The correct order of bond dissociation enthalpy is:
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Remember the F$_2$ anomaly in bond enthalpy. Due to its small size and strong lone pair-lone pair repulsions, the F-F bond is unexpectedly weak, weaker than both Cl-Cl and Br-Br bonds. This is a frequently tested exception in p-block chemistry.
Bond dissociation enthalpy is the energy required to break one mole of a specific bond in the gaseous state.
The general trend for halogens is that the bond enthalpy decreases down the group from Cl to I. This is because as the atomic size increases, the bond length increases, and the overlap between the atomic orbitals becomes less effective, resulting in a weaker bond.
This gives the order: Cl$_2$>Br$_2$>I$_2$.
However, fluorine (F$_2$) is an exception to this trend. Although fluorine is the smallest and most electronegative halogen, its bond dissociation enthalpy is lower than that of chlorine and even bromine.
This anomaly is due to the small size of the fluorine atom. The F-F bond length is very short, which brings the non-bonding electron pairs (lone pairs) on the two fluorine atoms very close to each other. This leads to significant inter-electronic repulsion (lone pair-lone pair repulsion), which weakens the F-F covalent bond.
The experimental values for bond dissociation enthalpies (in kJ/mol) are approximately:
Cl$_2$: 242
Br$_2$: 193
F$_2$: 159
I$_2$: 151
Therefore, the correct decreasing order is Cl$_2$>Br$_2$>F$_2$>I$_2$.
This matches option (C). (Note: The provided answer key may state option (D), but this contradicts experimental data. The correct order is C).
