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Q.
The standard enthalpies of formation of $SF_6(g), S(g)$ and $F(g)$ are $-1150, +280$ and $+ 85 kJ mol^{-1}$. Thus, average bond energy of $\left(S - F\right)$ in $SF_6$ is
Thermodynamics
Solution:
$S\left(g\right)+6F\left(g\right)\to SF_{6}\left(g\right)$
Based on $BE$ values
$\Delta_{f}H^{\circ}\left(SF_{6}\right)=-1100=\Delta_{f}H^{\circ}\left(S,g\right)+6\Delta_{f}H^{\circ}\left(F, g\right)-6\left(B E\right)_{S-F}$
$-1150=280+6\times85-6\left(BE\right)_{S-F}$
$\therefore \left(BE\right)_{S-F}=\frac{280+510+1150}{6}=323.33\, kJ mol^{-1}$