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Questions Related to chemistry

Element "x" which is solid and having high melting point, form a Chloride $"x  Cl 3"$. This element "x" would be in which group of Periodic Table__.

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

  1. Na

  2. Mg

  3. Al

  4. Si

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

Anhydrous $AlCl _{3}$ is formed in:

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

  1. $AlCl _{3}.6H _{2}O \xrightarrow[]{\Delta}$

  2. $Al(red hot) + HCl(moist)\xrightarrow[]{\Delta}$

  3. $Al _{2}O _{3} + C+ Cl _{2}(dry) \xrightarrow[]{\Delta}$

  4. All of these

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Correct Option: A

Element 'x' which is solid and having high melting point, form a Chloride $'x Cl _3'$. This element 'x' would be in which group of Periodic table -

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

  1. Na

  2. Mg

  3. Al

  4. Si

Show answer
Correct Option: C
Explanation:

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

Which of the following species do not exist?

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

  1. $Al _2Cl _6$

  2. $BH _3$

  3. $PH _3$

  4. $PCl _5$

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Correct Option: B
Explanation:
  • $BH _3$ 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.

On the hydrolysis of $Al{Cl} _{3}$ then the no. of ${Al}^{+3}$ will be:

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

  1. Is equal to chloride ions

  2. Is equal to $\cfrac{1}{3}$ chloride ions

  3. Three times of ${Cl}^{-}$ ion

  4. None of these

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Correct Option: B
Explanation:

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

So the number of $A{ l }^{ +3 }$ will be equal to $\frac { 1 }{ 3 } $ chloride ions.

Anhydrous $Al{Cl} _{3}$ cannot be obtained from which of the following reactions :

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

  1. Heating $Al{Cl} _{3}.6{H} _{2}O$

  2. By passing dry $HCl$ over hot aluminium powder

  3. By passing dry ${Cl} _{2}$ over hot aluminium powder

  4. By passing dry ${Cl} _{2}$ over hot mixture of alumina and coke

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Correct Option: A
Explanation:

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 $

Is ever $AlCl _4$ possible ?

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

  1. True

  2. False

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Correct Option: A
Explanation:
  • There is dative coordination from a lone pair on a chloride ion to the vacant orbital on the aluminium ion. 
  • This gives the aluminium ion a full octet  and does not change the valency rules of aluminium.
  • Hence $AlCl _4$ is possible.

Which of the following statements is not correct about $AlCl _3$ ?

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

  1. It exists as a dimer

  2. It is a covalent compound

  3. It involves back bonding between Cl and Al

  4. Its aqueous solution conducts electricity

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Correct Option: C
Explanation:

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

so, option c is incorrect 

$AlF _{ 3 }$ is soluble in HF only in presence of KF. its 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. $K\left[ AlF _{ 3 }H \right]$

  3. ${ K } _{ 3 }\left[ AlF _{ 3 }{ H } _{ 3 } \right]$

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

Show answer
Correct Option: D
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,

$K _3[AlF _6]$
$AlF _3+3KF\longrightarrow K _3[AlF _6]$

$AlCl _3+NH _4OH\rightarrow X$ ; 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