Atomic radil of fluorine and neon in Angstrom units are respectively given by

Options

A. `0.72,1.60`
B. `1.60, 1.60`
C. `0.72, 0.72`
D. `1.60, 0.72`

Explanation

To solve the question regarding the atomic radii of fluorine and neon in Angstrom units, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Atomic Radii**: Atomic radii can be defined in several ways, including covalent radii and Van der Waals radii. For fluorine, we typically refer to its covalent radius, while for neon, we refer to its Van der Waals radius since neon is a noble gas.

2. **Periodic Trends**: As you move from left to right across a period in the periodic table, the atomic size generally decreases due to increasing nuclear charge, which pulls the electrons closer to the nucleus. Neon, being to the right of fluorine in the periodic table, will have a larger atomic radius when considering Van der Waals radii.

3. **Values of Atomic Radii**:
- The covalent radius of fluorine (F) is approximately 0.72 Å.
- The Van der Waals radius of neon (Ne) is approximately 1.60 Å.

4. **Comparison of Radii**:
- The atomic radius of fluorine (0.72 Å) is less than that of neon (1.60 Å). This is consistent with the trend where nonmetals (like fluorine) have smaller radii compared to noble gases (like neon).

5. **Conclusion**: Based on the values and the comparison, the atomic radii of fluorine and neon in Angstrom units are respectively 0.72 Å and 1.60 Å.

### Final Answer:
The atomic radii of fluorine and neon in Angstrom units are respectively 0.72 Å and 1.60 Å.