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The gas liberated by heating potassium permanganate (\[KMn{{O}_{4}}\]) is?
(A) Hydrogen
(B) Oxygen
(C) Nitrogen
(D) Methane

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Last updated date: 21st Sep 2024
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Answer
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Hint: Whenever we are heating any chemical then the chemical converts from one form to other forms by liberating some gas. Potassium permanganate (\[KMn{{O}_{4}}\]) is a chemical that exists in the solid-state at room temperature.

Complete step by step solution:
-The structure of the potassium permanganate is as follows.
 
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-The reaction of heating of potassium permanganate is as follows.
  \[KMn{{O}_{4}}\xrightarrow{\Delta }{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}\]
-On heating, potassium permanganate gives potassium manganate, Manganese dioxide and oxygen as the products.
-Coming to given options, option A Hydrogen. In the molecular formula of potassium permanganate (\[KMn{{O}_{4}}\]), there is no hydrogen. So, on heating potassium permanganate, hydrogen gas is not going to form as one of the product. Hence option A is wrong.
-Coming option B, Oxygen. In the molecular formula of potassium permanganate (\[KMn{{O}_{4}}\]), there is a presence of oxygen. So, on heating potassium permanganate, oxygen gas is going to form as one of the product. Hence option B is correct.
-Coming to option C, Nitrogen. In the molecular formula of potassium permanganate (\[KMn{{O}_{4}}\]), there is no nitrogen. So, on heating potassium permanganate, Nitrogen gas is not going to form as one of the product. Hence option C is wrong.
-Coming to option D, Methane. In the molecular formula of potassium permanganate (\[KMn{{O}_{4}}\]) there is no Carbon and hydrogen. So, on heating potassium permanganate, Methane gas is not going to form as one of the product. Hence option D is wrong.
-Therefore on heating potassium permanganate, oxygen gas is going to form as one of the product.

So, the correct option is (B).

Note: The oxidation state of Manganese in potassium permanganate (\[KMn{{O}_{4}}\]) is +7. On heating potassium permanganate (\[KMn{{O}_{4}}\]) gives potassium manganate (\[{{K}_{2}}Mn{{O}_{4}}\]). The oxidation state of manganese in \[{{K}_{2}}Mn{{O}_{4}}\] is +6.