The oxidation number of Cr in `K_2Cr_2O_7` is

Options

A. `+6`
B. `-7`
C. `+2`
D. `-2`

Explanation

To find the oxidation number of chromium (Cr) in potassium dichromate, \( K_2Cr_2O_7 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the compound and its components**:
The compound is \( K_2Cr_2O_7 \), which consists of potassium (K), chromium (Cr), and oxygen (O).

2. **Assign known oxidation states**:
- The oxidation state of potassium (K) is +1 (since it is an alkali metal).
- The oxidation state of oxygen (O) is typically -2.

3. **Set up the equation**:
Let the oxidation number of chromium (Cr) be \( x \). Since there are 2 chromium atoms in the formula, the contribution from chromium will be \( 2x \).
The total oxidation state of the compound must equal 0 because it is neutral. Therefore, we can write the equation:
\[ 2(\text{oxidation state of K}) + 2(\text{oxidation state of Cr}) + 7(\text{oxidation state of O}) = 0 \]
Substituting the known values:
\[ 2(+1) + 2x + 7(-2) = 0 \]

4. **Simplify the equation**:
\[ 2 + 2x - 14 = 0 \]
Combine like terms:
\[ 2x - 12 = 0 \]

5. **Solve for \( x \)**:
\[ 2x = 12 \]
\[ x = \frac{12}{2} = 6 \]

6. **Conclusion**:
The oxidation number of chromium (Cr) in \( K_2Cr_2O_7 \) is +6.