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 a halogen and has 7 valence electrons. Since there are three iodine atoms in \( I_3^- \), we first calculate the total number of valence electrons contributed by the iodine atoms.

\[ \text{Total valence electrons from iodine} = 3 \times 7 = 21 \]

### Step 2: Account for the Negative Charge
The \( I_3^- \) ion has an additional electron due to the negative charge. Therefore, we add 1 more electron to our total.

\[ \text{Total valence electrons} = 21 + 1 = 22 \]

### Step 3: Draw the Lewis Structure
In the Lewis structure of \( I_3^- \), we connect the three iodine atoms with single bonds. Each bond uses 2 electrons, so we use 2 electrons for the bonds between the iodine atoms.

\[ \text{Electrons used in bonds} = 2 \text{ (for one bond)} \times 2 \text{ (two bonds)} = 4 \]

### Step 4: Calculate Remaining Electrons
Now, we subtract the number of electrons used in bonding from the total number of valence electrons.

\[ \text{Remaining electrons} = 22 - 4 = 18 \]

### Step 5: Distribute Remaining Electrons as Lone Pairs
Each iodine atom will try to achieve an octet. Each iodine atom will have 3 lone pairs (6 electrons) and will share 1 electron with the other iodine atoms in the bond.

- Each iodine atom has 3 lone pairs, which means:
\[ \text{Lone pairs per iodine} = 3 \]
- For three iodine atoms:
\[ \text{Total lone pairs} = 3 \times 3 = 9 \]

### Conclusion
Thus, the total number of lone pairs of electrons in the \( I_3^- \) ion is 9.

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

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