# Large difference in boiling points of tetrafluoromethane and tetrachloromethane

Why is carbon tetrachloride $$\ce{CCl4}$$ is seen to posses liquid state (b.p. $$\pu{76.72 °C}),$$ whereas carbon tetrafluoride $$\ce{CF4}$$ is in gaseous state at room temperature (b.p. $$\pu{−127.8 °C})?$$

• Compare the bond strength of C-F and C-Cl bond. – Nilay Ghosh Dec 6 '19 at 7:14

The vapour pressure can be estimated by equating the rate of leaving the liquid surface with that returning to it from the vapour. This leads to an expression of the form $$\displaystyle p\sim n_{l}RTe^{-\Delta E/RT}$$ where $$\Delta E$$ is the energy needed to remove a molecule from the liquid to the vapour phase and $$n_{l}$$ is the number density (number / unit volume) of the liquid.
The energy $$\Delta E$$ will clearly depend on the type of molecules involved. CCl$$_4$$ has large easily polarisable chlorine atoms. This means that these electrons are easily influenced by nearby atoms and this leads to an increased intermolecular interaction compared to CF$$_4$$ because the F atoms, having fewer electrons, and ones close to the nucleus, are only slightly polarisable. The increased $$\Delta E$$ (CCl$$_4$$ vs CF$$_4$$) then leads to a lower vapour pressure for CCl$$_4$$ and hence a higher boiling point which is defined at the temperature when the vapour pressure reaches one bar, or one atm. if you prefer.
• It is really just rewriting $\Delta G^o = -RT \ln K$ for the liquid-vapor equilibrium, where K=p. Since the $\Delta S$ term is approx a constant for the two compounds it can be ignored, and then if $\Delta U = \Delta H$ which it approx is for a liquid, you have your result. – Buck Thorn Dec 7 '19 at 11:50