1.

Anhydrous `AlCl_(3)` is covalent. From the date given below , predict whether it would remain covalent or became ionic in aqueous solution (Ionizatiion energy for `AlCl_(3) = 5137 kJ mol^(-1), Delta H_("hydration")` for `Al^(3+) = - 4665 kJ mol^(-1), DeltaH_("hydration")` for `Cl^(-)= - 381 KJ mol^(-1))`.

Answer» `AlCl_(3) (S) + aqto AlCl_(3) (aq)^(3+) (aq) + 3Cl^(-) (aq)`
Total energy released on hydration of 1 mole of `Al^(3+)` ions and 3 moles of `Cl^(-)` ions
`= 4665+3xx381 KJ = 5808 mol ^(-1)`
Energy required for ionization = 5137 kJ `mol^(-1)`
As energy released is greater than the energ required the compound will ionize in aqueous
solution .


Discussion

No Comment Found

Related InterviewSolutions