The average S - F bond energy in kJ mol$$^{-1}$$ of $$SF_6$$ is ______. (Rounded off to the nearest integer)
[Given: The values of standard enthalpy of formation of $$SF_6(g)$$, $$S(g)$$ and $$F(g)$$ are $$-1100$$, 275 and 80 kJ mol$$^{-1}$$ respectively.]

We need to find the average S-F bond energy in $$SF_6$$. The formation of $$SF_6$$ from gaseous atoms involves forming 6 S-F bonds.

The reaction for formation of $$SF_6(g)$$ from gaseous atoms is: $$S(g) + 6F(g) \rightarrow SF_6(g)$$.

The enthalpy of this reaction equals $$-6 \times (\text{average S-F bond energy})$$ since 6 bonds are formed (bond formation is exothermic).

Using Hess's law: $$\Delta H = \Delta_f H^\circ(SF_6) - \Delta_f H^\circ(S(g)) - 6 \times \Delta_f H^\circ(F(g))$$.

Substituting: $$\Delta H = (-1100) - (275) - 6 \times (80) = -1100 - 275 - 480 = -1855$$ kJ/mol.

Since $$\Delta H = -6 \times (\text{bond energy})$$, the average S-F bond energy is $$\frac{1855}{6} = 309.17$$ kJ/mol.

Rounded to the nearest integer, the average S-F bond energy is $$309$$ kJ/mol.