Select from each group the species which has the smallest radius stating appropriate reason
`(a) O, O^(-) , O^(2-) (b) K^(+) , Sr^(2+) , Ar ( C) Si, P,Cl`

Explanation

To solve the problem of selecting the species with the smallest radius from each group, we need to analyze the species based on their atomic or ionic characteristics. Let's break it down step by step.

### Step 1: Analyze Group (a) - O, O^(-), O^(2-)

1. **Identify the species**: We have three species: O (neutral oxygen atom), O^(-) (oxide ion), and O^(2-) (peroxide ion).
2. **Understand the charge effect**: The radius of an atom or ion is influenced by its charge. Generally, as the negative charge increases, the size of the ion increases due to the addition of electrons which leads to electron-electron repulsion.
3. **Comparison**:
- O has 8 protons and 8 electrons (neutral).
- O^(-) has 8 protons and 9 electrons (one extra electron).
- O^(2-) has 8 protons and 10 electrons (two extra electrons).
4. **Conclusion**: The neutral oxygen atom (O) has the smallest radius because it has the least number of electrons and the same number of protons compared to the anions.

**Answer for Group (a)**: O (smallest radius).

### Step 2: Analyze Group (b) - K^(+), Sr^(2+), Ar

1. **Identify the species**: We have K^(+) (potassium ion), Sr^(2+) (strontium ion), and Ar (argon).
2. **Understand the charge effect**: The ionic size is inversely related to the positive charge. The more positive charge an ion has, the smaller its radius due to increased effective nuclear charge pulling the electrons closer.
3. **Comparison**:
- K^(+) has 19 protons and 18 electrons.
- Sr^(2+) has 38 protons and 36 electrons.
- Ar has 18 protons and 18 electrons (neutral atom).
4. **Conclusion**: Among these, Sr^(2+) has the highest positive charge and thus the smallest radius.

**Answer for Group (b)**: Sr^(2+) (smallest radius).

### Step 3: Analyze Group (c) - Si, P, Cl

1. **Identify the species**: We have Si (silicon), P (phosphorus), and Cl (chlorine).
2. **Understand the periodic trend**: As we move across a period from left to right, the atomic radius decreases due to the increase in nuclear charge without a significant increase in shielding effect.
3. **Comparison**:
- Si has 14 protons.
- P has 15 protons.
- Cl has 17 protons.
4. **Conclusion**: Cl, being the furthest right in the period, has the smallest atomic radius among these three elements.

**Answer for Group (c)**: Cl (smallest radius).

### Final Summary of Answers:
- (a) O
- (b) Sr^(2+)
- (c) Cl