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. 800 KJ `"mol"^(-1)`
B. 100 KJ `"mol"^(-1)`
C. 200 KJ `"mol"^(-1)`
D. 400 KJ `"mol"^(-1)`

Answer

A

Explanation

The reaction for `Delta_(f)H^(@)(XY)`
`(1)/(2)X_(2)(g) +(1)/(2)Y_(2)(g) to XY(g)`
Bond energies of `X_(2), Y_(2) and XY` are `X, (X)/(2),X` respectively
`therefore DeltaH=((X)/(2)+(X)/(4))-X=-200`
On solving , we get
`implies -(X)/(2)+(X)/(4)=-200`
`implies X=800 ` KJ/mole