Which among the following factors is the most important in making fluorine, the strongest oxidation halogen?

Options

A. Bond dissociation energy
B. Ionisation enthalpy
C. Hydration enthalpy
D. Electron affinity

Answer

To determine why fluorine is the strongest oxidizing halogen, we can analyze several factors that contribute to its oxidation potential. Here’s a step-by-step breakdown of the reasoning:

### Step 1: Understanding the Electronic Configuration
Fluorine has the electronic configuration of \(1s^2 2s^2 2p^5\). This configuration indicates that fluorine is one electron short of achieving the stable noble gas configuration of neon (\(1s^2 2s^2 2p^6\)).

**Hint:** Remember that elements tend to gain, lose, or share electrons to achieve a stable electronic configuration.

### Step 2: Gaining an Electron
When fluorine gains one electron, it achieves the stable noble gas configuration. This process releases energy, making it energetically favorable for fluorine to act as an oxidizing agent.

**Hint:** Consider the stability associated with achieving a full outer shell of electrons.

### Step 3: Analyzing Hydration Enthalpy
Hydration enthalpy is the energy released when an ion is solvated by water. For fluorine, when it gains an electron to become \(F^-\), it is highly exothermic due to the strong interaction between the negatively charged ion and polar water molecules. This large release of energy significantly contributes to its oxidation potential.

**Hint:** Think about how the interaction between ions and water molecules can affect energy changes in a reaction.

### Step 4: Considering Bond Dissociation Energy
The bond dissociation energy of fluorine is relatively low because of its small atomic size and the presence of three lone pairs of electrons that repel each other. This weakens the bond between the two fluorine atoms in \(F_2\), making it easier to break.

**Hint:** Remember that smaller atoms can have stronger electron-electron repulsions, affecting bond strength.

### Step 5: Evaluating Ionization Enthalpy
Ionization enthalpy is the energy required to remove an electron from an atom. For fluorine, the ionization enthalpy is low because it is easier to remove an electron from the \(2p\) orbital, which is already close to being filled.

**Hint:** Consider how the position of an element in the periodic table can influence its ionization energy.

### Conclusion
Among the factors discussed (hydration enthalpy, bond dissociation energy, and ionization enthalpy), hydration enthalpy plays the most significant role in making fluorine the strongest oxidizing halogen. The large negative value of hydration enthalpy when fluorine gains an electron leads to a significant increase in its oxidation potential.

**Final Answer:** The most important factor in making fluorine the strongest oxidizing halogen is **hydration enthalpy**.