What is the correct order of bond dissociatio | vedclass

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(B) $N_2$ contains a triple bond $(N \equiv N)$,which results in the highest bond dissociation enthalpy.$O_2$ has a double bond $(O = O)$,giving it a

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$F_2 < Cl_2 < O_2 < N_2$

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(B) $N_2$ contains a triple bond $(N \equiv N)$,which results in the highest bond dissociation enthalpy.$O_2$ has a double bond $(O = O)$,giving it a higher bond dissociation enthalpy than the halogens.Due to the small size of the $F$ atom,there is significant inter-electronic repulsion between the lone pairs in $F_2$,making its bond weaker than that of $Cl_2$.Therefore,the correct order of bond dissociation enthalpy is $F_2 < Cl_2 < O_2 < N_2$.

What is the correct order of bond dissociation enthalpy for $N_2, O_2, F_2$ and $Cl_2$?

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