Concept:A disproportionation reaction is a specific type of redox reaction where the exact same element is simultaneously oxidized and reduced.
Solution:- Look at the oxidation state of Iodine (I) on the reactant side: \( I_2 \) is in its elemental state, so its oxidation number is \( 0 \).
- On the product side, Iodine exists in two distinct compounds.
- In Sodium Iodate (\( NaIO_3 \)): \( +1 + x + 3(-2) = 0 \implies x = +5 \) (Oxidation).
- In Sodium Iodide (\( NaI \)): \( +1 + x = 0 \implies x = -1 \) (Reduction).
- Because Iodine is both oxidized and reduced simultaneously, this is definitively a redox reaction.
Why other options are incorrect:There is no insoluble solid formed (so not precipitation). There are no free radicals involved. It is an internal electron transfer, not a simple physical substitution.
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