Estimate the average `S-F` bond enthalpy in `SF_(6)`. The values of standard enthalpy of formation of `SF_(6)(g),S(g)` and `F(g)` are :-`1100, 274` and `80 kJ mol^(-1)` respectively.

Explanation

To estimate the average S-F bond enthalpy in SF₆, we can follow these steps:

### Step 1: Write the reaction for the formation of SF₆
The formation of sulfur hexafluoride (SF₆) from its elements can be represented as:
\[ S(g) + 3F₂(g) \rightarrow SF₆(g) \]

### Step 2: Use the standard enthalpy of formation values
We are given the following standard enthalpy of formation values:
- \( \Delta H_f^\circ(SF₆(g)) = -1100 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ(S(g)) = 274 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ(F(g)) = 80 \, \text{kJ/mol} \)

### Step 3: Write the equation for the enthalpy of formation
Using Hess's law, the enthalpy change for the reaction can be expressed as:
\[ \Delta H_f^\circ(SF₆) = \Delta H_f^\circ(S) + 6 \times \Delta H_f^\circ(F) \]
This can be rearranged to find the bond enthalpy:
\[ \Delta H_f^\circ(SF₆) = \Delta H_b(S-F) \times 6 \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ -1100 \, \text{kJ/mol} = 274 \, \text{kJ/mol} + 6 \times 80 \, \text{kJ/mol} \]

### Step 5: Calculate the total enthalpy of formation for the reactants
Calculate the total enthalpy of formation for the reactants:
\[ 6 \times 80 = 480 \, \text{kJ/mol} \]
So,
\[ \Delta H_f^\circ(S) + 6 \times \Delta H_f^\circ(F) = 274 + 480 = 754 \, \text{kJ/mol} \]

### Step 6: Rearranging to find the bond enthalpy
Now we can rearrange the equation to find the bond enthalpy:
\[ -1100 = 754 + 6 \times \Delta H_b(S-F) \]
\[ -1100 - 754 = 6 \times \Delta H_b(S-F) \]
\[ -1854 = 6 \times \Delta H_b(S-F) \]

### Step 7: Solve for the average S-F bond enthalpy
Now, divide by 6 to find the average S-F bond enthalpy:
\[ \Delta H_b(S-F) = \frac{-1854}{6} = -309 \, \text{kJ/mol} \]

### Final Answer
The average S-F bond enthalpy in SF₆ is:
\[ \Delta H_b(S-F) = 309 \, \text{kJ/mol} \]