In which of the following pairs of species,the size of the first species is not more than the second species?

Options

A. `Na^(+),F^(-)`
B. `Fe^(2+),Fe^(3+)`
C. Li,F
D. S,O

Explanation

To solve the question, we need to determine in which of the given pairs of species the size of the first species is not more than the size of the second species. This means we need to find pairs where the size of the first species is either smaller than or equal to the size of the second species.

### Step-by-Step Solution:

1. **Understanding Ionic and Atomic Size Trends**:
- As we move from left to right across a period in the periodic table, the atomic and ionic sizes decrease due to an increase in effective nuclear charge.
- As we move down a group, the atomic and ionic sizes increase due to the addition of electron shells.

2. **Analyzing Each Pair**:
- **Option A: Na⁺ and F⁻**
- Na⁺ (sodium ion) has lost one electron and has 11 protons and 10 electrons.
- F⁻ (fluoride ion) has gained one electron and has 9 protons and 10 electrons.
- Both are isoelectronic (same number of electrons), but Na⁺ is a cation and F⁻ is an anion. The size of cations is smaller than that of anions. Therefore, Na⁺ < F⁻.

- **Option B: Fe²⁺ and Fe³⁺**
- Fe²⁺ has lost 2 electrons and has a larger size than Fe³⁺, which has lost 3 electrons. Thus, Fe²⁺ > Fe³⁺.

- **Option C: Li and F**
- Lithium (Li) is in Group 1 and has a larger atomic size compared to fluorine (F), which is in Group 17. Therefore, Li > F.

- **Option D: S and O**
- Sulfur (S) is below oxygen (O) in Group 16. Therefore, S > O.

3. **Conclusion**:
- From the analysis, the only pair where the size of the first species is not more than the second species is **Option A: Na⁺ and F⁻**.

### Final Answer:
The correct answer is **Option A: Na⁺ and F⁻**.

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