What is the oxidation state of Vanadium in V2O5 ?

In Vanadium pentoxide (V2O5), the oxidation state of Vanadium can be determined using the principle of electroneutrality, where the sum of oxidation states in a neutral molecule is zero.

• Oxygen (O) almost always carries an oxidation state of -2 in its compounds.
• Let the oxidation state of Vanadium (V) be x.
• The chemical formula V2O5 indicates there are 2 atoms of Vanadium and 5 atoms of Oxygen.
• The calculation is: 2(x) + 5(-2) = 0.
• This simplifies to: 2x - 10 = 0, which means 2x = 10.
• Therefore, x = +5.

Vanadium reaches its maximum oxidation state of +5 in this compound. V2O5 is widely known as an industrial catalyst, most notably in the Contact process for the production of sulfuric acid.