p-Block Elements-I - Result Question 29

####29. Anhydrous $AlCl _3$ is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution.

(Ionisation energy for $Al=5137 kJ mol^{-1}$

$ \begin{aligned} & \Delta H _{\text {hydration }} \text { for } Al^{3+}=-4665 kJ mol^{-1} \ & \Delta H _{\text {hydration }} \text { for } Cl^{-}=-381 kJ mol^{-1} \end{aligned}) $

(1997, 2M)

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Solution:

  1. The total hydration energy of $AlCl _3$

$=$ Hydration energy of $Al^{3+}+3 \times$ Hydration energy of $Cl^{-}$

$=-4665+3(-381) kJ / mol$

$=-5808 kJ / mol$

The above hydration energy is more than the energy required for ionisation of $AlCl _3$ into $Al^{3+}$ and $3 Cl^{-}$.

Due to this reason, $AlCl _3$ becomes ionic in aqueous solution. In aqueous solution, it is ionised completely as

$$ AlCl _3+6 H _2 O \longrightarrow\left[Al\left(H _2 O\right) _6\right]^{3+}+3 Cl^{-} $$



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