What is the correct increasing order of ionic atomic radii in the following ?

Options

A. `Si^(4+)ltP^(5+)ltS^(6+)ltCl^(7+)`
B. `P^(5+)ltSi^(4+)ltCl^(7+) lt S^(6+)`
C. `Cl^(7+) lt S^(6+)ltP^(5+) lt Si^(4+)`
D. `S^(6+)ltP^(5+)ltCl^(7+)lt Si^(4+)`

Answer

To determine the correct increasing order of ionic atomic radii for the elements silicon (Si), phosphorus (P), sulfur (S), and chlorine (Cl), we need to consider the trends in ionic radii as we move across a period in the periodic table.

### Step-by-Step Solution:

1. **Identify the Elements and Their Ionic Forms**:
- The elements given are silicon (Si), phosphorus (P), sulfur (S), and chlorine (Cl).
- We need to consider their common ionic forms:
- Silicon typically forms Si⁴⁺.
- Phosphorus typically forms P³⁻.
- Sulfur typically forms S²⁻.
- Chlorine typically forms Cl⁻.

2. **Understand the Trend Across a Period**:
- As we move from left to right across a period in the periodic table, the effective nuclear charge increases. This means that the nucleus has a stronger pull on the electrons.
- As a result, the ionic radius tends to decrease because the electrons are pulled closer to the nucleus.

3. **Compare the Ionic Radii**:
- Si⁴⁺ has the smallest radius because it has lost four electrons, resulting in a high effective nuclear charge and a strong attraction to the remaining electrons.
- P³⁻ has a larger radius than Si⁴⁺ because it has gained three electrons, leading to increased electron-electron repulsion, but it still has a relatively high effective nuclear charge.
- S²⁻ has a larger radius than P³⁻ for the same reason; it has gained two electrons, increasing repulsion.
- Cl⁻ has the largest radius because it has gained one electron, resulting in the least effective nuclear charge compared to the others.

4. **Establish the Increasing Order**:
- Based on the analysis, the increasing order of ionic radii is:
- Si⁴⁺ < P³⁻ < S²⁻ < Cl⁻

5. **Final Answer**:
- Therefore, the correct increasing order of ionic atomic radii is:
- **Si⁴⁺ < P³⁻ < S²⁻ < Cl⁻**