The standard heat of formation values of $S{F}_6\left(g\right),S\left(g\right)\text{ and }F\left(g\right)$ are: -1100, 275 and 80 kJ $mo{l}^{-1}$ respectively. Then the average $S - F$ bond energy  (in kJ $mo{l}^{-1}$ ) in $S{F}_6$ is ___ (Round off to nearest integer)

$S\left(g\right)+6F\left(g\right)\to S{F}_6\left(g\right) ; \Delta H_f=-1100kJmo{l^{-}}^1$

$S\left(s\right)\to S\left(g\right) ; \Delta H=+275kJmo{l^{-}}^1$

$\mathrm{½}{F}_2\left(g\right)\to F\left(g\right) ; \Delta H=80kJmo{l^{-}}^1$

Therefore heat of formation = Bond energy of reaction - Bond energy of product 

$-1100=[275+6\times 80]-[6\times S-F]$

Thus bond energy of $S-F=309kJmo{l}^{-1}$