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

Answer

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

### Step 1: Write the formula for CrO5
The compound CrO5 consists of one chromium atom and five oxygen atoms.

### Step 2: Identify the oxidation states of oxygen
In CrO5, there are two types of oxygen:
- Two oxygen atoms are in the peroxide form (O-O), which each have an oxidation state of -1.
- The remaining three oxygen atoms are in the oxide form, which each have an oxidation state of -2.

### Step 3: Set up the equation for the oxidation state of chromium
Let the oxidation state of chromium be \( X \). The total charge of the compound must equal zero because it is neutral. Therefore, we can set up the equation as follows:

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

### Step 4: Simplify the equation
Calculating the contributions from the oxygen atoms:
- The three oxide oxygen atoms contribute \( 3 \times -2 = -6 \).
- The two peroxide oxygen atoms contribute \( 2 \times -1 = -2 \).

Now substituting these values into the equation:

\[ X - 6 - 2 = 0 \]

This simplifies to:

\[ X - 8 = 0 \]

### Step 5: Solve for X
Now, solving for \( X \):

\[ X = +8 \]

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