If the bond dissociation energy of XY, `x_ (2) AND y_(2)` (all diatomic molecules) are in the ratio of `1 : 1 : 0.5 and Delta_(f)H` for the formation of XY is `- 200 kJ mol^(-1)`. The bond dissociation energy of `X_(2)` will be :

Options

A. `100 kJ mol^(-1)`
B. `200 kJ mol^(-1)`
C. `800 kJ mol^(-1)`
D. `400 kJ mol^(-1)`

Answer

C

Explanation

Let the bond dissociation energies of XY, `X_(2) and Y_(2)` be x, x and `0.5 x kJ mol^(-1)` respectively.
`(1)/(2) X_(2) + (1)/(2) Y_(2) rarr XY Delta H_(f) = - 200 kJ mol^(-1)`
`Delta H = (1)/(2) Delta H (X_(2)) + (1)/(2) Delta H (Y_(2)) - Delta H (XY) - 200 = (1)/(2) x + (0.5 x)/(2) - x` or `- 0.25 x = - 200`
`x = (200)/(0.25) = 800 kJ mol^(-1)`