Question

Bond of ${F_2},C{l_2}{\text{ and }}B{r_2}$ follow the order:
(A) $F - F > Cl - Cl > Br - Br$
(B) $Cl - Cl > Br - Br > F - F$
(C) $Br - Br > Cl - Cl > F - F$
(D) $Cl - Cl > F - F > Br - Br$

Answer 1

In the question, we have to find the order for the given elements,Now we all know that In chemistry, bond energy (BE) is also called the mean bond enthalpy or average bond enthalpy. Bond Energy is defined by the sum of all the bonds broken minus the sum of all of the bonds formed.

Complete step by step answer:
Firstly, as we know that these compounds which are given in the question belong to the same group so your go down in the group bond dissociation energy generally increases because there is increase in the size. But here ${F_2}$ has low bond energy as compared to $C{l_2}$ and $B{r_2}$ because there are interelective repulsions present in florins due to its small size. So, the Bond energy order is- ${F_2} < B{r_2} < C{l_2}$.
Therefore,
The order of bond energy is:
$Cl - Cl > Br - Br > F - F$
So, fluorine has high bond energy because it is the smallest among these molecules so it is bound very tightly to other fluorine atoms; it is just due to greater nuclear charge.
As we know that, bond dissociation enthalpy decreases as the bond distance increased from ${F_2}\;{\text{to }}{{\text{I}}_{\text{2}}}$ due to just increase in size of an atom, as we know from F to I. Therefore,
$F - F$ bond dissociation enthalpy is smaller than $Cl - Cl$ and even smaller than $Br - Br$. This is just because the F atom is very small and has large electron-electron repulsion.
And hence option (B) is the correct one that is $Cl - Cl > Br - Br > F - F.$

Note:
In molecular geometry, bond length or bond distance is defined as the average distance between nuclei of two bonded atoms in a molecule. It is a transferable property of a bond between atoms of fixed types, relatively independent of the rest of the molecule.