What NEET Asks
- Identification: Recognize reactions where a single element undergoes both oxidation and reduction.
- Oxidation State Calculation: Crucial for determining if an element can disproportionate.
- Examples: Be familiar with common examples like H₂O₂, Cl₂, P₄, S in specific mediums.
Key Points
- A disproportionation reaction is a type of redox reaction.
- The same element in a single reactant is simultaneously oxidized and reduced.
- The element must exist in an intermediate oxidation state to disproportionate.
- It must be able to achieve both higher and lower oxidation states from its initial state.
- Elements in their highest or lowest possible oxidation state cannot disproportionate.
- Examples include H₂O₂ (oxygen), Cl₂ (chlorine) in alkaline solutions, P₄ (phosphorus) in alkali.
Must-Know Formula / Reaction
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
Here, Oxygen in H₂O₂ has an oxidation state of -1.
In H₂O, Oxygen is -2 (reduced).
In O₂, Oxygen is 0 (oxidized).
Common Mistakes
- Students often confuse disproportionation with intermolecular redox reactions where different elements change oxidation states.
- Don't forget to check the oxidation state of the same element in all products.
- Overlooking the reaction conditions (acidic vs. basic medium) which can dictate whether disproportionation occurs and what products are formed.
Rapid Revision
Disproportionation: one element, one reactant, simultaneously oxidized and reduced. Requires intermediate oxidation state. Look for elements like O in H₂O₂, Cl in Cl₂ (base), P₄ (base), S (base). Cannot happen if element is in max/min O.S.