Question

An inorganic compound $A$ shows the following reaction:
(i) It is white solid, exists as dimer and fumes in the wet air.
(ii)It sublimes at ${180}^{\circ }C$ and forms monomer if heated to ${400}^{\circ }C$.
(iii) Its aqueous solution turns blue litmus to red and gives a white precipitate with $AgN{O}_3$ solution, which is soluble in $N{H}_{4}OH$.
Addition of $N{H}_{4}OH$ and NaOH separately to the solution of $A$ gives a gelatinous precipitate which in however soluble in excess of $NaOH$.
The compound $A$ is:

• A. $Al(OH{)}_3$
• B. $A{l}_{2}C{l}_6$
• C. $A{l}_2{O}_3$
• D. $A{l}_2(S{O}_4{)}_3$

Answer 1

The correct option is B $A{l}_{2}C{l}_6$

(B) Compound $B$ is ${Al}_2{Cl}_6$.

(i) ${Al}_2{Cl}_6$ is a white solid, exists as dimer and fumes in wet air.

(ii) It sublimes at 180°C and converts into $Al{Cl}_3$ at 400°C.

(iii) Its aqueous solution is acidic, so it turns blue litmus to red.

${Al}_2{Cl}_6+6H_{2}O⟶\underset{\text{(weak base)}}{2Al{\left(OH\right)}_3}+\underset{\text{(strong acid)}}{6HCl}$.

It gives white ppt. with $AgN{O}_3$ which is soluble in $N{H}_{4}OH$.

(iv)$Al{Cl}_3+3NaOH⟶\underset{\text{(white ppt.)}}{2Al{\left(OH\right)}_3}+3NaCl$

$Al{\left(OH\right)}_3+\underset{\text{(excess)}}{NaOH}⟶\underset{\text{(soluble)}}{NaAl{O}_2}+2H_{2}O$

But in case of $N{H}_{4}OH$, the ppt. of $Al{\left(OH\right)}_3$ does not dissolve in excess of $N{H}_{4}OH$.