`Delta H_(f)^(@)` value (in KJ `"mol"^(-1)`) for graphite, diamond and `C_(60)` are respectively

Options

A. `0, 1.9, 38.1`
B. `1.8, 1.9, 38.1`
C. `0, 0, 21.4`
D. `1.8, 1.9, 2.0`

Answer

A

Explanation

`:'` Graphite is the most stable state of carbon and its `Delta H_(f)^(@)` is considered as zero (`:'` has more van der Waal's force) alsi `DeltaH_(f)^(@)` for `C_(60) gt Delta II_(f)^(@)` for diamond.
`DeltaH_(f)^(@)` values of graphite, diamond and `C_(60)` are `=0`, 1.9 and `38.1" kJ mol"^(-1)` and `Delta H_(f)^(@)` values are in order :
Diamond `lt` Fullerence `lt` Graphite
Hence, option (a) is the correct answer.