To solve the question of how \( Cl_2 \), \( Br_2 \), or \( I_2 \) reacts with hot concentrated alkali solution, we can break down the reactions step by step.
### Step 1: Identify the Reactants
The reactants in this case are halogens (\( Cl_2 \), \( Br_2 \), or \( I_2 \)) and a hot concentrated alkali solution, which we can represent as sodium hydroxide (\( NaOH \)).
### Step 2: Write the General Reaction
For each halogen, the general reaction with sodium hydroxide can be written as:
\[
\text{Halogen} + \text{NaOH (hot, concentrated)} \rightarrow \text{Halide} + \text{Halate} + \text{Water}
\]
### Step 3: Specific Reactions
1. **For Chlorine (\( Cl_2 \))**:
\[
Cl_2 + 6 NaOH \rightarrow 5 NaCl + NaClO_3 + 3 H_2O
\]
- Products: Sodium chloride (\( NaCl \)), sodium chlorate (\( NaClO_3 \)), and water (\( H_2O \)).
2. **For Bromine (\( Br_2 \))**:
\[
Br_2 + 6 NaOH \rightarrow 5 NaBr + NaBrO_3 + 3 H_2O
\]
- Products: Sodium bromide (\( NaBr \)), sodium bromate (\( NaBrO_3 \)), and water (\( H_2O \)).
3. **For Iodine (\( I_2 \))**:
\[
I_2 + 6 NaOH \rightarrow 5 NaI + NaIO_3 + 3 H_2O
\]
- Products: Sodium iodide (\( NaI \)), sodium iodate (\( NaIO_3 \)), and water (\( H_2O \)).
### Step 4: Summary of Products
From the reactions above, we can summarize the products formed when \( Cl_2 \), \( Br_2 \), or \( I_2 \) reacts with hot concentrated alkali:
- **Halide Ion**: \( NaCl \) (for \( Cl_2 \)), \( NaBr \) (for \( Br_2 \)), \( NaI \) (for \( I_2 \))
- **Halate**: \( NaClO_3 \) (for \( Cl_2 \)), \( NaBrO_3 \) (for \( Br_2 \)), \( NaIO_3 \) (for \( I_2 \))
- **Water**: \( H_2O \)
### Final Answer
The reaction of \( Cl_2 \), \( Br_2 \), or \( I_2 \) with hot concentrated alkali solution forms:
- **Halide Ion + Halate + Water**
