As far as I know, the more the polar nature of the organic compound, the more will be the energy required to break the intermolecular bonds and thus a higher boiling point. Here in this case, $\ce{R-F}$ is more polar in nature than $\ce{R-I}$. Then, how does $\ce{R-I}$ have a higher boiling point than $\ce{R-F}$?
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3 Answers
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The main factor affecting the boiling point is the molecular weight, as can be pictured from the noble gases boiling point in the periodic table (the noble gases are those which have the weakest interactions due to their complete outer shell). But note that if you want to consider all the effects altogether, you must use Kelvins instead of Celsius. So going from 44°C to 88°C is not twice as much but a mere 14% increase.
The other factor affecting boiling point is the nature and strength of intermolecular interactions, usually in the order ionic > hydrogen bond > dipole-dipole > Van der Waals.
In the comparison of RF vs RI boiling point, molecular weight is definitely in favor of higher boiling points for RI, at least for small molecules. It is true that the C-F bond is much more ionized than C-I so it will create stronger dipole-dipole interactions but not sufficiently to compensate for the effect of the mass increase.
HTH
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Assuming they have the same alkyl groups i.e. methyl fluoride, methyl chloride, methyl bromide, and methyl iodide.
In this case I considered the molecular mass of these compounds. The one with the highest molecular mass, which was methyl iodide was supposed to have the highest boiling point (BP) since BP increases with increase in molecular weight.
The one with the highest molecular weight has more electrons which creates more temporary bonds with other molecules. The more the bonds the more the energy needed to break them, hence higher boiling point.
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Polarity consideration is important when dealing with interactions between molecules. A hydrogen bond (H-bond) is a good example. Water has polar $\ce{H-O}$ bonds, as a result partially positive $\ce{H}$ interacts with partially negative $\ce{O}$ from a dofferent molecule as a result water is liquid at $T_\mathrm r$ and solidifies at $0\ \mathrm{^\circ C}$, while methane of about the same molecular weight is a gas up until $-168\ \mathrm{^\circ C}$.
The reason for $\ce{HF}$ to be liquid and $\ce{HI}$ to be gas is different. $\ce{H}$ and $\ce{F}$ have similar sized orbitals and the molecule is polar. As a result intramolecular H-bonds are strong. In $\ce{HI}$ the overlap is very poor (minute $\ce{H}$ orbitals don't overlap well with huge diffuse orbitals of $\ce{I}$. As a result $\ce{HI}$ doesn't form strong intermolecular $\ce{H}$ bonds and the bonding between $\ce{H}$ and $\ce{I}$ in the molecule is much weaker than in $\ce{HF}$. Thus, $\ce{HI}$ is both gas and a very strong acid (much stronger than $\ce{HF}$).
I will assume that by $\ce{R}$ you mean an alkyl group. In this situation neither $\ce{R-F}$, nor $\ce{R-I}$ can form hydrogen bonds. For example, the polarization of $\ce{$\mathbf{H{-}C}$H2F}$ is much closer to that in methane ($\ce{CH4}$) than in water, not to mention $\ce{HF}$. Some polarization exists but its effects are small. Much greater effect is the molecular weight and density that translate into much stronger London forces in $\ce{R-I}$ compared to $\ce{R-F}$. Both weight and density favor $\ce{R-I}$ to be less volatile than $\ce{R-F}$. The effect of polarization is not large enough to compensate weight difference.
So, how large is this polarization effect in $\ce{R-F}$? Well, compare bp of ethane ($\ce{CH3CH3}$) and fluoromethane ($\ce{CH3F}$) and iodomethane ($\ce{CH3I}$). $T_\mathrm{bp}(\text{ethane})=-88\ \mathrm{^\circ C}$ and $T_\mathrm{bp}(\ce{CH3F})=-77\ \mathrm{^\circ C}$, $T_\mathrm{bp}(\ce{CH3I})=+44\ \mathrm{^\circ C}$. The effect is there but it is very small.
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2$\begingroup$ You should expand the third paragraph into description of dispersion forces and why they are increased with move from fluorine to iodine. As of now, your answer does describe nicely, why polarity does not play a role here, but what does ? $\endgroup$ Commented Jul 21, 2016 at 16:42
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$\begingroup$ I would go with "molecular weight is the main contributor" as the rule of thumb. Sure, there are dispersion fores, but consider this. CH3I weigh 142 g/mol and boils at 44°C. C10H22 weighs 142 g/mol and boils at 174.1°C. If I go into dispersion forces I would have to explain why C10H22 has so much higher boiling point without dispersion forces. I can, but that would be a convoluted off-topic story that TC didn't ask for. $\endgroup$ Commented Jul 21, 2016 at 16:51
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3$\begingroup$ I don't buy "molecular weight" as the main contributor. It absolutely has to be about the forces between the molecules. Weight only matters through gravity, and I assure you it isn't gravity that keeps a CH3I molecule in the condensed phase at the same temperature a CH3F molecule escapes into the gas. $\endgroup$– Jason B.Commented Jul 21, 2016 at 17:00
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$\begingroup$ MoF6 boils at 34 C while undeniably heavier .WF6 boils at 19 C. Consequetly, general reference to 'weight' is not an acceptable answer. $\endgroup$ Commented Jul 21, 2016 at 17:03
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$\begingroup$ I think high boiling point of alkyl iodide in comparison to alkyl fluoride is because of the alkyl iodide's higher molecular mass which in turn results in increase of van der Waals' force. $\endgroup$ Commented Jul 21, 2016 at 17:43
4delectrons, which do contribute to the polarizability. But you make a great point, that isn't made on that page, that it is the size of these orbitals that really impacts the polarizability, and therefore the magnitude of the dispersion forces. $\endgroup$