Which of the following species has four lone..

Options

A. `He`
B. `O^(-)`
C. `I`
D. `Cl^(-)`

Explanation

To determine which of the given species has four lone pairs, we will analyze each species by drawing their Lewis dot structures and counting the lone pairs of electrons.

### Step-by-Step Solution:

1. **Identify the species**: The species we need to analyze are He (Helium), O⁻ (Oxygen ion), Cl⁻ (Chlorine ion), and I (Iodine).

2. **Draw the Lewis dot structure for Helium (He)**:
- Helium has 2 electrons.
- Lewis structure: He: ••
- Lone pairs: 1 (since it has 2 electrons, they form 1 lone pair).

3. **Draw the Lewis dot structure for Oxygen ion (O⁻)**:
- Oxygen has 6 valence electrons. With a negative charge, it gains 1 additional electron, giving it 7 electrons.
- Lewis structure: ••:O:•• (2 electrons in lone pairs, 1 bond with another atom, and 3 lone pairs).
- Lone pairs: 3.

4. **Draw the Lewis dot structure for Chlorine ion (Cl⁻)**:
- Chlorine has 7 valence electrons. With a negative charge, it gains 1 additional electron, resulting in 8 electrons.
- Lewis structure: ••:Cl:•• (4 lone pairs).
- Lone pairs: 4.

5. **Draw the Lewis dot structure for Iodine (I)**:
- Iodine has 7 valence electrons.
- Lewis structure: ••:I:•• (similar to chlorine, but it typically has fewer lone pairs due to bonding).
- Lone pairs: 3 (if bonded to another atom).

6. **Count the lone pairs**:
- He: 1 lone pair
- O⁻: 3 lone pairs
- Cl⁻: 4 lone pairs
- I: 3 lone pairs

7. **Conclusion**: The species that has 4 lone pairs is **Cl⁻ (Chlorine ion)**.

### Final Answer:
The species with four lone pairs is **Cl⁻ (Chlorine ion)**.

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