To determine the correct order of ionic radii among the given options, we need to analyze the effective nuclear charge and how it affects the size of the ions. Let's break down the analysis step by step.
### Step 1: Understanding Ionic Radii
Ionic radii refer to the size of an ion in a crystal lattice. The size of an ion is influenced by its effective nuclear charge (Z_eff), which is the net positive charge experienced by an electron in a multi-electron atom. A higher effective nuclear charge typically results in a smaller ionic radius because the nucleus pulls the electrons closer.
### Step 2: Analyzing the First Option (H⁻, H, H⁺)
- **H⁻ (Hydride ion)**: Has 1 proton and 2 electrons. The effective nuclear charge is low (1 - 2 = -1), leading to a larger size.
- **H (Neutral hydrogen)**: Has 1 proton and 1 electron. The effective nuclear charge is 1 (1 - 1 = 0), leading to a medium size.
- **H⁺ (Hydrogen ion)**: Has 1 proton and 0 electrons. The effective nuclear charge is high (1 - 0 = 1), leading to the smallest size.
**Order of sizes**: H⁺ < H < H⁻ (smallest to largest)
### Step 3: Analyzing the Second Option (Na⁺, F⁻, O²⁻)
- **Na⁺ (Sodium ion)**: Has 11 protons and 10 electrons. The effective nuclear charge is 1 (11 - 10 = 1), leading to a smaller size.
- **F⁻ (Fluoride ion)**: Has 9 protons and 10 electrons. The effective nuclear charge is less (9 - 10 = -1), leading to a larger size than Na⁺.
- **O²⁻ (Oxide ion)**: Has 8 protons and 10 electrons. The effective nuclear charge is even lower (8 - 10 = -2), leading to the largest size.
**Order of sizes**: Na⁺ < F⁻ < O²⁻ (smallest to largest)
### Step 4: Analyzing the Third Option (F⁻, O²⁻, Na⁺)
This option is similar to the previous one but incorrectly orders the ions. The correct order should be Na⁺ < F⁻ < O²⁻.
### Step 5: Analyzing the Fourth Option (N³⁻, Mg²⁺, Al³⁺)
- **N³⁻ (Nitride ion)**: Has 7 protons and 10 electrons. The effective nuclear charge is low (7 - 10 = -3), leading to a large size.
- **Mg²⁺ (Magnesium ion)**: Has 12 protons and 10 electrons. The effective nuclear charge is higher (12 - 10 = 2), leading to a smaller size.
- **Al³⁺ (Aluminum ion)**: Has 13 protons and 10 electrons. The effective nuclear charge is even higher (13 - 10 = 3), leading to the smallest size.
**Order of sizes**: Al³⁺ < Mg²⁺ < N³⁻ (smallest to largest)
### Conclusion
After analyzing all the options, the correct order of ionic radii is found in the fourth option: **N³⁻ > Mg²⁺ > Al³⁺**.
### Final Answer
The correct order of ionic radii is given by option D: **N³⁻ > Mg²⁺ > Al³⁺**.
