`AB, A_(2) and B_(2)` are diatomic molecules. If the bond enthalpies of `A_(2), AB, and B_(2)` are in the ratio 1 : 1 : 0.5 and the enthalpy of formation of AB from `A_2` is `-100 kJ mol^(-1)`, what is the bond enthalpy of `A_(2)` in kJ ?

Options

A. `400 kJ mol^(-1)`
B. `1650 kJ mol^(-1)`
C. `1200 kJ mol^(-1)`
D. `200 kJ mol^(-1)`

Explanation

`(1)/(2)A_(2)+(1)/(2)B_(2)rarr AB, " " Delta H=-100`
`(1)/(2)BE_(A-A)+(1)/(2)BE_(B-B)-BE_(A-B)=-100`
Let A - A bond energy be equal to 'x', then `(x)/(2)+(x)/(4)-x=-100`
or `x=+400 kJ mol^(-1)`
`(1)/(2)A_(2)+(1)/(2)O_(2)rarr AO , Delta H_(C )=-1200 kJ mol^(-1)`
`(1)/(2)(400)+(1)/(2)BE_(O-O)-BE_(A-O)=-1200 kJ mol^(-1)`
`BE_(A-O)=200+250+1200=1650 kJ mol^(-1)`
The correct answer is