Question

Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of${{F}_{2}}$ and $C{{l}_{2}}$.

Answer 1

Hint: Oxidizing power of any molecule is based on its ability to donate or gain electrons. Bond dissociation enthalpy is the measure of energy in breaking a bond to form atoms; electron gain enthalpy is the ability to gain an electron by an isolated gaseous atom to complete its octet, while hydration enthalpy is the ability to form ions in solution. These all affect the oxidizing power.

Complete answer:
The oxidizing power of any atom is its ability to readily gain electrons. The smaller atoms having high electron gain affinity are capable of good oxidizing agents. But the oxidizing power can be affected by three factors like bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Due to these factors, ${{F}_{2}}$ has more oxidizing power than $C{{l}_{2}}$. Chlorine atoms have more negative electron gain enthalpy, so it should be a better oxidizing agent, but the bond dissociation enthalpy and the hydration enthalpies of fluorine atom makes it a more powerful oxidizing agent.
 ${{F}_{2}}$ consist of a very small size, so it has lesser bond dissociation energy, also due to its small size it readily gets dissolved, so it has a higher hydration energy, these two factors contribute to overshadow the less negative electron gain enthalpy of ${{F}_{2}}$, making it a more powerful oxidizing agent than $C{{l}_{2}}$.
So, due to a higher bond dissociation and hydration enthalpy, ${{F}_{2}}$, is a more powerful oxidizing agent than $C{{l}_{2}}$.

Note:
In halogens the oxidizing power of the atoms decreases down the group. The bond dissociation energy decreases down the group, but fluorine has lower bond dissociation energy, this is because fluorine has a smaller size so there is greater inter electronic repulsion among the lone pairs that decreases its bond dissociation energy.