Tag: structure of some compounds

Questions Related to structure of some compounds

Anhydrous $AlCl _3$ fumes in air. What is the reason for it?

  1. It is hygroscopic in nature

  2. It gives out chlorine when exposed to air.

  3. It is hydrolysed in moist air giving out fumes of $HCl$.

  4. It loses water when exposed to moist air.


Correct Option: C
Explanation:
White fumes are due to the presence of $HCl$ gas.

Anhydrous aluminium chloride is hydrolysed partially with the moisture in the atmosphere to give $HCl$ gas. 

This HCl combines with the moisture in the air and appears white in colour.

$Al{Cl} _3 + 3H _2O \; \longrightarrow \; Al{(OH)} _3 + 3HCl \text{(white fumes)}$

Hence, the correct answer is option $\text{C}$.

Which reaction cannot give anhydrous $AlCl _{3}$

  1. Heating of $AlCl _{3}.6H _{2}O$

  2. Passing dry $HCl$ over heated aluminium powder

  3. Passing dry $Cl _{2}$ over heated aluminium powder

  4. Heating a mixture of alumina and coke in a current of dry $Cl _{2}$


Correct Option: A
Explanation:
Option A is correct because heating of $AlCl _{3}.6H _{2}O$ do not produce anhydrous $AlCl _{3}$ on heating

$AlCl _{3}.6H _{2}O\rightarrow Al _{2}O _{3}+HCl+H _{2}O$

Which of the following is a Lewis acid?

  1. $AlCl _3$

  2. $MgCl _2$

  3. $CaCl _2$

  4. $BaCl _2$


Correct Option: A
Explanation:
$Al$ needs two more electrons to complete its octet, it has the tendency to accept a pair of electrons to achieve stable electronic configuration. By definition, those which accepts electrons are called lewis acids. 

Thus, $Al{Cl} _{3}$ is a Lewis acid.


Hence, the correct answer is option $\text{A}$.

Which of the following statement about anhydrous aluminium chloride is correct?

  1. Its anhydrous form is deliquescent and fumes in air.

  2. It is not easily hydrolyzed.

  3. It sublimes at  $20^0C$ under vaccum.

  4. It is a strong Lewis base.


Correct Option: A
Explanation:

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).

It is  easily hydrolyzed.

It do not sublimes at $20^0C$ under vacuum.

It is a strong Lewis acid.(it accepts electrons)

So option A only is correct

Hence option A is correct statement.

The solubility of anhydrous $AlCl _{3}$ and hydrous $AlCl _{3}$ in diethyl ether are $S _{1}$ and $S _{2}$ respectively. Then -

  1. $S _{1}=S _{2}$

  2. $S _{1}> S _{2}$

  3. $S _{1}< S _{2}$

  4. $S _{1}< S _{2}$ but not $S _{1}=S _{2}$


Correct Option: B
Explanation:

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.


So, 
$S _{1}> S _{2}$.

Which of the following methods can be used for the preparation of anhydrous aluminium chloride?

  1. Heating ${Al Cl _3.6 H _2 O}$

  2. Heating a mixture of alumina and coke in a current of dry chlorine

  3. Passing dry HCl gas over heated aluminium powder

  4. Passing dry chlorine over heated aluminium


Correct Option: B,C,D
Explanation:

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


(C) Passing dry HCl gas over heated aluminium powder.
$2Al+6HCl \rightarrow 2AlCl _3+3H _2$

(D) Passing dry chlorine over heated aluminium.
$2Al+3Cl _2 \rightarrow 2AlCl _3$

It can not be obtained by heating of $AlCl _3.6H _2O$, because $AlCl _3$ is covalent and hence when the water of crystallisation is evaporated, $AlCl _3$ will react with the water to form the hydroxide.

Which one of the following processes does not yield anhydrous $ AlCl _3 $?

  1. Heating of $ AlCl _3 . 6H _2O $.

  2. Passing dry HCl over heated aluminium powder.

  3. Passing dry $ Cl _2 $ over heated aluminium powder.

  4. Heating a mixture of alumina and coke in a current of dry $ Cl _2 $.


Correct Option: A
Explanation:

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

Anhydrous $AlCl _3$ is obtained from:

  1. dilute $HCl$ and $Al$

  2. $Al$ and $Cl$ gas

  3. $HCl$ gas and $Al$ metal

  4. none


Correct Option: C
Explanation:

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.

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 $AgNO _3$ solution, which is soluble in $NH _4OH$.
Addition of $NH _4OH$ and NaOH separately to the solution of $A$ gives a gelatinous precipitate which in however soluble in excess of $NaOH$.
The compound $A$ is:

  1. $Al(OH) _3$

  2. $Al _2Cl _6$

  3. $Al _2O _3$

  4. $Al _2(SO _4) _3$


Correct Option: B
Explanation:

(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.
${Al} _{2}{Cl} _{6} + 6{H} _{2}O \; \longrightarrow \; \underset{\text{(weak base)}}{2Al{(OH)} _{3}} + \underset{\text{(strong acid)}}{6HCl}$.
It gives white ppt. with $AgN{O} _{3}$ which is soluble in $N{H} _{4}OH$.
(iv)$Al{Cl} _{3} + 3NaOH \; \longrightarrow \; \underset{\text{(white ppt.)}}{2Al{(OH)} _{3}} + 3NaCl$
$Al{(OH)} _{3} + \underset{\text{(excess)}}{NaOH} \; \longrightarrow \; \underset{\text{(soluble)}}{NaAl{O} _{2}} + 2{H} _{2}O$
But in case of $N{H} _{4}OH$, the ppt. of $Al{(OH)} _{3}$ does not dissolve in excess of $N{H} _{4}OH$.

Substance which readily undergo hydrolysis are :

  1. $AlCl _3$

  2. $BCl _3$

  3. $SiCl _4$

  4. $PbCl _4$


Correct Option: A,B,C,D
Explanation:
All the given compounds readily undergo hydrolysis.
$B{ Cl } _{ 3 }+3{ H } _{ 2 }O\longrightarrow B{ \left( OH \right)  } _{ 3 }+3HCl$
$Al{ Cl } _{ 3 }+3{ H } _{ 2 }O\longrightarrow Al{ \left( OH \right)  } _{ 3 }+3HCl$
$Si{ Cl } _{ 4 }+2{ H } _{ 2 }O\longrightarrow Si{ O } _{ 2 }+4HCl$
$Pb{ Cl } _{ 4 }+2{ H } _{ 2 }O\longrightarrow Pb{ O } _{ 2 }+4HCl$