# dipole–dipole interactions: OCR Sample Paper wrong? [closed]

I was doing OCR A Level Chemistry A, H432/01 Periodic table, elements and physical Sample Question Paper.

Question 6 was:

The boiling point of hydrogen bromide is –67 ºC.

The boiling point of hydrogen iodide is –34 ºC.

The different boiling points can be explained in terms of the strength of bonds or interactions.

Which bonds or interactions are responsible for the higher boiling point of hydrogen iodide?

I put answer C: permanent dipole–dipole interactions ? The given answer was D: induced dipole–dipole interactions

[Revision post down-vote]

My learning to date is that:

• dipole-dipole interactions could only be one or the other i.e. permanent or induced.

• iodine is more electronegative than hydrogen, therefore the interactions are permanent.

Please would someone explain where how the induced interactions are caused?

• Your understanding is wrong, the types are not mutually exclusive. Moreover, the induced dipoles are always there, no matter what molecules it is. – Ivan Neretin May 29 '19 at 17:54
• – Mathew Mahindaratne May 29 '19 at 17:58
• Ivan & Mathew - many thanks for the clarification. By way of explanation, I was asking on behalf of my daughter, interpreting what she had told me to the best of my understanding. Funnily enough I had noticed the 'trend in the boiling point of the halides' question and pointed it out to her, thinking it might be germane. She had dismissed that article but, in light of your information, I will redirect her to it - though, in my ignorance, I'm unsure if it's at a suitable level for an English A level student. Thanks both. – EndUzr May 29 '19 at 18:49
• Possible duplicate of Why are the dispersion forces in CS2 stronger than the dipole-dipole forces in COS? – M.A.R. May 29 '19 at 18:56
• I think that questions like this are tricky, and somewhat unfair: both molecules have dipoles and somehow the student is supposed to know that they happen to be similar. The induced dipole-dipole interaction depends on the polarisability which is proportional to an atom's volume and so number of electrons. This is greater for HI than HBr but to know that it counteracts the difference in dipole is tricky as I have said. I cannot see that an A level student could reasonably work this out as the question is written. – porphyrin May 30 '19 at 10:35

Permanent Dipoles occur when two atoms bonded to each other have significantly different electronegativities. When the atomic number of an element increases the number of electrons present in them also increases. These electrons are constantly moving around in the atom. At some point, there are more electrons on one side of the atom than the other. This creates an imbalance of charge, creating a small charge of $$\delta ^+$$ and $$\delta ^-$$. These small charges induce its neighbors to create additional charges of $$\delta ^+$$ and $$\delta ^-$$, thus creating a chain of induced dipoles. These induced dipoles are temporary as the electrons can move, and hence can be reverted. They are, therefore, known as induced dipole forces, or better, known as London Dispersion Forces. In reality, they are not really forces, but rather attraction between molecules.

Naturally, more the number of electrons in the atom, more the amount of London dispersion forces, and hence greater the attraction. The greater the attraction, the more energy you have to put to break the bonds apart. Between the considered hydrogen halides ($$\ce{HBr}$$ and $$\ce{HI}$$), iodine has more electrons than bromine, and hence $$\ce{HI}$$ has a greater attraction force between molecules than $$\ce{HBr}$$. As a consequence, $$\ce{HI}$$ has a greater boiling point between them. Appropriate values are listed in following table:

$$\begin{array}{cccccc} \hline \ce{H-X} & \text{BP}^\text{Ref.1} & \ce{X}\text{'s EN} & \mu ^\text{Ref.2} & \text{Polarizability }(\pu{10^{-24} cm^3}) & (r_\ce{H-X})^\text{Ref.3} \\ \hline \ce{HCl} & \pu{-85.1 ^\circ C} & 3.16 & \pu{1.09 D} & 2.63 & \pu{1.27 Å} \\ \ce{HBr} & \pu{-67.1 ^\circ C} & 2.96 & \pu{0.82 D} & 3.61 & \pu{1.42 Å} \\ \ce{HI} & \pu{-35.1 ^\circ C} & 2.66 & \pu{0.45 D} & 5.44 & \pu{1.60 Å} \\ \hline \end{array}$$

Moreover, a related question you can find here. Orthocresol has given very reliable answer, which included two equations to calculate the strength of permanent dipole interactions ($$U_\mathrm{dipole}$$) and the strength of dispersion interactions ($$U_\mathrm{dispersion}$$). The author has stated that:

If you plug all the numbers in for the hydrogen halides $$\ce{HX}$$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $$\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$$.

For example, for $$\ce{HCl}$$, dispersion forces contribute $$86\%$$ to the intermolecular attractions, and for $$\ce{HI}$$, they contribute $$99\%$$.

References:

1. J. F. Ogilvie, W. R. Rodwell, R. H. Tipping, “Dipole moment functions of the hydrogen halides,” J. Chem. Phys. 1980, 73(10), 5221 – 5229 (https://doi.org/10.1063/1.439950).
2. N. N. Greenwood, A. Earnshaw, “Chapter 17 – The Halogens: Fluorines, Chlorinee, Bromine, Iodine, and Astatine,” In Chemistry of the Elements; 2nd Edn.; Elsevier: Boston, MA, 1997, pp. 789–887.
3. K. Fajans, N. Bauer, “Electronic Structure and Stability of Hydrogen Halides and of Complex Ions $$\ce{XO4}$$,” J. Chem. Phys. 1942, 10(7), 410 – 415 (https://doi.org/10.1063/1.1723742).