The standard heat of formation values of `SF_(6)(g), S(g)`, and `F(g)` are `-1100, 275`, and `80 kJ mol^(-1)`, respectively. Then the average `S-F` bond enegry in `SF_(6)`

Options

A. `310 kJ mol^(-1)`
B. `220 kJ mol^(-1)`
C. `309 kJ mol^(-1)`
D. `280 kJ mol^(-1)`

Answer

C

Explanation

`S(g) +6F(g) rarr SF_(6)(g), DeltaH =- 1100 kJ mol^(-1)`
`S(s) rarr S(g) ,DeltaH =+ 275 kJ mol^(-1)`
`(1)/(2) F_(2)(g) rarr F(g),DeltaH = 80 kJ mol^(-1)`
Therefore, heta of formation =Bond enegry of reaction -Bond enegry of product
`- 1100 = (275 +6 xx 80) - (6 xx S -F)`
Thus, bond enegry of `S-F = 309 kJ mol^(-1)`