What is the oxidation state of Vanadium in V₂O₅?
- (a)+2
- (b)+4
- (c)+3
- (d)+5
Correct — D, plus five. In V2O5 each oxygen carries its usual oxidation state of minus two, so the five oxygen atoms contribute a total of minus ten. Since the compound is electrically neutral, the two vanadium atoms together must balance this with plus ten, giving each vanadium an oxidation state of plus five. This is why the compound is named vanadium pentoxide and represents vanadium in its highest common oxidation state.
- (a)+2 — Plus two is the oxidation state of vanadium in the monoxide VO, not in V2O5; here the arithmetic (two vanadium balancing minus ten from oxygen) forces plus five.
- (b)+4 — Plus four is vanadium's state in the dioxide VO2, but in V2O5 the oxygen total of minus ten requires each vanadium to be plus five, not plus four.
- (c)+3 — Plus three is vanadium's state in the sesquioxide V2O3; it does not balance the charge in V2O5, where each metal atom must be plus five.
The oxidation state of an element in a neutral compound is found by fixing the states of the other atoms and making the total zero. Oxygen is almost always minus two, so in any oxide the metal's oxidation state is set by how many oxygens balance how many metal atoms. Vanadium is a transition metal that shows several oxidation states, reaching a maximum of plus five in V2O5.
Working from oxygen, five oxygens at minus two is minus ten, which the two vanadium atoms must offset by plus ten, so each vanadium is plus five. A neat check is that vanadium's four oxides map to four states, VO is plus two, V2O3 is plus three, VO2 is plus four and V2O5 is plus five, so the distractors are all genuine vanadium states in other oxides.
- Oxygen takes an oxidation state of minus two in almost all oxides (peroxides and OF2 are exceptions).
- For a neutral compound the oxidation states of all atoms sum to zero.
- Vanadium pentoxide, V2O5, is an orange-yellow solid used as the catalyst in the Contact process that makes sulphuric acid.
- Vanadium's common oxidation states are plus two, plus three, plus four and plus five.
- Reading '5' in the formula as the oxidation state instead of computing it.
- Forgetting that the two vanadium atoms share the plus-ten balance, so each is plus five.
- Mixing up which vanadium oxide (VO, V2O3, VO2, V2O5) corresponds to which state.
Exams give a formula such as V2O5, KMnO4 or K2Cr2O7 and ask the oxidation state of the metal — assign oxygen minus two, potassium plus one, and solve for the rest.
No directly related past PYQ was found.
- practice — not a real PYQ
What is the oxidation state of manganese in KMnO4?
- (a)+2
- (b)+4
- (c)+6
- (d)+7
Answer(d) +7 — potassium is +1 and four oxygens give -8, so manganese must be +7 to make the compound neutral.
- practice — not a real PYQ
In which one of the following compounds does vanadium show the oxidation state +4?
- (a)VO
- (b)V2O3
- (c)VO2
- (d)V2O5
Answer(c) VO2 — two oxygens give -4, so the single vanadium is +4; VO is +2, V2O3 is +3 and V2O5 is +5.