The reaction, Xe(excess) $+{\mathrm{F}}_2⟶{\mathrm{XeF}}_2$ is conducted at

The reaction of the formation of xenon and fluorine forming xenon difluoride gas is as follows. 

 ${\mathrm{Xe}}_{(\mathrm{g},\mathrm{excess})}+{\mathrm{F}}_{2\left(\mathrm{g}\right)}\underset{1 \mathrm{bar}}{\overset{673\mathrm{K}}{\to }}{\mathrm{XeF}}_{2\left(\mathrm{s}\right)}$

Above this temperature and pressure xenon tetrafluoride is formed.

Hence option C is the correct answer.