The oxidation state of `Cr` in `Cr_(2)O_(7)^(2-)` is

Options

A. `+2`
B. `+6`
C. `-6`
D. `-4`

Answer

To determine the oxidation state of chromium (Cr) in the dichromate ion \( \text{Cr}_2\text{O}_7^{2-} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the formula and oxidation states**:
The formula we are working with is \( \text{Cr}_2\text{O}_7^{2-} \). We know that the oxidation state of oxygen (O) is typically \(-2\).

2. **Set up the equation**:
Let the oxidation state of chromium (Cr) be \( x \). Since there are 2 chromium atoms in the formula, the contribution from chromium will be \( 2x \). The contribution from the 7 oxygen atoms will be \( 7 \times (-2) = -14 \).

3. **Account for the overall charge**:
The overall charge of the dichromate ion is \(-2\). Therefore, we can set up the following equation:
\[ 2x + (-14) = -2 \]

4. **Simplify the equation**:
Rearranging the equation gives:
\[ 2x - 14 = -2 \]

5. **Solve for \( x \)**:
Adding 14 to both sides:
\[ 2x = -2 + 14 \]
\[ 2x = 12 \]
Now, divide both sides by 2:
\[ x = \frac{12}{2} = 6 \]

6. **Conclusion**:
The oxidation state of chromium in \( \text{Cr}_2\text{O}_7^{2-} \) is \( +6 \).

### Final Answer:
The oxidation state of Cr in \( \text{Cr}_2\text{O}_7^{2-} \) is \( +6 \).

---