Question

Which of the following gasses does not give a precipitate with an ammonical solution of silver nitrate but decolorizes \[KMn{{O}_{4}}\] (neutral or slightly alkaline)
A. Ethane
B. Methane
C. Ethene
D. Acetylene

Answer 1

Hint: Potassium permanganate does not undergo decolorization in presence of ethane and methane and these two gases do not react with ammoniacal cuprous chloride.
Acetylene decolorizes the permanganate solution and also provides a red color precipitate with ammoniacal cuprous chloride.

Complete step by step solution:Here in this question, we have to find a solution that decolorizes the KMnO4 solution that provides no precipitate with ammoniacal cuprous chloride.
Let us know about potassium permanganate first.
Alkaline potassium permanganate is utilized as an oxidizing agent.
It means it has a preference to reduce itself and oxidize other compounds.
It accepts electrons from other chemicals.
It is a purple solution.
The Mn possesses a +7 oxidation state which is responsible for its reducing character.
When an unsaturated compound is treated with KMnO4, it undergoes decolorization.
For alkene, the below reaction occurs.
\[2KMn{{O}_{4}}+{{H}_{2}}O\to 2KOH+2Mn{{O}_{2}}+3\left[ O \right]\]
\[-C=C-+{{H}_{2}}O\to -C\left( OH \right)-C\left( OH \right)-+\text{ }KOH\]
Alkanes do not decolorize this solution.
So, options A. Ethane and B. Methane are not correct.
So, ethene and acetylene decolorize this solution.
Let us know about the ammoniacal solution of silver nitrate.
This is also called Tollens reagent \[\left[ Ag{{\left( N{{H}_{3}} \right)}_{2}} \right]N{{O}_{3}}\] and is utilized to test for the existence of aldehydes.
\[R-CHO+2A{{g}^{+}}+2O{{H}^{-}}\to R-COOH+2Ag+{{H}_{2}}O\]
It reacts with the aldehyde to form a carboxylic acid. The silver ions in tollens reagent get transformed into metallic silver which creates a silver mirror underneath the test tube.
Acetylene when treated with ammoniacal silver nitrate solution forms a white ppt of silver acetylide (\[{{C}_{2}}A{{g}_{2}}\]).
\[{{C}_{2}}{{H}_{2}}+2AgN{{O}_{3}}+2N{{H}_{4}}OH\to {{C}_{2}}A{{g}_{2}}+2N{{H}_{4}}N{{O}_{3}}+2{{H}_{2}}O\]
So, acetylene decolorizes KMnO4 and also forms a silver precipitate with an ammoniacal solution of silver nitrate.
So, D is incorrect.
So, ethene is the gas that does not provide a precipitate with an ammoniacal solution of silver nitrate but decolorizes \[KMn{{O}_{4}}\] (neutral or slightly alkaline).

So, option C is correct.

Note: Acetylene reacts with ammoniacal silver nitrate solution because the two hydrogen atoms attached to the carbon atoms are acidic.
These atoms can be replaced easily by silver atoms forming silver acetylide. It's because the s-character of the sigma electron density of the C-H bond in acetylene is \[50%\].
The electrons possess less energy near the nucleus and the s-orbital is spherical.