Question

Among the following halogens \(F_2, \, Cl_2, \, Br_2, \, \text{and} \, I_2\)
Which can undergo disproportionation reaction?

• A. Only \(I_2\)
• B. \(Cl_2, \, Br_2, \, \text{and} \, I_2\)
• C. \(F_2, \, Cl_2, \, \text{and} \, Br_2\)
• D. \(F_2 \, \text{and} \, Cl_2\)

Answer 1

The Correct Option is B

The question asks which of the halogens \( F_2, \, Cl_2, \, Br_2, \) and \( I_2 \) can undergo a disproportionation reaction. Let's explore the concept of disproportionation and apply it to these halogens to identify which can undergo such reactions.

Concept of Disproportionation: A disproportionation reaction is a specific type of redox reaction in which a single element is simultaneously oxidized and reduced. In the context of halogens, this means that the halogen element will both gain and lose electrons, forming two different products with different oxidation states.

To determine which halogens can undergo disproportionation, let's analyze each:

1. \(F_2\): Fluorine is the most electronegative element and does not undergo disproportionation reactions because it cannot be oxidized further (it is in its highest oxidation state already as fluoride, \(F^-\)\)).
2. \(Cl_2\): Chlorine can undergo disproportionation as it can be both oxidized to form hypochlorite \((ClO^-)\) and reduced to chloride \((Cl^-)\). An example of such a disproportionation reaction is: \(2Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O\)
3. \(Br_2\): Bromine can also undergo disproportionation into bromate \((BrO_3^-)\) and bromide \((Br^-)\) in an alkaline medium.
4. \(I_2\): Iodine undergoes disproportionation to form iodate \((IO_3^-)\) and iodide \((I^-)\).

From the analysis above, it is clear that

• \(Cl_2\)
• \(Br_2\)
• \(I_2\)

can undergo disproportionation reactions, while \(F_2\) cannot. Therefore, the correct answer is \(Cl_2, \, Br_2, \, \text{and} \, I_2\).

Answer 2

To identify which halogens can undergo a disproportionation reaction, we must understand what disproportionation involves. Disproportionation is a type of redox reaction where a single substance is simultaneously oxidized and reduced, giving two different products.

Now, let's analyze each halogen:

• Fluorine \((F_2)\): Fluorine is the most electronegative element and the strongest oxidizing agent among the halogens. It cannot undergo disproportionation because it doesn't have a higher oxidation state than \(0\) to be oxidized further.
• Chlorine \((Cl_2)\): Chlorine can undergo disproportionation. An example is the reaction with water: \(2Cl_2 + 2H_2O \rightarrow HCl + HOCl\) Here, chlorine is both reduced to \(\text{HCl}\) and oxidized to \(\text{HOCl}\).
• Bromine \((Br_2)\): Bromine can also undergo disproportionation. Example reaction with \(\text{OH}^-\) is: \(3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O\) Here, bromine is simultaneously reduced to \(\text{Br}^-\) and oxidized to \(\text{BrO}_3^-\).
• Iodine \((I_2)\): Iodine also undergoes disproportionation. An example with basic solution: \(I_2 + 2OH^- \rightarrow I^- + IO^- + H_2O\) In this reaction, iodine is reduced to \(\text{I}^-\) and oxidized to \(\text{IO}^-\).

Thus, the halogens \(Cl_2\), \(Br_2\), and \(I_2\) can undergo disproportionation reactions because they can exist in intermediate oxidation states that allow both oxidation and reduction from the neutral molecule.

Therefore, the correct option is: Cl₂, Br₂, and I₂.