Question

Given ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ can be converted to anhydrous ${\text{AlC}}{{\text{l}}_3}$ by heating:
(A) Hydrated ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ with ${\text{C}}{{\text{l}}_{\text{2}}}$ gas
(B) ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ with aqueous ${\text{HCl}}$
(C) ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ with ${\text{NaCl}}$ in solid state
(D) A mixture of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and carbon in dry ${\text{C}}{{\text{l}}_{\text{2}}}$ gas

Answer 1

We should look carefully that the question asks to form an anhydrous product. Alumina (${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$) is amphoteric in nature and it can react with very strong acid or strong base.

Complete step by step answer:
(A) Reaction of Hydrated ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ with ${\text{C}}{{\text{l}}_{\text{2}}}$ gas will not produce ${\text{AlC}}{{\text{l}}_3}$. Hence, option (A) is wrong.
(B) ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is an amphoteric compound. It can react with concentrated acids to produce water. Here, the given acid is not concentrated. Thus, no reaction will occur.
Hence, option (B) is wrong.
(C) When ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and ${\text{NaCl}}$ both are in solid state, no reaction can occur.
Hence, option (C) is wrong.
(D) When dry chlorine gas is passed over alumina and hot coke (carbon), anhydrous aluminium chloride will be formed:
${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 3{\text{C}} + 3{\text{C}}{{\text{l}}_2}\xrightarrow{\Delta }2{\text{AlC}}{{\text{l}}_3} + 3{\text{CO}}$.
Hence, the correct option is d.

Additional Information:
Anhydrous aluminium chloride is hygroscopic in nature, which means it has a high affinity for water. It fumes in moist air as with water molecules, it reacts to form hexahydrate ${\text{[Al(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_6}{\text{]C}}{{\text{l}}_3}$. Heating this solid will not produce the anhydrous form but it produces aluminium hydroxide as:
${\text{[Al(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_6}{\text{]C}}{{\text{l}}_3}\xrightarrow{\Delta }{\text{Al(OH}}{{\text{)}}_{\text{3}}}{\text{ + 3 HCl + 3 }}{{\text{H}}_{\text{2}}}{\text{O}}$.
Solid ${\text{AlC}}{{\text{l}}_3}$ is a sheet-like cubic close packed layers. When aluminium trichloride is in its melted state, it exists as the dimer ${\text{A}}{{\text{l}}_2}{\text{C}}{{\text{l}}_6}$. ${\text{A}}{{\text{l}}_2}{\text{C}}{{\text{l}}_6}$ dimers are also found in the vapour phase. At higher temperatures, the ${\text{A}}{{\text{l}}_2}{\text{C}}{{\text{l}}_6}$ dimers dissociate into trigonal planar ${\text{AlC}}{{\text{l}}_3}$.

Note: Alumina reacts with hot and concentrated hydrochloric acid to produce aluminium chloride as:
${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 6{\text{HCl (conc}}{\text{.)}}\xrightarrow{\Delta }2{\text{AlC}}{{\text{l}}_3} + 3{{\text{H}}_2}{\text{O}}$.
But in option (B), instead of concentrated hydrochloric acid, aqueous ${\text{HCl}}$ is mentioned.