Aluminium chloride exists as dimer, ${A l}_{2} {C l}_{6}$ , in solid state as well as in solution or non-polar solvents such as benzene. When dissolved in water, it gives:

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

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{[A l (H_{2} O_{6})]}^{3 +} + 3 {C l}^{-}

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{[A l ({H O}_{6})]}^{3 +} + H C l

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{A l}_{2} O_{3} + 6 H C l

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Solution

The correct option is A ${[A l (H_{2} O_{6})]}^{3 +} + 3 {C l}^{-}$
$A l {C l}_{3}$ is covalent but in water, it becomes ionic due to large hydration energy of ${A l}^{3 +}$ .

$A l {C l}_{3} + 6 H_{2} O \to {[A l (H_{2} O_{6})]}^{3 +} + 3 {C l}^{-}$

Hence, option $B$ is correct.

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Similar questions

Aluminium chloride exists as dimer, $A l_{2} C l_{6}$ in solid state as well as in solution of non-polar solvent such as benzene. When dissolved in water it gives:

Aluminium chloride exists as a dimer, $A l_{2} C l_{3}$ , in solid state as well as in solution of no-polar solvents such as benzene. When dissolved in water it gives.

Aluminium chloride exists as a dimer ( $A l_{2} C l_{6}$ ) in solid state as well as in solution of non-polar solvents such as $C_{6} H_{6}$ . When dissolved in water it gives:

Q. Aluminimum chloride exists 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 :

Q. Aluminum chloride exists as a dimer

A l_{2} C l_{6}

solid state as well as in solution of non-polar solvents such as

C_{5} H_{5}

. When dissolved in water, it gives:

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Aluminium chloride exists as dimer, Al2Cl6, in solid state as well as in solution or non-polar solvents such as benzene. When dissolved in water, it gives:

The correct option is A [Al(H2O6)]3++3Cl−AlCl3 is covalent but in water, it becomes ionic due to large hydration energy of Al3+.AlCl3+6H2O→[Al(H2O6)]3++3Cl−Hence, option B is correct.

Aluminium chloride exists as dimer, ${A l}_{2} {C l}_{6}$ , in solid state as well as in solution or non-polar solvents such as benzene. When dissolved in water, it gives:

The correct option is A ${[A l (H_{2} O_{6})]}^{3 +} + 3 {C l}^{-}$
$A l {C l}_{3}$ is covalent but in water, it becomes ionic due to large hydration energy of ${A l}^{3 +}$ .

$A l {C l}_{3} + 6 H_{2} O \to {[A l (H_{2} O_{6})]}^{3 +} + 3 {C l}^{-}$

Hence, option $B$ is correct.