Question

Identify \[\left( {\text{X}} \right){\text{ to }}\left( {\text{Z}} \right)\]
 \[\left( {\text{X}} \right)\xrightarrow{{{\text{KOH}}}}\left( {\text{Y}} \right)\left( {{\text{gas turns red litmus blue}}} \right) + \left( {\text{Z}} \right)\xrightarrow{{{\text{Zn}} + {\text{KOH}}}}\left( {\text{Y}} \right)\left( {{\text{gas}}} \right)\]
\[\left( {\text{X}} \right)\xrightarrow{\Delta }\left( {{\text{does not support combustion}}} \right)\]

A.\[{\text{X}} = {\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_2}{\text{, Y}} = {\text{N}}{{\text{H}}_3}{\text{, Z }} = {\text{ KN}}{{\text{O}}_2}\]
B.\[{\text{X}} = {\left[ {{\text{N}}{{\text{H}}_4}} \right]_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}{\text{, Y}} = {\text{N}}{{\text{H}}_3}{\text{, Z }} = {\text{ C}}{{\text{r}}_2}{{\text{O}}_3}\]
C.\[{\text{X}} = {\left[ {{\text{N}}{{\text{H}}_4}} \right]_2}{\text{S}}{{\text{O}}_4}{\text{, Y}} = {\text{N}}{{\text{H}}_3}{\text{, Z }} = {\text{ }}{{\text{K}}_2}{\text{S}}{{\text{O}}_4}\]
D.\[{\text{X}} = {\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_3}{\text{, Y}} = {\text{N}}{{\text{H}}_3}{\text{, Z }} = {\text{ KN}}{{\text{O}}_3}\]

Answer 1

Hint: The gas Y that is evolved has a very pungent smell and is highly soluble in water. The gas formed out of heating X is a diatomic gas having bond order 3.

Complete step by step solution:
The compound X is ammonium nitrite whose formula is \[{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_2}\]. When we react ammonium nitrite with potassium hydroxide the compound formed is ammonia, potassium nitrite and water. The reaction occurs as: \[{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_2} + {\text{KOH}} \to {\text{N}}{{\text{H}}_3} + {\text{KN}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}\]
The gas Y is ammonia. Ammonia is basic in nature because of the presence of lone pairs of electrons. A base turns red litmus to blue and so ammonia will turn red litmus to blue.
When we react, potassium nitrite with zinc and potassium hydroxide, the ammonia gas is formed again. The reaction occurs as follow: \[{\text{KN}}{{\text{O}}_2}\xrightarrow{{{\text{Zn}} + {\text{KOH}}}}{\text{N}}{{\text{H}}_3} + {{\text{K}}_2}{\text{Zn}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}\]
Hence the same gas is evolved that is ammonia. So the gas Y is ammonia. The compound Z is \[{\text{KN}}{{\text{O}}_2}\]
All the above reaction is given by \[{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_3}\] but the difference is that the combustion of \[{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_3}\] that is ammonium nitrate gives dinitrogen oxide which is combustible in nature. The \[{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_2}\] on combustion gives us dinitrogen gas which is highly stable due to very strong bond in between two nitrogen atoms and hence is non combustible. So all the properties are fulfilled by \[{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_2}\].

Hence the correct option is A.

Note: Ammonia is a Lewis base. A Lewis base is a base which has the tendency to donate electrons to others. Nitrogen has 5 electrons in its valence shell. After forming 3 bonds with hydrogen in ammonia the nitrogen still has one pair of electrons. These lone pairs are available for donation and hence ammonia is a Lewis base.