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Q67
(NDA-I/2024)
Science & Technology › Basic Science (Physics, Chemistry, Biology)
Official Key
What is the oxidation state of Vanadium in V2O5 ?
Result
Your answer:
—
·
Correct:
D
Explanation
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.
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SIMILAR QUESTIONS
In KMn04 molecule, the oxidation states of the elements potassium (K), manganese (Mn) and oxygen (O) are respectively
Which one of the following nitrogen oxides has the highest oxidation state of nitrogen?
A) NO
B) NO₂
C) N₂O
D) N₂O₅