What is the correct order of enthalpy of dissociation of halogens?

Cl₂ > Br₂ > F₂ > I₂

The correct answer is option 3. Cl₂ > Br₂ > F₂ > I₂.

The enthalpy of dissociation of a halogen molecule refers to the energy required to break the bond between the two halogen atoms. Here's a breakdown of the factors affecting the trend:

Atomic Size: As you move down the halogen group \((F, Cl, Br, I)\), the atomic size increases. This leads to a larger distance between the halogen atoms in the molecule. Larger distances translate to weaker electrostatic attractions between the nuclei and shared electrons, making the bond easier to break. This explains why \(Br_2\), \(I_2\) have progressively lower dissociation enthalpies compared to \(Cl_2\).

Electron Configuration and d-orbitals: Fluorine \((F)\) is the first element in the group. Unlike heavier halogens \((Cl, Br, I)\), it doesn't have d-orbitals in its valence shell. These d-orbitals can participate in a type of bonding called pi \((\pi )\) bonding, which strengthens the bond between the halogen atoms in heavier elements. The absence of d-orbitals in \(F\) weakens the \(F-F\) bond compared to the \(Cl-Cl\) bond, even though \(F\) is smaller.

Therefore, the combined effect of these factors leads to the following trend in enthalpy of dissociation:

\(Cl_2\, \  (highest)\, \  >\, \  Br_2\, \  >\, \  F_2\, \  >\, \  I_2\, \  (lowest)\)

Here's a table summarizing the key points: