What is the oxidation state of chromium in K2Cr2O7?

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Q: What is the oxidation state of chromium in K2Cr2O7?

Solution: In K2Cr2O7, the total oxidation state of the two potassium ions and seven oxygen ions is -12, leading to +6 for each chromium.

Steps: 9

Step 1: Identify the compound K2Cr2O7. It contains potassium (K), chromium (Cr), and oxygen (O).
Step 2: Determine the oxidation state of potassium (K). Potassium is in Group 1 of the periodic table, so its oxidation state is +1.
Step 3: Since there are 2 potassium ions in K2Cr2O7, the total oxidation state from potassium is 2 * (+1) = +2.
Step 4: Determine the oxidation state of oxygen (O). Oxygen typically has an oxidation state of -2.
Step 5: Since there are 7 oxygen ions in K2Cr2O7, the total oxidation state from oxygen is 7 * (-2) = -14.
Step 6: Now, add the total oxidation states from potassium and oxygen: +2 (from K) + x (from Cr) + (-14) (from O) = 0, because the compound is neutral.
Step 7: Rearrange the equation: x - 12 = 0, where x is the total oxidation state of chromium.
Step 8: Solve for x: x = +12.
Step 9: Since there are 2 chromium atoms in K2Cr2O7, divide +12 by 2 to find the oxidation state of each chromium atom: +12 / 2 = +6.