Why does sulphur reacting with concentrated nitric acid give sulphuric acid instead of sulphur dioxide?
(A) It has greater reactivity with acid.
(B) End valency for oxidation $ 6. $
(C) End valency for oxidation $ 4. $
(D) All of the above.
Answer
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Hint :Concentrated nitric acid has a chemical formula as $ HN{O_3} $ . It is a very strong oxidising agent and can oxidise the other substance up to its highest oxidation state. While sulphur is a p-block element and has an atomic number as sixteen. It is considered non-metal. We will see in the solution, the reaction between the sulphur and concentrated nitric acid.
Complete Step By Step Answer:
We know that concentrated nitric acid has a chemical formula as $ HN{O_3} $ . Concentrated nitric acid is a very strong oxidising agent and can oxidise the other substance up to its highest oxidation state.
About sulphur we know that it is a p-block element and has an atomic number as sixteen. It is considered non-metal.
So, the reaction of sulphur with concentrated nitric acid is given as:
$ S + 6HN{O_3} \to {H_2}S{O_4} + 2{H_2}O + 6N{O_2} $
As we can see through the chemical equation that sulphuric acid is formed and not sulphur dioxide. This happens because in sulphur dioxide the oxidation state of sulphur is four and oxidation state in sulphuric acid is six.
Nitric acid being a strong oxidising agent, oxidises sulphur to its highest oxidation state which is $ + 6. $
Hence, the correct option is (B) End valency for oxidation $ 6. $
Note :
There are many uses of concentrated nitric acid such as it is used to produce ammonium nitrate and many fertilizers. It is also used to produce the explosives such as nitroglycerin, trinitrotoluene (TNT). It is also used as a strong oxidising agent in many chemical reactions and synthesis.
Complete Step By Step Answer:
We know that concentrated nitric acid has a chemical formula as $ HN{O_3} $ . Concentrated nitric acid is a very strong oxidising agent and can oxidise the other substance up to its highest oxidation state.
About sulphur we know that it is a p-block element and has an atomic number as sixteen. It is considered non-metal.
So, the reaction of sulphur with concentrated nitric acid is given as:
$ S + 6HN{O_3} \to {H_2}S{O_4} + 2{H_2}O + 6N{O_2} $
As we can see through the chemical equation that sulphuric acid is formed and not sulphur dioxide. This happens because in sulphur dioxide the oxidation state of sulphur is four and oxidation state in sulphuric acid is six.
Nitric acid being a strong oxidising agent, oxidises sulphur to its highest oxidation state which is $ + 6. $
Hence, the correct option is (B) End valency for oxidation $ 6. $
Note :
There are many uses of concentrated nitric acid such as it is used to produce ammonium nitrate and many fertilizers. It is also used to produce the explosives such as nitroglycerin, trinitrotoluene (TNT). It is also used as a strong oxidising agent in many chemical reactions and synthesis.
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