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)/3Å`
C. 0.5 Å
D. `(1.54)/2Å`

Answer

To calculate the covalent radius of the fluorine atom (R_F) using the given bond distances, we can follow these steps:

### Step 1: Write down the given bond distances
- The bond distance for C-F in CF₄ is 1.33 Å.
- The bond distance for Si-F in SiF₄ is 1.54 Å.
- The bond distance for C-Si is 1.87 Å.

### Step 2: Set up the equations based on bond lengths
From the bond distances, we can set up the following equations:

1. For the C-F bond:
\[ R_C + R_F = 1.33 \, \text{Å} \quad \text{(Equation 1)} \]

2. For the Si-F bond:
\[ R_{Si} + R_F = 1.54 \, \text{Å} \quad \text{(Equation 2)} \]

3. For the C-Si bond:
\[ R_C + R_{Si} = 1.87 \, \text{Å} \quad \text{(Equation 3)} \]

### Step 3: Solve for R_C and R_Si in terms of R_F
From Equation 1:
\[ R_C = 1.33 - R_F \quad \text{(Substituting into Equation 3)} \]

From Equation 2:
\[ R_{Si} = 1.54 - R_F \quad \text{(Substituting into Equation 3)} \]

### Step 4: Substitute R_C and R_Si into Equation 3
Now substitute R_C and R_Si into Equation 3:
\[ (1.33 - R_F) + (1.54 - R_F) = 1.87 \]

### Step 5: Simplify the equation
Combine the terms:
\[ 1.33 + 1.54 - 2R_F = 1.87 \]
\[ 2.87 - 2R_F = 1.87 \]

### Step 6: Solve for R_F
Rearranging gives:
\[ 2R_F = 2.87 - 1.87 \]
\[ 2R_F = 1.00 \]
\[ R_F = \frac{1.00}{2} = 0.50 \, \text{Å} \]

### Conclusion
The covalent radius of the fluorine atom (R_F) is **0.50 Å**.

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