After looking for proper reasons for boiling point orders, nobody could even explain why $\ce{CCl4}$ has a higher boiling point than $\ce{SiCl4}$, but after looking for patterns, in lots and cases of $\ce{CH2X2}$, $\ce{CH3X}$, $\ce{CHX3}$ has more boiling point than similar silicon derivatives.

In some places some explanation related to electronegativity difference is given, but how is that relevant and since Si is less electronegative hence there's more partial charge on Cl which results in more polarisability and stronger London forces; which are quite powerful as evident from various examples like $\ce{HCl < HBr < HI}$.

Order of boiling point (leaving HF due to Hydrogen bonding) London forces even dominate dipole-dipole , which I observed in various cases like the above one and they increase rapidly with polarisability and number of polarisable electrons.

Now, in comparing $\ce{CH2F2,CH3F,CHF3,CHF4}$ we can see that it follows dipole's magnitude order as polarization is negligible.

So my questions are

  1. Is there a good way with no real calculation to compare polarisability/effectiveness of polarisable electrons of molecule to compare magnitude of vanderwall forces or any other way to do it? And does polarising power of central atom play a role?

  2. Why do halomethanes have more boiling point than their respective silicon derivatives?

  3. While comparing the boiling points of $\ce{CH4}$ and $\ce{CF4}$, as fluorine is less polarisable, methane should have more boiling point as both are non-polar (my reasoning works for higher homologous for $n\geq 2$ in $\ce{C_{n}H_{2n +2}}$ and $\ce{C_{n}F_{2n + 2}}$)

If somebody could include nitrogen family too, that would be great.

  • 1
    $\begingroup$ There was already such question: chemistry.stackexchange.com/questions/43334/… but is unanswered $\endgroup$
    – Mithoron
    Jan 11 '16 at 20:55
  • $\begingroup$ could you tell me how would I compare london forces in two molecules in general with an example? like same surrounding atoms then same central atom would be sufficient (exceptions like this excluded) $\endgroup$
    – Mrigank
    Jan 11 '16 at 21:02
  • $\begingroup$ I think shape of molecules can be important here. $\endgroup$
    – Mithoron
    Jan 11 '16 at 21:13

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