7
$\begingroup$

The boiling points of the chlorinated silanes and methanes are given below:

$$\begin{array}{ccc} \hline \text{Species} & \text{Boiling point (X = Si) / }\mathrm{^\circ C} & \text{Boiling point (X = C) / }\mathrm{^\circ C} \\ \hline \ce{XH3Cl} & -30.4 & -24.2 \\ \ce{XH2Cl2} & 8 & 39.6 \\ \ce{XHCl3} & 31.8 & 69.2 \\ \ce{XCl4} & 57.6 & 76.7 \\ \hline \end{array}$$

The chlorosilanes are expected to have larger dipole moments and stronger dispersion forces. Why do the silanes instead have lower boiling points than the methanes?

|improve this question
$\endgroup$
  • $\begingroup$ Out of curiosity, I checked the bromides (most data from this source plus Wikipedia), and the same pattern holds. I also added silane and methane for comparison, which challenges the trend: $$\begin{array}{ccc} \hline \text{Species} & \text{Boiling point (X = Si) / }\mathrm{^\circ C} & \text{Boiling point (X = C) / }\mathrm{^\circ C} \\ \hline \ce{XH4} & -112 & -161.5 \\ \hline \ce{XH3Br} & 1.9 & 4 \\ \ce{XH2Br2} & 66 & 96-98 \\ \ce{XHBr3} & 109 & 147-151 \\ \ce{XBr4} & 153 & 189.7 \\ \hline \end{array}$$ $\endgroup$ – Nicolau Saker Neto Jan 6 '16 at 4:38
  • $\begingroup$ And here are the fluorides, which seem to be more complicated (data less certain): $$\begin{array}{ccc} \hline \text{Species} & \text{Boiling point (X = Si) / }\mathrm{^\circ C} & \text{Boiling point (X = C) / }\mathrm{^\circ C} \\ \hline \ce{XH3F} & ? & -78.4 \\ \ce{XH2F2} & -77.8 & -52 \\ \ce{XHF3} & -97.5? & -82.2 \\ \ce{XF4} & -86 & -127.8 \\ \hline \end{array}$$ $\endgroup$ – Nicolau Saker Neto Jan 6 '16 at 4:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.