Study of aluminium chloride - class-XII

Solution:- (A) $1, ; 2$ and $3$

Anhydrous aluminium chloride cannot be obtained by heating its hydrated form because hydrolysis yields alumina and hydrochloric acid instead. It is industrially prepared by passing dry chlorine gas over a heated mixture of alumina and carbon.

Element aluminium has atomic number 13. Electronic configuration is 2, 8, 3
$Valency = 3$
$2Al + 3Cl _2 \rightarrow 2AlCl _3$

• $BH _3$ molecule is electron deficient in nature since 'B' atom has only 6 electrons around it. So it exists in the form of dimer that is $B _2H _6.$
  
• Hence option B is correct answer.
  

The hydrolysis reaction will be $AlC{ l } _{ 3 }+{ 3H } _{ 2 }O\rightarrow Al(OH{ ) } _{ 3 }+3HCl$

On heating $AlCl _{ 3 }.6{ H } _{ 2 }O$ hydrolysis takes place to produce $Al\left( OH \right) _{ 3 }$ and $HCl$

and on further heating $Al\left( OH \right) _{ 3 }$ converts to $Al _{ 2 }O _{ 3 }$.

By passing dry $HCl$ or ${ Cl } _{ 2 }$ over heated aluminium,
$2Al+3C{ l } _{ 2 }\longrightarrow 2AlC{ l } _{ 3 }$

$2Al+6HC{ l }\longrightarrow 2AlC{ l } _{ 3 }+3{ H } _{ 2 }$

Macah's method:$Al _{ 2 }O _{ 3 }+3C+3{ Cl } _{ 2 }(dry)\xrightarrow [ \Delta  ]{  } 2AlC{ l } _{ 3 }+3CO\uparrow $

Due to high size of chlorine ,there is no back bonding in$AlCl _3$

Silicon tetrachloride, boron trifluoride, and aluminium chloride all readily undergo hydrolysis due to vacant d-orbitals or electron deficiency allowing water molecules to coordinate. Sulfur hexafluoride is sterically hindered and lacks accessible coordination sites, making it inert to water at room temperature.

When anhydrous aluminum chloride dissolves in water, it undergoes hydration to form the hexaaquaaluminum(III) ion, [Al(H2O)6]3+, and releases chloride ions into the solution.

Reaction:

$AlCl _3$ on hydrolysis gives $Al(OH) _3$

Heating an aqueous solution of AlCl3 causes hydrolysis where the aluminum ion reacts with water to form aluminum hydroxide, which then dehydrates to aluminum oxide (Al2O3) upon heating to dryness.

 Aluminium trichloride is a dimer, $Al _2Cl _6$.

$AlCl _{3}$ on hydrolysis gives $Al(OH) _{3}$
$AlCl _{3}+3H _{2}O\rightarrow Al(OH) _{3}+3HCl$
Hence option B is correct.

In the liquid state aluminium trichloride exists as the dimer with tetracoordinate aluminium. This change in structure is related to the lower density of liquid phase. These dimers are also found in the vapor phase. At higher temperature dimers dissociate into trigonal planar. The melt conducts electricity poorly.

Aluminium chloride reacts with water to form  hexahydrate. The anhydrous phase cannot be regained on heating as HCl is lost leaving aluminium hydroxide or alumina.
$Al\left( { H } _{ 2 }O \right) _{ 6 }{ Cl } _{ 3 }\rightarrow { Al\left( OH \right)  } _{ 3 }+3HCl+{ 3H } _{ 2 }O$

Both BCl3 and AlCl3 are Lewis acids because they have an incomplete octet. BCl3 is a stronger Lewis acid than AlCl3 because the boron atom is smaller, making the vacant p-orbital more accessible for electron donation.

Anhydrous aluminum chloride is a covalent compound due to the high charge density of the Al3+ ion, which polarizes the electron cloud of the chloride ion (Fajans' rule).

Similar to question 430733, the dissolution of AlCl3 in water results in the formation of the octahedral hexaaquaaluminum(III) complex and chloride ions.

We knoe $AlCl _3$ exists as a dimer $Al _2Cl _6$

Aluminium chloride is hygroscopic, having a very pronounced affinity for water. It fumes in moist air and hisses when mixed with liquid water as the Cl ions are displaced with $H _2O$ molecules in the lattice to form the hexahydrate $AlCl _36H _2O$ (also white to yellowish in color).

Anhydrous AlCl3 will dissolve the diethyl ether because anhydrous ​AlCl3 is a good lewis acid. It is electron deficient. The lone pairs of diethyl ether will be donated to the anhydrous ​AlCl3 due to which it is soluble. On the other hand, hydrous ​AlCl3 have the water molecules which make it poor lewis acid.

Anhydrous aluminium chloride can be prepared by the following methods:
(B) Heating a mixture of alumina and coke in a current of dry chlorine.
$Al _2O _3+3C+3Cl _2 \rightarrow 2AlCl _3+3CO\uparrow$

$ 2AlCl _3 . 6H _2O \xrightarrow {\Delta} Al _2O _3 + 6HCl + 9H _2O $ .

When $Al$ metal is reacted with acid $HCl$ it libetrates hydrogen gas and  $AlCl _3$
$6HCl+2Al\rightarrow 2AlCl _3+3H _2$
Hence option $C$ is correct.

(B) Compound $B$ is ${Al} _{2}{Cl} _{6}$.