To determine the charge of the vanadium ion in the compound V2O5, we first need to consider the charges of the oxide ions. In this compound, oxygen typically has a charge of -2.
Since there are five oxygen atoms, the total negative charge contributed by the oxygen ions is:
5 (oxygen) x -2 (charge of each oxygen) = -10
This means that the total positive charge from the vanadium ions must equal +10 to balance out the -10 charge from the oxygen.
There are two vanadium ions in the compound, so to find the charge of one vanadium ion, we divide the total positive charge by the number of vanadium ions:
+10 (total charge from vanadium) / 2 (vanadium ions) = +5
Therefore, the charge of the vanadium ion in V2O5 is +5. This indicates that vanadium in this compound is in the +5 oxidation state.