To determine which ion has the largest radius among the given options (Se²⁻, F⁻, O²⁻, and Rb⁺), we can follow these steps:
### Step 1: Understand the Concept of Ionic Radius
The ionic radius is influenced by the number of protons and electrons in an ion. Generally, as you move down a group in the periodic table, the ionic radius increases due to the addition of electron shells. Conversely, moving across a period from left to right typically leads to a decrease in ionic radius due to increased nuclear charge attracting the electrons more strongly.
**Hint:** Remember that more protons in the nucleus lead to a stronger attraction to the electrons, which can decrease the size of the ion.
### Step 2: Identify the Ions and Their Electron Configurations
- **Se²⁻ (Selenium)**: Atomic number 34, has gained 2 electrons (total 36 electrons).
- **F⁻ (Fluoride)**: Atomic number 9, has gained 1 electron (total 10 electrons).
- **O²⁻ (Oxide)**: Atomic number 8, has gained 2 electrons (total 10 electrons).
- **Rb⁺ (Rubidium)**: Atomic number 37, has lost 1 electron (total 36 electrons).
**Hint:** Write down the total number of electrons for each ion to compare their sizes effectively.
### Step 3: Compare the Positions in the Periodic Table
- **Selenium (Se)** is in the 4th period and 16th group.
- **Fluoride (F)** is in the 2nd period and 17th group.
- **Oxide (O)** is in the 2nd period and 16th group.
- **Rubidium (Rb)** is in the 5th period and 1st group.
**Hint:** The position in the periodic table can give clues about the size; elements in lower periods generally have larger radii.
### Step 4: Analyze the Effective Nuclear Charge
- For **Se²⁻**, with 34 protons and 36 electrons, the effective nuclear charge is relatively low, leading to a larger radius.
- For **F⁻** and **O²⁻**, both have 9 and 8 protons respectively but have gained electrons, which leads to a smaller radius due to higher effective nuclear charge.
- For **Rb⁺**, it has 37 protons and 36 electrons, which means it has a strong effective nuclear charge, leading to a smaller radius compared to Se²⁻.
**Hint:** The effective nuclear charge is crucial in determining the size of the ion; the more protons compared to electrons, the smaller the ion tends to be.
### Step 5: Conclusion
Based on the analysis, **Se²⁻** has the largest radius because it has the lowest effective nuclear charge among the ions considered, allowing for a larger size compared to the others.
### Final Answer
**Se²⁻ (Selenium ion) has the largest radius.**
---
