Tag: study of aluminium chloride

Questions Related to study of aluminium chloride

$AlCl _3+NH _4OH\rightarrow X$ ; x is 

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. A white gelatinous precipitate

  2. Soluble in excess of $NH _4OH$

  3. Soluble in excess of $NaOH$

  4. Amphoteric in nature

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Correct Option: A

 ${ AlF } _{ 3 }$ is soluble in $HF$ only in presence of $KF$. It is due to the formation of:

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. ${ AlH } _{ 3 }$

  2. ${ K[AlF } _{ 3 }H]$

  3. ${ { K } _{ 3 }[AlF } _{ 3 }{ H } _{ 3 }]$

  4. ${ { K } _{ 3 }[AlF } _{ 6 }]$

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Correct Option: A
Explanation:
$AlF _3$  is insoluble in the anhydrous $HF$ because the $F^-$ ions are not available in intermolecular hydrogen bonded HF, but it becomes soluble in the presence of $KF$ due to the formation of soluble complex, $K _3[AlF _6].$
$AlF _3+3KF⟶K _3[AlF _6]$
Hence option D is correct answer.

Which have Vont Hoff factor same as $K _4 [Fe(CN) _6]$ 

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. $Al _2(SO _4) _3$

  2. $Mg(NO _3) _2$

  3. $CaCl _2$

  4. $NaNO _3$

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Correct Option: A
Explanation:

$i$ for $K _4 [Fe(CN) _6]$ is $5$ and equal to $i$ of $Al _2(SO _4) _3$ 

The molecule which is NOT hydrolysed by water at $25^oC$ is

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. $AICI _3$

  2. $SiCI _4$

  3. $BF _3$

  4. $SF _6$

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Correct Option: A

${Al} _{2}{O} _{3}$ is reduced by electrolysis at low potential and high currents. If $4.0\times {10}^{4}$ amperes of current is passed through molten ${Al} _{2}{O} _{3}$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, At. mass of $Al=27g{mol}^{-1}$)

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. $1.3\times {10}^{4}g$

  2. $9.0\times {10}^{3}g$

  3. $8.1\times {10}^{4}g$

  4. $2.4\times {10}^{5}g$

Show answer
Correct Option: C
Explanation:

$Al _2O _3(l) \rightarrow2Al^{3+}(l) + 3O^{2-}(l) $

Reduction equation of $Al^{3+}: $
$Al^{3+}(aq) + 3e^- \rightarrow Al(s) $

As,

Ampere = Coulomb / sec 

Time unit should be second. 
$6 hr \times 60 \dfrac{min}{ hr} \times 60 \dfrac{sec}{ min} = 21600 sec $
Coulomb = Ampere x second $= 40000 \times 21600 = 8.64\times10^8 $

1 mol $e^-$ carries a charge of 1 faraday, or 96485 coulombs. 
Then the mole of $e^-$ used in the electrolysis will be: 

$\dfrac{8.64\times10^8   \text{coulombs}}{ 96485 -\text{coulombs/mol}}- of -e^- = 8954.8  \text{mol of} -e^- $

According to the reduction equation, 3 moles of e- are needed to reduce 1 mol of Al^3+ to Al metal. 
Then , the mole of Al to be deposited at the cathode will be; 
$\dfrac{8954.8 \text{mol e-} }{ \dfrac{3 -\text{mol of e-}}{\text{mol Al}}} = 2985\text{ mol Al} $

Molar mass of $Al : 26.98 g/mol $
Mass of Al produced$: 2985 mol \times 26.98 g/mol = 80535 g $
$Al = 8.1\times10^4 g$

Alluminium chloride nexists as dimer, $A{l _2}C{l _6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives:

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. ${{\text{[Al(OH}}{{\text{)}} _{\text{6}}}{\text{]}}^{{\text{3 - }}}}{\text{ + 3HCl}}$

  2. ${{\text{[Al(}}{{\text{H}} _2}{\text{O}}{{\text{)}} _{\text{6}}}{\text{]}}^{{\text{3 + }}}}{\text{ + 3HC}}{{\text{l}}^ - }$

  3. $A{l^{3 + }} + 3C{l^ - }$

  4. ${\text{A}}{{\text{l}} _{\text{2}}}{{\text{O}} _{\text{3}}}{\text{ + 6HCl}}$

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Correct Option: B

Aluminium chloride exists as dimer, $Al _2Cl _6$, in solid state as well as in solution of non-polar solvents such as $C _6H _6$.


 When dissolved in water it gives :

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. $Al _2O _3+6HCl$

  2. $[Al(H _2O) _6]^{3+}+3Cl^-$

  3. $[Al(OH) _6]^{3+}+6Cl^-$

  4. $Al^{3+}+3Cl^-$

Show answer
Correct Option: B
Explanation:

Reaction:

$AlCl _3 $ is covalent but in water it becomes ionic due to large hydration energy oh $Al^{3+}$.

$AlCl _3 + 6H _2O \rightarrow [Al(H _2O) _6]^{3+} + 3Cl^-$.


Hence,option B is correct.

$AlCl _3$ on hydrolysis gives :

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. $Al _2O _3\cdot H _2O$

  2. $Al(OH) _3$

  3. $Al _2O _3$

  4. $AlCl _3\cdot 6H _2O$

Show answer
Correct Option: B
Explanation:

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

Hydrolysis is breaking with the help of water.
$AlCl _3 + 3 H _2O\rightarrow Al(OH) _3+ 3 HCl$

Hence option B is correct.

Aqueous solution of $AIC{I _3}$ on heating to dryness will give 

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. $AIC{I _3}$

  2. $A{l _2}{O _3}$

  3. $Al(OH)C{L _2}\,$

  4. $A{L _2}C{l _6}$

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Correct Option: B

Aluminium (III) chloride forms a dimer because aluminium :

chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

  1. cannot form a trimer.

  2. has high ionization energy.

  3. belongs to third group.

  4. can have higher coordination number.

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Correct Option: D
Explanation:

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

This is because: 
$AlCl _3$ forms a dimer as there is unsaturation in coordination. Also it enables Al atoms to complete their octets.

Aluminium is more like a metalloid than a metal.The electronegativity of aluminium is higher than the electronegativity of sodium .