Bond distance `C-F` in `(CF_(4))` & `Si-F` in `(SiF_(4))` are respective `1.33 Å` & `1.54 Å. C-SI` bond is `1.87 Å`. Calculate the colvalent radius of F atom ignoring the electronegativity differences.

Options

A. `0.64 Å`
B. `(1.33 +1.54 +1.8)/(3) Å`
C. `0.5 Å`
D. `(1.54)/(2) Å`

Explanation

To calculate the covalent radius of the fluorine atom (F) in the compounds CF₄ and SiF₄, we can use the bond length data provided and the formula for bond lengths in covalent compounds.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Bond distance C-F in CF₄ = 1.33 Å
- Bond distance Si-F in SiF₄ = 1.54 Å
- Bond distance C-Si in SiF₄ = 1.87 Å

2. **Set Up the Equations:**
- The bond length formula states that:
\[ \text{Bond Length (AB)} = r_A + r_B - 0.09 \times \Delta EN \]
- Since we are ignoring the electronegativity differences, we can simplify this to:
\[ \text{Bond Length (AB)} = r_A + r_B \]

3. **Write the Equations for CF₄ and SiF₄:**
- For CF₄:
\[ r_C + r_F = 1.33 \quad \text{(Equation 1)} \]
- For SiF₄:
\[ r_Si + r_F = 1.54 \quad \text{(Equation 2)} \]

4. **Add the Two Equations:**
- Adding Equation 1 and Equation 2:
\[ (r_C + r_F) + (r_Si + r_F) = 1.33 + 1.54 \]
\[ r_C + r_Si + 2r_F = 2.87 \quad \text{(Equation 3)} \]

5. **Use the C-Si Bond Length:**
- From the bond length of C-Si:
\[ r_C + r_Si = 1.87 \quad \text{(Equation 4)} \]

6. **Substitute Equation 4 into Equation 3:**
- Substitute \( r_C + r_Si \) from Equation 4 into Equation 3:
\[ 1.87 + 2r_F = 2.87 \]

7. **Solve for \( r_F \):**
- Rearranging gives:
\[ 2r_F = 2.87 - 1.87 \]
\[ 2r_F = 1.00 \]
\[ r_F = 0.5 \, \text{Å} \]

### Final Answer:
The covalent radius of the fluorine atom (F) is **0.5 Å**.