SATHEE: p-Block Elements-I - Result Question 58

p-Block Elements-I - Result Question 58

####29. Anhydrous $A l C l_{3}$ is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution.

(Ionisation energy for $A l = 5137 k J m o l^{- 1}$

$\begin{aligned} Δ H_{hydration} for A l^{3 +} = - 4665 k J m o l^{- 1} & Δ H_{hydration} for C l^{-} = - 381 k J m o l^{- 1} \end{aligned}$ )

(1997, 2M)

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Solution:

The total hydration energy of $A l C l_{3}$

$=$ Hydration energy of $A l^{3 +} + 3 \times$ Hydration energy of $C l^{-}$

$= - 4665 + 3 (- 381) k J / m o l$

$= - 5808 k J / m o l$

The above hydration energy is more than the energy required for ionisation of $A l C l_{3}$ into $A l^{3 +}$ and $3 C l^{-}$ .

Due to this reason, $A l C l_{3}$ becomes ionic in aqueous solution. In aqueous solution, it is ionised completely as

$A l C l_{3} + 6 H_{2} O ⟶ {[A l {(H_{2} O)}_{6}]}^{3 +} + 3 C l^{-}$

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p-Block Elements-I - Result Question 58

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