If the bond dissociation energies of `XY`,`X_(2)` and `Y_(2)` are in the ratio of `1:1:0.5` and `DeltaH_(f)` for the formation of `Xy` is `-200 KJ//mol`. The bond dissociation energy of `X_(2)` will be `:-`

Options

A. 200 kJ `mol^(-1)`
B. 100 kJ `mol^(-1)`
C. 800 kJ `mol^(-1)`
D. 300 kJ `mol^(-1)`

Answer

C

Explanation

Given `(1)/(2)X_(2)+(1)/(2)(Y)/(2)rarr XY , Delta H=-200 kJ mol^(_1)`
Let the bond dissociation energy of XY = a
The ratio of bond dissociation energies of `X_(2), Y_(2)` and XY be `a:(a)/(2):a` (the ratio given)
For given equation : `Delta H=(B.E.)_(R )-(B.E.)_(P)`
`-200=[(1)/(2)xx a+(1)/(2)xx(a)/(2)]-a`
`therefore a=800 kJ mol^(-1)`