Question

Lattice enthalpies of `BeF_(2),MgF_(2),CaF_(2) and BaF_(2)` are `-2906,-2610,-2459 and -2367 kJ mol^(-1)` resoectively. Hydration enthalpies of `Be^(2+),Ca^(2+),Ba^(2+) and F^(Θ)` are `-2194,-1921,-1577,-130 and -457 kJ mol^(-1)` respectively. Which of the fluorides has the highest solublility in water?

Answer 1

For `BeF_(2)`,
`Delta_(hyd)H^(Θ)(BeF_(2))=Delta_(hyd)H^(Θ)(Be^(2+))+2Delta_(hyd)G^(Θ)(F^(Θ))`
`=[-2494+2(-457)]kJ mol^(-1)`
`=-3408 kJ mol^(-1)`
`Delta_(U)H^(Θ)(BeF_(2))=-2906 kJ mol^(-1)`
`Delta_("Soln")H^(Θ)(BeF_(2))=Delta_(hyd)H^(Θ)(BeF_(2))-Delta_(U)H^(Θ)(BeF_(2))`
`=[-3408-(-2906)]kJ mol^(-1)`
`=-502 kJ mol^(-1)`
For `MgF_(2)`,
`Delta_(hyd)H^(Θ)(MgF_(2))=Delta_(hyd)H^(Θ)(Mg^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))`
`=[-1921+2(-457)]kJ mol^(-1)`
`=-2835 kJ mol^(-1)`
`Delta_(U)H^(Θ)(MgF_(2))=-2610 kJ Mol^(-1)`
`Delta_("soln")H^(Θ)(MgF_(2))=Delta_(hyd)H^(Θ)(MgF_(2))-Delta_(U)H^(Θ)(MgF_(2))`
`=[-2835-(-2610)]kJ mol^(-1)=-225 kJ mol^(-1)`
For `CaF_(2)`,
`Delta_(hyd)H^(Θ)(CaF_(2))=Delta_(hyd)H^(Θ)(Ca^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))`
`=[-1577+2(-457)] kJ mol^(-1)`
`=-2491 kJ mol^(-1)`
`Delta_(U)H^(Θ)(CaF_(2))=-2459 kJ mol^(-1)`
`Delta_("soln")H^(Θ)(CaF_(2))=Delta_(hyd)H^(Θ) (CaF_(2))-Delta_(U)H^(Θ)(CaF_(2))`
`=[-2491-(-2459)] kJ mol^(-1)`
`=-32 kJ mol^(-1)`
For `BaF_(2)`,
`Delta_(hyd)H^(Θ)(BaF_(2)=Delta_(hyd)H^(Θ)(Ba^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))`
`=-1305+2(-457)`
`=-2219 kJ mol^(-1)`
`Delta_(U)H^(Θ)(BaF_(2))=-2367 kJ mol^(-1)`
`Delta_("soln")H^(Θ)(BaF_(2))=Delta_(hyd)H^(Θ)(BaF_(2))-Delta_(U)H^(Θ)(BaF_(2))`
`=[-2219-(-2367)] kJ mol^(-1)`
`=148 kJ mol^(-1)`
Since enthalpy of solution `Delta_("soln")H^(Θ)` is maximum for `BeF_(2)`, it has highest solubility in water.