Halogen that absorbas visible radiation of least wave length

Options

A. `Cl_(2)`
B. `F_(2)`
C. `Br_(2)`
D. `I_(2)`

Answer

To determine which halogen absorbs visible radiation of the least wavelength, we need to analyze the bond strength of the halogen molecules. The bond strength is directly related to the energy required to break the bond, and this energy is inversely related to the wavelength of the radiation absorbed.

### Step-by-Step Solution:

1. **Understanding Bond Strength and Energy**:
- The energy required to break a bond is related to the bond strength. The stronger the bond, the more energy is needed to break it.
- The relationship between energy (E) and wavelength (λ) is given by the equation:
\[ E = \frac{hc}{\lambda} \]
- Here, \(h\) is Planck's constant and \(c\) is the speed of light.

2. **Relating Energy to Wavelength**:
- From the equation, we can see that as energy increases, the wavelength decreases:
\[ E \propto \frac{1}{\lambda} \]
- Therefore, to find the halogen that absorbs radiation of the least wavelength, we need to find the one with the highest bond strength.

3. **Analyzing Halogen Bonds**:
- The halogens in question are F₂ (fluorine), Cl₂ (chlorine), Br₂ (bromine), and I₂ (iodine).
- The bond types are:
- F₂: 2p-2p overlapping
- Cl₂: 3p-3p overlapping
- Br₂: 4p-4p overlapping
- I₂: 5p-5p overlapping

4. **Comparing Bond Strengths**:
- The extent of overlapping decreases as the size of the atoms increases. Therefore:
- F₂ has the strongest bond due to the effective 2p-2p overlap.
- Cl₂ has a weaker bond than F₂ due to 3p-3p overlap.
- Br₂ and I₂ have even weaker bonds due to larger p orbitals (4p and 5p, respectively).

5. **Conclusion**:
- Since F₂ has the strongest bond, it requires the most energy to break. Consequently, it absorbs radiation of the least wavelength.
- Therefore, the halogen that absorbs visible radiation of the least wavelength is **F₂ (fluorine)**.

### Final Answer:
The halogen that absorbs visible radiation of least wavelength is **F₂ (fluorine)**.