The radii of `F,F^(-),O` and `O^(-2)` are in the order of

Options

A. `O^(2-) gt O gt F^(-)F`
B. `F^(-) gt O^(2-) gt F gt O`
C. `O^(2-) gt F^(-) gt O gt F`
D. `O^(2) gt F - gt F gt O`

Answer

To determine the order of the radii of F, F^(-), O, and O^(-2), we need to consider the effects of atomic structure and charge on atomic and ionic radii.

### Step-by-Step Solution:

1. **Understanding Atomic Radii**:
- Atomic radii generally decrease as you move from left to right across a period in the periodic table. This is due to the increase in nuclear charge, which pulls the electrons closer to the nucleus.

2. **Identifying the Elements**:
- We have fluorine (F) and oxygen (O). Fluorine is to the right of oxygen in the periodic table.
- Therefore, the atomic radius of fluorine (F) will be smaller than that of oxygen (O).

3. **Considering Ionic Charges**:
- When an atom gains electrons to become negatively charged (anions), its size increases because the added electrons repel each other and the nucleus has less pull on the outer electrons.
- For F^(-), fluorine has gained one electron, which will increase its size compared to neutral F.
- For O^(-2), oxygen has gained two electrons, which will further increase its size compared to neutral O.

4. **Comparing the Sizes**:
- The order of sizes based on the above points is:
- O^(-2) (largest, due to gaining two electrons)
- F^(-) (next largest, due to gaining one electron)
- O (neutral, smaller than F^(-))
- F (smallest, neutral atom)

5. **Final Order**:
- Therefore, the order of the radii from largest to smallest is:
- O^(-2) > F^(-) > O > F

### Answer:
The radii of F, F^(-), O, and O^(-2) are in the order of:
**O^(-2) > F^(-) > O > F**