The `IE` values of `Al_((g)) = Al^(+) +e` is `577.5 kJ mol^(-)` and `DeltaH` for `Al_((g)) = Al^(3+) +3e` is `5140 kJ mol^(-1)`. If second and third IE values are in the ratio 2:3. Calculate `IE_(2)` and `IE_(3)`.

Answer

`[IE_(2) = 1825 kJ//"mole", IE_(3) = 2737.5 kJ//"mol"]`

Explanation

`[Al(g)+Al^(+) +e^(-), IE_(1) = 577.5`
`I.E_(1) +IE_(2) +IE_(3) = 5140`
`(IE_(2))/(IE_(3)) = (2)/(3), I.E_(2) = (2)/(3),IE_(2) = (2)/(3)IE_(3)]`