Question

# The solution of meta aluminate on boiling with ammonium chloride gives a white precipitate of:a.) $A{{l}_{2}}{{O}_{3}}$b.) $AlC{{l}_{3}}$c.) $Al{{\left( OH \right)}_{3}}$d.) $NaAl{{\left( OH \right)}_{4}}$

Hint: To find the answer to this question we have to keep few things in mind which are as follows:
The products formed from a thermodynamic reaction are always stable and lower in energy than the reactants.
If any gaseous product is possible it is mostly formed because it can escape from the reacting mixture and drive the reaction forward.
If any neutralization product or salt which is neutral is possible to form is product it is mostly formed because it is stable.

The solution of sodium meta aluminate on boiling with ammonium chloride gives a white precipitate of aluminium hydroxide $Al{{\left( OH \right)}_{3}}$. This reaction is a simple non redox reaction as there is no change in oxidation state of any of the species.
$NaAl{{O}_{2}}+N{{H}_{4}}Cl+{{H}_{2}}O\to Al{{\left( OH \right)}_{3}}+NaCl+N{{H}_{3}}$
Hence, the correct answer is the solution of meta aluminate on boiling with ammonium chloride gives a white precipitate of $Al{{\left( OH \right)}_{3}}$