2
$\begingroup$

Which of the following compounds exhibits dipole-dipole forces as its strongest attraction between molecules?

a) $\ce{CO2}$
b) $\ce{CH3NH2}$
c) $\ce{Kr}$
d) $\ce{H2}$
e) $\ce{HBr}$

From what I understand, $\ce{CH3NH2}$ has an H-bond and it's the strongest form of dipole-dipole. So I thought that would be the answer.

I also knew that $\ce{HBr}$ shows the greatest electronegativity difference between $\ce{H}$ and $\ce{Br}$ atom, it would be a polar covalent bond (?!).

In that case, $\ce{HBr}$ may have stronger attraction between molecules than $\ce{CH3NH2},$ but it is not dipole-dipole. Isn't that correct? Then shouldn't the answer be $\ce{CH3NH2}$ instead of $\ce{HBr}?$

I must have some concept issues here.

$\endgroup$
  • 1
    $\begingroup$ Mistaking bonds for intermolecular interactions is a serious "concept issue". $\endgroup$ – Mithoron Jul 18 at 15:15
  • $\begingroup$ Wait, I thought there would be H-bonds between CH3NH2? And each HBr molecule is attracted to other HBr molecules by a mixture of permanent dipole-dipole and dispersion forces. This is intermolecular bonding. So Hbond vs dipole-dipole, why is HBr stronger than CH3NH2? I AM trying to compare intermolecular interactions, not the bonds? Please tell me where I am making a mistake. I am going in circles and very confuse. Thank you. $\endgroup$ – Molly_K Jul 18 at 15:29
  • $\begingroup$ Well, that's what this terrible title suggests, so edit it. $\endgroup$ – Mithoron Jul 18 at 15:32
  • $\begingroup$ Thank you for the suggestion, I edited and made it specific. $\endgroup$ – Molly_K Jul 18 at 15:41
1
$\begingroup$

EDIT (after title edited): $\ce{HBr}$ doesn't have stronger interaction than $\ce{CH2NH2}$, but it has dipole-dipole interaction as the strongest forces between it's molecules, which is obviously weaker than H-bonding.


Hydrogen bonding is the strongest intermolecular attraction. It is a type of dipole-dipole interaction1, but it is specific to Hydrogen.

In general, dipole-dipole interactions are considered weaker than H-bonding. The relative strength of forces are as follows,

Hydrogen bonding > Van der Waals dipole-dipole interactions > Van der Waals dispersion forces

As per your question, (a), (c) and (d) have dispersion forces (as they're non-polar), while (b) has H-bonding as it's strongest interaction (not dipole-dipole).

Finally, as only (e) has dipole-dipole as it's strongest attraction between molecules, hence it's the answer.


Notes:

1: Dipoles are created due to difference in electronegativity. Below image shows that there is sufficient electronegativity difference to create a dipole. H-bonding and dipole-dipole interactions have same origins.

enter image description here

![H-BONDING and dipole-dipole interactions](https://cdn.masterorganicchemistry.com/wp-content/uploads/2019/11/3-van-der-waals-dipole-dipole-interactions-acetone-methyl-acetate-propyl-fluoride.gif)

(Source: Master Organic Chemistry)

| improve this answer | |
$\endgroup$
  • $\begingroup$ So H-bond is not a dipole-dipole? But it's said to be the strongest dipole-dipole? $\endgroup$ – Molly_K Jul 18 at 15:22
  • $\begingroup$ @Molly_K: H-bond is the strongest dipole-dipole interaction. But, dipole-dipole forces are said to be other interactions than H-bond, in general $\endgroup$ – Rahul Verma Jul 18 at 16:28
  • 1
    $\begingroup$ @YiKo: 1) have a look at this image (src: masterorganicchemistry) 2) H-bonding is theoretically separated from other dipole-dipole interactions. But, in essence both of them have same origin, as said in image. 3) intermolecular forces aren't significant in determining melting point, packing fraction and crystal structure are used to compare MP. 4) intermolecular forces are significant in comparison of BP, BP(hbr) = -60, BP(ch3nh2) = -6 $\endgroup$ – Rahul Verma Jul 19 at 3:01
  • 1
    $\begingroup$ Thank you, this helps a lot. $\endgroup$ – Molly_K Jul 19 at 3:03

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.