Which ion has the largest radius ?

Options

A. `Se^(2-) `
B. `F^(-)`
C. `O^(2-)`
D. `Rb^(+)`

Answer

To determine which ion has the largest radius among the given options (selenium 2- (Se²⁻), fluorine (F⁻), oxygen 2- (O²⁻), and rubidium (Rb⁺)), we can follow these steps:

### Step 1: Understand the Electronic Configurations
- **Selenium 2- (Se²⁻)**: Gains 2 electrons, achieving the electronic configuration of krypton (Kr).
- **Fluorine (F⁻)**: Gains 1 electron, achieving the electronic configuration of neon (Ne).
- **Oxygen 2- (O²⁻)**: Gains 2 electrons, also achieving the electronic configuration of neon (Ne).
- **Rubidium (Rb⁺)**: Loses 1 electron, achieving the electronic configuration of krypton (Kr).

### Step 2: Compare the Noble Gas Configurations
- Both Se²⁻ and Rb⁺ have the electronic configuration of Kr, while F⁻ and O²⁻ have the configuration of Ne.
- The atomic radius of Kr is larger than that of Ne.

### Step 3: Analyze the Effective Nuclear Charge (Z_eff)
- **Selenium 2-**: When selenium gains 2 electrons, the number of protons (Z) remains the same, but the effective nuclear charge (Z_eff) decreases because there are more electrons than protons. This leads to a larger radius.
- **Rubidium 1+**: When rubidium loses an electron, the number of protons increases relative to the number of electrons, leading to an increase in Z_eff, which causes the radius to decrease.

### Step 4: Conclusion
- Since selenium 2- has a lower effective nuclear charge compared to rubidium 1+, it will have a larger ionic radius.
- Therefore, the ion with the largest radius is **selenium 2- (Se²⁻)**.

### Final Answer
The ion that has the largest radius is **selenium 2- (Se²⁻)**.

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