NEET 2024 Chemistry Question 80 Solution
Question
Which reaction is NOT a redox reaction?
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Which reaction is NOT a redox reaction?
A redox reaction requires changes in oxidation states! Precipitation of BaSO4 from BaCl2 + Na2SO4 has ZERO change in oxidation numbers (Ba²⁺, Cl⁻, Na⁺, SO4²⁻ throughout) -> NOT a redox reaction!
A redox (reduction-oxidation) reaction is a chemical process in which one or more elements undergo a change in oxidation number resulting from an electron transfer between reacting species:
Assigning oxidation numbers to each atom in BaCl2 + Na2SO4 -> BaSO4 + 2 NaCl: - Barium: +2 in BaCl2 and +2 in BaSO4 (Δ = 0) - Chlorine: -1 in BaCl2 and -1 in NaCl (Δ = 0) - Sodium: +1 in Na2SO4 and +1 in NaCl (Δ = 0) - Sulfur: +6 in Na2SO4 and +6 in BaSO4 (Δ = 0) - Oxygen: -2 in Na2SO4 and -2 in BaSO4 (Δ = 0) Because no element undergoes oxidation or reduction, it is purely a precipitation (ionic partner exchange) reaction, NOT a redox reaction.
Checking the other options to confirm they are indeed redox reactions: - Reaction (1): H increases from 0 to +1 (oxidation); Cl decreases from 0 to -1 (reduction). - Reaction (3): Zn increases from 0 to +2 (oxidation); Cu decreases from +2 to 0 (reduction). - Reaction (4): I increases from 0 to +5 in KIO3 (oxidation); Cl decreases from +5 in KClO3 to 0 in Cl2 (reduction).
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Combination redox reaction involving elemental hydrogen and chlorine.
Double displacement precipitation reaction with zero change in oxidation numbers of any element.
Classic single displacement redox reaction.
Redox displacement where iodine is oxidized to +5 and chlorine is reduced to 0.
Single displacement (like Zn + CuSO4) is ALWAYS redox. Double displacement (precipitation and neutralization) is NEVER redox because ions merely swap partners without electron transfer.
"Precipitation and neutralization reactions are non-redox double displacement reactions because oxidation numbers of all atoms remain unchanged."
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