Question

\[Al + {N_2}\xrightarrow{\Delta }(A)\xrightarrow{{{H_2}O}}(B)\xrightarrow{{{K_2}Hg{I_4}/O{H^ - }}}(D){\text{ (brown ppt}}{\text{.)}}\]
What is \[({\text{D)}}\] ?
A.\[{{\text{H}}_2}{\text{N}} - {\text{Hg}} - {\text{Hg}} - {\text{I}}\]
B.\[{{\text{H}}_2}{\text{N}} - {\text{Hg}} - {\text{O}} - {\text{Hg}} - {\text{I}}\]
C.\[{{\text{H}}_2}{\text{N}} - {\text{O}} - {\text{Hg}} - {\text{Hg}} - {\text{I}}\]
D.\[{{\text{H}}_2}{\text{N}} - {\text{Hg}} - {\text{I}} - {\text{Hg}} - {\text{O}}\]

Answer 1

Hint: We must know the properties of each reagent given in question. First process is nitration. Then addition of water occurs to form vapors of a colorless pungent smelling gas.

Complete step by step answer:
We will move step by step to find the product D.
Group 13, has 3 electrons in valence shell and nitrogen also needs 3 electrons to fulfill its octet. Aluminum is also a 13 group element and hence it will also react with nitrogen to form the respective nitride having the molecular formula, \[{\text{AlN}}\]. Hence the compound A is \[{\text{AlN}}\].
\[{\text{Al}} + {{\text{N}}_2}\xrightarrow{\Delta }{\text{ AlN}}\]
The reaction of \[{\text{AlN}}\] with water yields the vapors of ammonia. Hence the product B will be \[{\text{N}}{{\text{H}}_3}\].
\[{\text{AlN}}\xrightarrow{{{H_2}O}}{\text{ N}}{{\text{H}}_3}\]
We know ammonia undergo self-ionization and forms \[{\text{NH}}_4^ + {\text{ and NH}}_2^ - \]
\[{\text{N}}{{\text{H}}_3} \rightleftharpoons {\text{NH}}_4^ + {\text{ + NH}}_2^ - \]
The given reagent \[{{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}{\text{/O}}{{\text{H}}^ - }\] is Nessler’s reagent. Alkaline solution of \[{{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}{\text{/O}}{{\text{H}}^ - }\] is called Nessler’s reagent. It reacts with ammonium ions to form iodide of Millon’s base whose formula is :
\[{{\text{H}}_2}{\text{N}} - {\text{Hg}} - {\text{O}} - {\text{Hg}} - {\text{I}}\].

Hence the correct option is B.

Additional information:
 Nessler’s reagent is an inorganic compound and is yellow in colour. It is a yellow crystalline compound. It is odorless and very soluble in water. It is also soluble in alcohol, ether and acetone. The pale yellow colour of ammonia of Nessler reagent turns deeper yellow in the presence of ammonia. Hence this is used to identify ammonia.

Note:
The final product D is a derivative of Millon’s base and is an inorganic compound. Millon’s base is known as mercuric amidochloride. It consists of zigzag polymeric structure.