What is the oxidation state of vanadium in V2O5?

• A. +3
• B. +4
• C. +5
• D. +6

Answer: (C)

Oxidation state is found by assigning typical charges to atoms so the total matches the compound’s overall charge. In V2O5, each oxide ion carries -2, so five oxygens contribute -10. The compound is neutral, so the vanadium atoms must sum to +10. There are two vanadium atoms, and since they’re the same element in the same compound, they share that total equally, giving each V an oxidation state of +5. You can check with the equation 2(V) + 5(-2) = 0, which simplifies to 2V = 10, so V = +5. Only +5 makes the overall charge zero; the other options don’t balance to zero.