Final answer:
The oxidation number of vanadium (V) in the V4O12^4- ion is +5, which is calculated by balancing the oxygen's oxidation contribution with the overall charge of the ion.
Step-by-step explanation:
We can determine the oxidation number of vanadium (V) in the V4O124- ion using the known oxidation number of oxygen and the overall charge of the ion. The usual oxidation number for oxygen is -2 according to guideline 3. With four vanadium atoms leading to an overall charge of -4 on the ion, the calculation for the oxidation number of vanadium is as follows:
- Total charge contributed by oxygen = number of oxygen atoms × oxidation number of oxygen = 12 × (-2) = -24.
- The total charge of the ion is -4, and therefore, the total oxidation number of the 4 vanadium atoms together must counterbalance the -24 from oxygen to give the overall -4 charge.
- Let the oxidation number of vanadium be x. So, 4x - 24 = -4.
- Solving for x gives us x = (24 - 4) / 4 = 20 / 4 = +5.
Therefore, each vanadium atom has an oxidation number of +5 in the V4O124- ion.