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(A) The standard enthalpy of formation $(\Delta_f H^{\Theta})$ of the most stable allotrope of an element in its standard state is defined as $0 \ kJ/

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$i-b, ii-c, iii-a$

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(A) The standard enthalpy of formation $(\Delta_f H^{\Theta})$ of the most stable allotrope of an element in its standard state is defined as $0 \ kJ/mol$.$i$. Graphite is the most stable allotrope of carbon,so its $\Delta_f H^{\Theta} = 0 \ kJ/mol$ $(i-b)$.$ii$. Diamond is less stable than graphite,with $\Delta_f H^{\Theta} = 1.90 \ kJ/mol$ $(ii-c)$.$iii$. Fullerene $(C_{60})$ has a higher enthalpy of formation,$\Delta_f H^{\Theta} = 38.1 \ kJ/mol$ $(iii-a)$.Therefore,the correct match is $i-b, ii-c, iii-a$.

Allotrope	$\Delta_f H^{\Theta}$
$i$. Graphite	$b$. $0 \ kJ/mol$
$ii$. Diamond	$c$. $1.90 \ kJ/mol$
$iii$. Fullerene	$a$. $38.1 \ kJ/mol$

Match the following allotropes of carbon with their standard enthalpy of formation $(\Delta_f H^{\Theta})$:
Allotrope $\Delta_f H^{\Theta}$
$i$. Graphite $b$. $0 \ kJ/mol$
$ii$. Diamond $c$. $1.90 \ kJ/mol$
$iii$. Fullerene $a$. $38.1 \ kJ/mol$

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Match the following allotropes of carbon with their standard enthalpy of formation $(\Delta_f H^{\Theta})$:Allotrope$\Delta_f H^{\Theta}$$i$. Graphite$b$. $0 \ kJ/mol$$ii$. Diamond$c$. $1.90 \ kJ/mol$$iii$. Fullerene$a$. $38.1 \ kJ/mol$