For ionic compound `A^(2+) 2B^(-) & 2C^(+)D^(2-)`
(a) `A_(4)(S) +B_(2)(g) rarr AB_(2)(S) DeltaH_(F) =- 500`
Find the lattice energy of `AB_(2)`
(b) Find `DeltaH_(1)` for according `D_(3) (g) +C(s) rarr C_(2)D(s)`
Given on sublimation of above metal `(A_(4))` it dissociates into individual atom.
`{:("Given:Sublimation energy of "A_(4) is 1600,,,"Sublimation energy of" C is 100),("Dissociation energy of" B_(2) is 200,,,"Dissociation energy of" D_(3) is 90):}`
`{:(A(g) rarr A^(+)(g),,,DeltaH = 50),(A^(+)(g)rarr A^(2+)(g),,,DeltaH = 150),(B^(-)(g)rarrB(g),,,DeltaH = 260),(B^(-)(g)rarrB^(2-)(g),,,DeltaH = 250),(D(g)rarrD^(-)(g),,,DeltaH =- 50),(D^(-)(g)rarrD^(2-)(g),,,DeltaH =100),(C^(+)(g) rarrC(g),,,DeltaH =- 150),(C_(2)D(s)rarr2C^(+)(g)+D^(2-)(g),,,DeltaH = 1000):}`
All the values are in `kJ//mol`

Answer

`[-780, - 420, AB gt CD]`

Explanation

`[A_(4)(s) +B_(2)(g) rarr AB_(2)(s)`
`AB_(2)` (ionic compound) having `A^(2+)//2B^(-)`
`(DeltaH)_("sublimation")` of `A_(4) = 1600`
`(DeltaH)_("sublimation")` of `A = (1600)/(4) = 400 KJ`
`- 500 = 400 +50 +150 - 260 xx 2 +200 +(DeltaH_(L.E)`
`(DeltaH)_(L.E) =- 500 - 400 - 50 - 150 +520 - 200 = 780]`