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# The increasing order of bond dissociation enthalpy of the following substances is: ${{F}_{2}}$ < $C{{l}_{2}}$< $B{{r}_{2}}$ < ${{I}_{2}}$[A] True[B] False

Last updated date: 09th Aug 2024
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Hint: To answer this, consider the factors that affect bond dissociation enthalpy. Here, consider the factor of atomic radius as bond dissociation enthalpy is indirectly proportional to atomic radii. Do not forget to consider the effect of inter-electronic repulsion in case of an highly electronegative atom.

Before answering the question, let us understand the meaning of the term bond dissociation enthalpy.
Bond dissociation enthalpy is basically the energy required for the breaking a solid crystal into its corresponding ions.
There are factors that affect the bond dissociation as well as formation. There are two factors that affect the lattice energy of any substance.
Firstly it is the charge on the ions on the ions. Higher is the charge on each ion, higher will be the lattice energy.
And the second factor is the radius. Lower the internuclear distances among the ions in an ionic crystal, higher will be their lattice energy i.e. closer the two ions are, higher is the force of attraction between them.

Now, let us see the molecules given to us. Firstly let us discuss their radius.
We know that upon moving down the group atomic radius increases. As the radius increases i.e. the size of the atom increases so the bond length between the molecules will increase and so lower energy will be required to break the bond and thus bond dissociation enthalpy will be lower.
Among the halogens, iodine has the largest atomic radii so the bond dissociation enthalpy is lowest for iodine molecules.
Now for bromine, fluorine and chlorine we must think that fluorine has the highest bond dissociation energy due to its smallest size but this does not happen.
Fluorine is highly electronegative and due to its smaller size there’s a strong inter-electronic repulsion in a fluorine molecule. This leads to the breaking of bond readily and a relatively lower bond dissociation energy.
So the highest bond dissociation energy is for chlorine followed by bromine, fluorine and iodine. The order is ${{F}_{2}}$ < $C{{l}_{2}}$< $B{{r}_{2}}$ < ${{I}_{2}}$.
So, the given order is incorrect.
So, the correct answer is “Option B”.

Note: In a crystalline solid, when the ions are combined, energy is released and this energy is known as lattice energy and the change in energy in this process is basically the lattice enthalpy.
We use the term lattice enthalpy for both bond formation as well as dissociation. However, to me more appropriate about the energy required for breaking of the crystal lattice and the formation of the crystal lattice, we use the terms bond dissociation enthalpy and bond formation enthalpy respectively.