Observe the following reaction: $$ 2KClO_3 (s) \xrightarrow{\Delta} 2KCl (s) + 3O_2 (g) $$ In this reaction:

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Step 1: Assign Oxidation Numbers   In potassium chlorate ($ KClO_3 $): - Potassium ($ K $) has an oxidation state of +1. - Oxygen ($ O $) in oxoanions typically has an oxidation state of -2. - Let the oxidation state of chlorine ($ Cl $) be $ x $. Using the sum rule for a neutral compound: $$ (+1) + x + 3(-2) = 0 $$ $$ x - 6 + 1 = 0 $$ $$ x = +5 $$ Thus, in $ KClO_3 $, $ Cl $ has an oxidation state of $ +5 $. In potassium chloride ($ KCl $): - $ K $ is still $ +1 $. - $ Cl $ must be $ -1 $ to balance the charge. Thus, the oxidation state of $ Cl $ changes from $ +5 $ in $ KClO_3 $ to $ -1 $ in $ KCl $, meaning $ Cl $ is reduced.  Step 2: Identify Oxidation of Oxygen   In $ KClO_3 $, oxygen is in the $-2$ oxidation state. In $ O_2 $ gas, oxygen exists in its elemental form, which has an oxidation state of $ 0 $. Since oxygen's oxidation state increases from $-2$ to $ 0 $, oxygen is oxidized.  Step 3: Verify the Correct Answer   - Chlorine is reduced (from $ +5 $ to $ -1 $). - Oxygen is oxidized (from $ -2 $ to $ 0 $). Thus, the correct answer is Option (2): Cl is reduced and O is oxidized.

Answer

Cl is oxidized and O is reduced

Answer

Cl is reduced and O is oxidized

Answer

K is oxidized and O is reduced

Answer

K is reduced and Cl is also reduced

Observe the following reaction: \[ 2KClO_3 (s) \xrightarrow{\Delta} 2KCl (s) + 3O_2 (g) \] In this reaction:

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The Correct Option is B

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