Question

What is the action of the following reagents on ammonia?
A. Nessler’s reagents
B. Sodium metal

Answer 1

Hint: Potassium iodomercurate (II) solution is known as Nessler’s reagent. When we combine sodium hydroxide and potassium iodide with mercuric chloride or mercury iodide. It is used to detect the presence of ammonia and ammonium compounds. Hydrogen gas is liberated by the reaction of sodium with ammonia .

Complete step by step answer:
Nessler’s reagent is an alkaline solution of potassium tetraiodo mercurate $\left( {{{{\rm I}{\rm I}}}} \right)$ complex and potassium hydroxide or sodium hydroxide.
Nessler’s reagent is added to the solution of ammonium salt$\left( {{{N}}{{{H}}_{{4}}}{{Cl}}} \right)$, the alkali present in Nessler’s reagent reacts with ammonium salt producing ammonia gas. Then ammonia reacts with tetraiodo mercurate $\left( {{{{\rm I}{\rm I}}}} \right)$ complex forming a brown precipitate of amino mercuric called iodide of Millon’s base
${{2N}}{{{H}}_{{4}}}{{C}}{{{l}}_{\left( {{{aq}}} \right)}}{{ + 2NaO}}{{{H}}_{\left( {{{aq}}} \right)}} \to {{2NaC}}{{{l}}_{\left( {{{aq}}} \right)}}{{ + 2}}{{{H}}_{{2}}}{{{O}}_{\left( {{l}} \right)}}{{ + 2N}}{{{H}}_{{3}}}_{\left( {{g}} \right)}$
${{2}}{{{K}}_{{2}}}{\left[ {{{Hg}}{{{I}}_{{4}}}} \right]_{\left( {{{aq}}} \right)}}{{ + 2N}}{{{H}}_{{{3}}\left( {{g}} \right)}} \to {{N}}{{{H}}_{{{4}}\left( {{{aq}}} \right)}}{{ + N}}{{{H}}_{{2}}}{{H}}{{{g}}_{{2}}}{{{I}}_{{{3}}\left( {{s}} \right)}}{{ + 4K}}{{{I}}_{\left( {{{aq}}} \right)}}$
Reaction of ammonia with sodium
Sodium amide is prepared by the reaction of sodium with ammonia gas in the presence of bit of Iron $\left( {{{{\rm I}{\rm I}{\rm I}}}} \right)$ catalyst
${{2Na + 2N}}{{{H}}_{{{3}}\left( {{g}} \right)}} \to {{2NaN}}{{{H}}_{{2}}}{{ + }}{{{H}}_{{2}}}$
Sodium, is oxidised from ${{0}}$oxidation state to ${{ + 1}}$ oxidation state
Hydrogen in Ammonia is reduced from ${{ + 1}}$ to ${{0}}$ oxidation state

Additional information:
Nessler’s reagent can be prepared when potassium iodide $\left( {{{KI}}} \right)$ solution is added slowly to the solution of mercuric chloride$\left( {{{HgC}}{{{l}}_{{2}}}} \right)$, a red precipitate of ${{Hg}}{{{I}}_{{2}}}$ and soluble ${{KCl}}$ are formed. If ${{KI}}$ solution is added in excess then ${{Hg}}{{{I}}_{{2}}}$ reacts with ${{KI}}$ forming soluble colourless complex called as potassium tetraiodo mercurate $\left( {{{{\rm I}{\rm I}}}} \right)$ complex
${{HgC}}{{{l}}_{{{2}}\left( {{{aq}}} \right)}}{{ + 2K}}{{{I}}_{\left( {{{aq}}} \right)}} \to {{2KC}}{{{l}}_{\left( {{{aq}}} \right)}}{{ + Hg}}{{{I}}_{{{2}}\left( {{s}} \right)\left( {{{red}}} \right)}}$
${{Hg}}{{{I}}_{{{2}}\left( {{s}} \right)}}{{ + 2K}}{{{I}}_{\left( {{{aq}}} \right)}} \to {{{K}}_{{2}}}{\left[ {{{Hg}}{{{I}}_{{4}}}} \right]_{\left( {{{aq}}} \right)}}$
Concentrated \[{{KOH}}\]or ${{NaOH}}$solution is added to the potassium tetraiodo mercurate inorder to make it alkaline. The alkaline solution of potassium tetraiodo mercurate $\left( {{{{\rm I}{\rm I}}}} \right)$ complex is called as Nessler’s Reagent

Note: The sodium metal gets dissolved in liquid ammonia in the absence of iron $\left( {{{{\rm I}{\rm I}{\rm I}}}} \right)$catalyst which results in the formation of blue coloured solution because of solvated electrons. In organic synthesis Sodium amide is widely used as strong base and has a molar mass of ${{39}}{{.01 gmo}}{{{l}}^{{{ - 1}}}}$.