What are the oxidation states of chromium in the following :
`CrO_(4)^-2`

Answer

To find the oxidation state of chromium in the chromate ion \( \text{CrO}_4^{2-} \), we can follow these steps:

### Step 1: Assign the oxidation state of oxygen
In most compounds, oxygen has an oxidation state of \(-2\). Since there are four oxygen atoms in the chromate ion, we can calculate the total contribution of oxygen to the charge.

\[ \text{Total oxidation state of oxygen} = 4 \times (-2) = -8 \]

### Step 2: Set up the equation for the total charge
Let the oxidation state of chromium be \( x \). The total charge of the chromate ion is \(-2\). Therefore, we can set up the following equation:

\[ x + \text{Total oxidation state of oxygen} = \text{Total charge} \]

Substituting the values we have:

\[ x - 8 = -2 \]

### Step 3: Solve for \( x \)
Now, we can solve for \( x \):

\[ x - 8 = -2 \]

Adding 8 to both sides gives:

\[ x = -2 + 8 \]

\[ x = +6 \]

### Conclusion
The oxidation state of chromium in the chromate ion \( \text{CrO}_4^{2-} \) is \( +6 \).

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