Oxidation state exhibited by Cr in `K_(2)CrO_(4)` is

Answer

To find the oxidation state of chromium (Cr) in potassium chromate (K₂CrO₄), we can follow these steps:

### Step 1: Identify the components of the compound
Potassium chromate (K₂CrO₄) consists of:
- 2 potassium (K) ions
- 1 chromium (Cr) ion
- 4 oxygen (O) ions

### Step 2: Assign known oxidation states
- The oxidation state of potassium (K) is +1.
- The oxidation state of oxygen (O) is -2.

### Step 3: Set up the equation based on the overall charge
Since potassium chromate is a neutral compound, the sum of the oxidation states must equal 0. We can express this with the following equation:

\[ \text{Total oxidation state} = (\text{oxidation state of K}) + (\text{oxidation state of Cr}) + (\text{oxidation state of O}) = 0 \]

### Step 4: Substitute the known values into the equation
Let the oxidation state of chromium be \( x \). The equation becomes:

\[ 2(+1) + x + 4(-2) = 0 \]

### Step 5: Simplify the equation
Calculating the contributions from potassium and oxygen:

\[ 2 + x - 8 = 0 \]

### Step 6: Solve for \( x \)
Now, simplify the equation:

\[ x - 6 = 0 \]

Adding 6 to both sides gives:

\[ x = +6 \]

### Conclusion
The oxidation state of chromium in potassium chromate (K₂CrO₄) is +6.

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