In the reaction `I^(-) +I_(2) rarr I_(3)^(-)`, the Lewis acid is ……….. .

Explanation

To determine the Lewis acid in the reaction \( I^- + I_2 \rightarrow I_3^- \), we can follow these steps:

### Step 1: Understand the Lewis Acid-Base Concept
A Lewis acid is defined as a substance that can accept a pair of electrons, while a Lewis base is a substance that can donate a pair of electrons.

**Hint:** Remember that Lewis acids are electron pair acceptors and Lewis bases are electron pair donors.

### Step 2: Analyze the Reactants
In the reaction, we have:
- \( I^- \) (iodide ion)
- \( I_2 \) (iodine molecule)

### Step 3: Identify the Electron Donor
The iodide ion \( I^- \) has an extra electron, giving it a total of 8 valence electrons. This extra electron allows \( I^- \) to donate a pair of electrons.

**Hint:** Look for the species that has an extra electron to donate.

### Step 4: Identify the Electron Acceptor
The iodine molecule \( I_2 \) consists of two iodine atoms bonded together. Each iodine atom has 7 valence electrons, and when they bond, they share a pair of electrons. However, \( I_2 \) can accept an additional pair of electrons from \( I^- \) to form the \( I_3^- \) ion.

**Hint:** Consider which species can accept electrons to form a new bond.

### Step 5: Formation of the Dative Bond
When \( I^- \) donates a pair of electrons to \( I_2 \), it forms a dative bond, resulting in the formation of \( I_3^- \). In this process, \( I_2 \) acts as the electron acceptor.

**Hint:** The formation of a new bond indicates which species is acting as the acid.

### Conclusion
Based on the analysis, the Lewis acid in the reaction \( I^- + I_2 \rightarrow I_3^- \) is \( I_2 \), as it accepts the pair of electrons from the iodide ion.

**Final Answer:** The Lewis acid is \( I_2 \).