Total number of lone pair of electrons in `I_3^-`, ion is:

Options

A. 3
B. 6
C. 9
D. 12

Explanation

To determine the total number of lone pairs of electrons in the \(I_3^-\) ion, we can follow these steps:

### Step 1: Determine the Valence Electrons
- Iodine (I) is in group 17 of the periodic table, which means it has 7 valence electrons.
- Since \(I_3^-\) has a negative charge, we need to add one extra electron to account for the charge.

**Calculation:**
- Total valence electrons from three iodine atoms: \(3 \times 7 = 21\)
- Adding the extra electron for the negative charge: \(21 + 1 = 22\) valence electrons.

### Step 2: Draw the Lewis Structure
- In the \(I_3^-\) ion, one iodine atom will be the central atom, and the other two iodine atoms will be bonded to it.
- Each bond between the central iodine and the outer iodines will use 2 electrons.

**Bonds:**
- There are 2 bonds (one to each outer iodine), using \(2 \times 2 = 4\) electrons.

### Step 3: Calculate Remaining Electrons
- Subtract the electrons used in bonding from the total number of valence electrons:

**Calculation:**
- Remaining electrons: \(22 - 4 = 18\) electrons.

### Step 4: Distribute Remaining Electrons as Lone Pairs
- The remaining 18 electrons will be distributed as lone pairs around the iodine atoms.
- Each lone pair consists of 2 electrons.

**Calculation:**
- Total lone pairs: \( \frac{18}{2} = 9\) lone pairs.

### Step 5: Count Lone Pairs on Each Iodine
- The central iodine atom has 3 lone pairs (6 electrons) and forms 2 bonds.
- Each of the two outer iodine atoms has 3 lone pairs (6 electrons) and forms 1 bond.

### Conclusion
- The total number of lone pairs of electrons in \(I_3^-\) is 9.

**Final Answer:** The total number of lone pairs of electrons in \(I_3^-\) is **9**.

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