Final answer:
In the reaction KClO₃ → KCl + O₂, the oxidation number for potassium (K) is always +1; for chlorine (Cl), it is +5 in KClO₃ (reactant state) and -1 in KCl (product state); and for oxygen (O), it is -2 in KClO₃ (reactant state) and 0 in O₂ (product state).
Step-by-step explanation:
To determine the oxidation number of each element in the given reaction, which is KClO₃ → KCl + O₂, we can use the rules for assigning oxidation states.
Potassium (K) is always +1 in its compounds, by rule 2.
Oxygen (O) is normally -2 in compounds, by rule 3.
Chlorine (Cl) can vary, but in the KClO₃ molecule, we can find Cl's oxidation number considering that the sum of oxidation states in a neutral molecule should be 0, by rule 6.
Let's calculate the oxidation number of chlorine in KClO₃:
1 K (+1) + 1 Cl (x) + 3 O (-2 each) = 0
+1 + x + (-6) = 0
x = +5
The oxidation number of chlorine in KClO₃ is +5.
In KCl, potassium (K) remains +1, and chlorine (Cl) is -1, which fits the standard pattern of halides formed with alkali metals. In the product O₂, oxygen is in its elemental state, so by rule 1, its oxidation number is 0.
Therefore, the oxidation numbers for the species in the reaction are:
Reactant state: K = +1, Cl = +5, O = -2.
Product state: K = +1, Cl = -1, O (in O₂) = 0.