Which of the following is the largest?

Options

A. `Cl^(-)`
B. `S^(2-)`
C. `Na^(+)`
D. `F^(-)`

Answer

To determine which of the given ions is the largest in size, we will analyze the electron configurations and the effective nuclear charge of each ion. The ions we are considering are Cl⁻, S²⁻, Na⁺, and F⁻.

### Step 1: Analyze the Electron Configuration
1. **Cl⁻ (Chloride Ion)**:
- Atomic number of Cl = 17
- Electrons in Cl⁻ = 18 (gains 1 electron)
- Electron configuration: 2 in the first shell, 8 in the second shell, and 8 in the third shell.

2. **S²⁻ (Sulfide Ion)**:
- Atomic number of S = 16
- Electrons in S²⁻ = 18 (gains 2 electrons)
- Electron configuration: 2 in the first shell, 8 in the second shell, and 8 in the third shell.

3. **Na⁺ (Sodium Ion)**:
- Atomic number of Na = 11
- Electrons in Na⁺ = 10 (loses 1 electron)
- Electron configuration: 2 in the first shell, 8 in the second shell.

4. **F⁻ (Fluoride Ion)**:
- Atomic number of F = 9
- Electrons in F⁻ = 10 (gains 1 electron)
- Electron configuration: 2 in the first shell, 8 in the second shell.

### Step 2: Determine the Number of Electrons and Shells
- Cl⁻ and S²⁻ both have 18 electrons and occupy 3 shells.
- Na⁺ and F⁻ have only 10 electrons and occupy 2 shells.

### Step 3: Compare the Effective Nuclear Charge
- **Effective Nuclear Charge (Z_eff)**: This is the net positive charge experienced by an electron in a multi-electron atom. It is calculated as:
\[ Z_{\text{eff}} = Z - S \]
where Z is the number of protons and S is the shielding constant (approximate number of inner electrons).

- For Cl⁻:
- Protons = 17, Electrons = 18
- Z_eff = 17 - (inner electrons shielding) = lower effective nuclear charge due to more electrons.

- For S²⁻:
- Protons = 16, Electrons = 18
- Z_eff = 16 - (inner electrons shielding) = lower effective nuclear charge due to more electrons.

- For Na⁺:
- Protons = 11, Electrons = 10
- Z_eff = 11 - (inner electrons shielding) = higher effective nuclear charge due to fewer electrons.

- For F⁻:
- Protons = 9, Electrons = 10
- Z_eff = 9 - (inner electrons shielding) = higher effective nuclear charge due to fewer electrons.

### Step 4: Conclusion
- The size of an ion increases with the number of electrons and decreases with the effective nuclear charge.
- Since Cl⁻ and S²⁻ have the same number of electrons but S²⁻ has fewer protons, S²⁻ will be larger than Cl⁻.
- Na⁺ and F⁻ are smaller than both Cl⁻ and S²⁻ due to having fewer electrons and higher effective nuclear charge.

Thus, the largest ion among the given options is **S²⁻**.