Question

The sublimation energy of a metal is `100kJmol^(-1)` and its Ist and Iind `IEs` are `4.0 eV` and `12.0 eV` respectively. The hydration energy of `X^(o+)` is `-380 kJ mol^(-1)` and `X^(2+)` is `-1280 kJ mol^(-1)`. Common the tability of two ions in water.

Answer 1

`1eV = 96.39 kJ`
`(1eV =1.602 xx10^(-10)J xx 6.023 xx 10^(-3) kJ mol^(-1))`
`[{:(4eV =385.92 kJ),(12eV =1157.76 kJ):}]`
`{:(X(s)rarrX(g),,,Delta_("sub")H =100kJ):}`
`[{:(X(g)rarrX^(o+)(g)+e^(-)IE =385.92 kJ),(X^(o+)(g)+aqrarrX^(o+)(aq)DeltaH =- 380kJ):}`
`ulbar(X(g)rarrX^(o+)(aq))`
`DeltaH = 385.92 - 380 = 5.92 kJ`
Similarly, for `X(g) rarr X^(2+)(aq)`
`DeltaH = 1157.76 - 1280 =- 122.24 kJ mol^(-1)`
Therefore, `X^(2+)(aq)` is more stable than `X^(o+)(aq)`, since `DeltaH` is `-ve`. More than -ve value, more stable is the ion, in aqueous state.