What is the oxidation state of chromium in K2Cr2O7? (2021) 2021

What is the oxidation state of chromium in K2Cr2O7? (2021) 2021

Step 1: Identify the compound K2Cr2O7. It contains potassium (K), chromium (Cr), and oxygen (O).
Step 2: Determine the oxidation state of potassium (K). Each potassium (K) has an oxidation state of +1. Since there are 2 potassium atoms, the total contribution from potassium is 2 * +1 = +2.
Step 3: Determine the oxidation state of oxygen (O). Each oxygen (O) has an oxidation state of -2. Since there are 7 oxygen atoms, the total contribution from oxygen is 7 * -2 = -14.
Step 4: Let the oxidation state of chromium (Cr) be represented as x. There are 2 chromium atoms, so their total contribution is 2x.
Step 5: Set up the equation based on the total charge of the compound. The equation is: 2x (from Cr) - 14 (from O) + 2 (from K) = 0.
Step 6: Simplify the equation: 2x - 14 + 2 = 0 becomes 2x - 12 = 0.
Step 7: Solve for x: 2x = 12, so x = 12 / 2 = 6.
Step 8: Conclude that the oxidation state of chromium (Cr) in K2Cr2O7 is +6.

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