Oxidation number of Cr in `CrO_5` is same as of S in

Options

A. `H_2SO_5`
B. `Na_2S_4O_6`
C. `Na_2S_2O_7`
D. `H_2SO_3`

Answer

To determine the oxidation number of chromium (Cr) in CrO5 and find a sulfur (S) compound with the same oxidation state, we will follow these steps:

### Step 1: Determine the oxidation state of Cr in CrO5
1. **Identify the oxidation states of oxygen**: In CrO5, there are 5 oxygen atoms. Four of these are involved in peroxy linkages, which means their oxidation state is -1. One oxygen is doubly bonded to chromium, which has an oxidation state of -2.

2. **Set up the equation**: Let the oxidation state of chromium be \( X \). The equation based on the oxidation states will be:
\[ X + 4(-1) + (-2) = 0 \]
This simplifies to:
\[ X - 4 - 2 = 0 \]
\[ X - 6 = 0 \]

3. **Solve for X**:
\[ X = +6 \]
Therefore, the oxidation state of Cr in CrO5 is +6.

### Step 2: Find a sulfur compound with the same oxidation state
Now, we need to find a sulfur compound where the oxidation state of sulfur is also +6.

1. **Consider H2SO7 (disulfur heptoxide)**:
- The oxidation state of sulfur can be calculated as follows:
- Let the oxidation state of sulfur be \( Y \). The equation will be:
\[ 2(+1) + 2Y + 7(-2) = 0 \]
This simplifies to:
\[ 2 + 2Y - 14 = 0 \]
\[ 2Y - 12 = 0 \]
\[ 2Y = 12 \Rightarrow Y = +6 \]

2. **Conclusion**: The oxidation state of sulfur in H2SO7 is also +6, which matches the oxidation state of chromium in CrO5.

### Final Answer:
The oxidation number of Cr in CrO5 is the same as that of S in H2SO7.