Solution

Given that: $${S_{(s)}} + {\text{ }}3{F_{2(g)}} \to {\text{ }}S{F_{6(g)}};{\text{ }}\Delta {\text{ }}H{\text{ }} = {\text{ }} - {\text{ }}1100{\text{ }}kJ \ \ \ (1)$$ $$S{_{(s)}} \to {\text{ }}{S_{(g)}};{\text{ }}\Delta {\text{ }}H{\text{ }} = {\text{ }}275{\text{ }}kJ{\text{ }} \ \ \ (2)$$ $$\frac{1}{2}{\text{ }}{F_{2(g)}} \to {\text{ }}{F_{(g)}};{\text{ }}\Delta {\text{ }}H{\text{ }} = {\text{ }}80{\text{ }}kJ \ \ \ (3)$$ To have $$S{F_{6(g)}} \to {\text{ }}{S_{(g)}} + {\text{ }}6{F_{(g)}}$$ we can proceed as $$\text{equation}\left( 2 \right) + 6 \times {\text{equation}}\left( 3 \right) - {\text{equation}}\left( 1 \right)$$ $$S{F_{6(g)}} \to {\text{ }}{S_{(g)}} + {\text{ }}6{F_{(g)}};{\text{ }}\Delta {\text{ }}H{\text{ }} = {\text{ }}1855{\text{ }}kJ$$ Thus average bond energy for $$S - F$$ bond$$ = \dfrac{{1855}}{6} = 309.16kJ$$.