In V₂O₅, oxygen has an oxidation state of −2. Let the oxidation state of vanadium be x. Then 2x + 5(−2) = 0, giving 2x = 10, so x = +5. Hence the oxidation state of vanadium in V₂O₅ is +5. This is also the highest oxidation state of vanadium, consistent with its group 5 position in the periodic table.
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