What is the primary oxidation state of vanadium in V2O5? (2023)

What is the primary oxidation state of vanadium in V2O5? (2023)

Step 1: Identify the compound V2O5. It consists of vanadium (V) and oxygen (O).
Step 2: Know that oxygen typically has an oxidation state of -2.
Step 3: Since there are 5 oxygen atoms in V2O5, calculate the total oxidation state contributed by oxygen: 5 oxygen x -2 = -10.
Step 4: Let the oxidation state of vanadium be x. There are 2 vanadium atoms, so the total oxidation state from vanadium is 2x.
Step 5: Set up the equation for the total oxidation states: 2x + (-10) = 0 (the compound is neutral).
Step 6: Solve the equation: 2x - 10 = 0, which simplifies to 2x = 10.
Step 7: Divide both sides by 2 to find x: x = 10 / 2 = +5.
Step 8: Conclude that the primary oxidation state of vanadium in V2O5 is +5.

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