: Identify the oxidation state of oxygen
In most compounds, the oxidation state of oxygen is -2. However, in CrO5, there are two types of oxygen atoms: those that are part of the peroxide (O-O) bonds and those that are in the oxide form.

### Step 3: Count the number of oxygen atoms
In CrO5, there are five oxygen atoms in total. Out of these, two are involved in the O-O bonds (peroxide) and three are in the oxide form.

### Step 4: Assign oxidation states to the oxygen atoms
- The two oxygen atoms in the O-O bonds have an oxidation state of -1 each.
- The three oxygen atoms in the oxide form have an oxidation state of -2 each.

### Step 5: Set up the equation for the oxidation state of chromium
Let the oxidation state of chromium be \( X \). The total charge of the compound is neutral (0), so we can write the equation:

\[ X + (3 \times -2) + (2 \times -1) = 0 \]

### Step 6: Simplify the equation
Substituting the values into the equation gives:

\[ X - 6 - 2 = 0 \]

This simplifies to:

\[ X - 8 = 0 \]

### Step 7: Solve for X
Now, solving for \( X \):

\[ X = +8 \]

### Step 8: Correct the oxidation state
However, we need to account for the fact that the oxygen atoms in the peroxide bonds have a different oxidation state. The correct equation should be:

\[ X - 6 = 0 \]

So, solving this gives:

\[ X = +6 \]

### Conclusion
The oxidation state of chromium in CrO5 is +6.