Question: Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the \({\rm{S - F}}\) bond in \({\rm{S}}{{\rm{F}}_{\rm{4}}}{\rm{(g)}}\) or in \({\rm{S}}{{\rm{F}}_{\rm{6}}}{\rm{(g)}}\)?

Bond energy is a measure of strength of the bond in a chemical bond.

The reactions with their enthalpy values are –

\(\begin{array}{l}{\rm{S}}{{\rm{F}}_{\rm{4}}}{\rm{(g)}} \to \frac{{\rm{1}}}{{\rm{8}}}{\rm{\;}}{{\rm{S}}_{\rm{8}}}{\rm{(s) + 2\;}}{{\rm{F}}_{\rm{2}}}{\rm{(g) \Delta }}{{\rm{H}}_{\rm{1}}}^{\rm{^\circ }}{\rm{ = - \Delta H}}_{{\rm{f}}\left( {{\rm{S}}{{\rm{F}}_{\rm{4}}}{\rm{(g)}}} \right)}^{\rm{^\circ }}\\\frac{{\rm{1}}}{{\rm{8}}}{\rm{\;}}{{\rm{S}}_{\rm{8}}}{\rm{(s)}} \to {\rm{S(g) \Delta H}}_{\rm{2}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{{\rm{f(S)(g)]}}}^{\rm{^\circ }}\\{\rm{2\;}}{{\rm{F}}_{\rm{2}}}{\rm{(g)}} \to {\rm{4\;F(g) 4\Delta }}{{\rm{H}}_{\rm{3}}}^{\rm{^\circ }}{\rm{ = 4\Delta H}}_{{\rm{f(F(g))}}}^{\rm{^\circ }}\end{array}\)

The net reaction is –

\({\rm{S}}{{\rm{F}}_{\rm{4}}}{\rm{(g)}} \to {\rm{S(g) + 4\;F(g)}}\)

\(\begin{array}{c}{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{\rm{1}}^{\rm{^\circ }}{\rm{ + \Delta H}}_{\rm{2}}^{\rm{^\circ }}{\rm{ + 4\Delta H}}_{\rm{3}}^{\rm{^\circ }}\\{{\rm{D}}_{{\rm{S}}{{\rm{F}}_{\rm{4}}}}}{\rm{ = \Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{298}}}{\rm{ = - \Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{f}}\left( {{\rm{S}}{{\rm{F}}_{\rm{4}}}{\rm{(g)}}} \right)}{\rm{ + \Delta H}}_{{\rm{f(S(g))}}}^{\rm{^\circ }}{\rm{ + 4\Delta H}}_{{\rm{f(F(g))}}}^{\rm{^\circ }}\end{array}\)

The bond energy is calculated as –

\(\begin{array}{c}{{\rm{D}}_{{\rm{S}}{{\rm{F}}_{\rm{4}}}}}{\rm{ = - 728}}{\rm{.43 + 278}}{\rm{.81 + 4(79}}{\rm{.4)}}\\{\rm{ = 1369}}{\rm{.7\;kJ}}\\{{\rm{D}}_{{\rm{S - F}}}}{\rm{ = }}\frac{{{\rm{1324}}{\rm{.84}}}}{{\rm{4}}}{\rm{ = 331}}{\rm{.21\;kJ}}\end{array}\)

If we proceed in the manner as described above, then \({\rm{ - \Delta Hf}}\left( {{\rm{S}}{{\rm{F}}_{\rm{6}}}{\rm{(\;g)}}} \right){\rm{ = 1220}}{\rm{.5 kJ}}\).

Now, since \({\rm{6\;F(g)}}\) and 1\({\rm{S(g)}}\) contribute \({\rm{755}}{\rm{.21\;kJ}}\), then

\({\rm{D}}{{\rm{F}}_{{\rm{S}}{{\rm{F}}_{\rm{6}}}}}{\rm{ = 1975}}{\rm{.71\;kJ}}\) and \({{\rm{D}}_{{\rm{S - F}}}}{\rm{ = }}\frac{{{\rm{1975}}{\rm{.71}}}}{{\rm{6}}}{\rm{ = 329}}{\rm{.3\;kJ}}\).

Therefore, based on the calculations it can be said that \({\rm{S - F}}\) bond is stronger.