$\Delta H_f^{\circ}$ values (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for graphite, diamond and $C_{60}$ are…
$\Delta H_f^{\circ}$ values (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for graphite, diamond and $C_{60}$ are respectively
$0 ; 1.9 ; 38.1$
$1.8 ; 1.9 ; 38.1$
$0 ; 0 ; 21.4$
$1.8 ; 1.9 ; 2.0$
Solution
$\because$ Graphite is the most stable state of carbon and its $\Delta H_f^{\circ}$ is considered as zero ( $\because$ has more van der Waals' force) also $\Delta H_f^{\circ}$ for $\mathrm{C}_{60}>\Delta H_f^{\circ}$ for diamond.
$\Delta H_f{ }^{\circ}$ values of graphite, diamond and $C_{60}$ are $=0$, 1.9 and $38.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta H_f^{\circ}$ values are in order :
Diamond < Fullerene < Graphite
Hence, option (1) is the correct answer.