3F.3

[Annie Ye]

For which of the following molecules will dipole–dipole inter- actions be important: (a) CH4; (b) CH3Cl; (c) CH2Cl2; (d) CHCl3; (e) CCl4?

[christabellej 1F]

In these types of questions, it could be helpful to first draw the Lewis structures. CCl4 and CH4 are symmetric and are non polar structures overall since the net moments cancel out; they will have London dispersion forces. CH3Cl, CH2Cl2, and CHCl3 will also have dipole dipole interactions, so those will be important for b, c, and d.

[Joanne Lee 1J]

Dipole-dipole interactions are only important for CH3Cl, CH2Cl2, and CHCl3 because these molecules are polar whereas CH4 and CCl4 are nonpolar. Only polar molecules have dipole-dipole moments so it is only significant for those three molecules.

[A Raab 1K]

I looked at the fact that b, c, and d were all uneven whereas a and e were even. This unevenness is due to the differing number of electrons on the hydrogen versus the chlorine.

[Jared Khoo 1G]

As CH₄ and CCl₄ are symmetrical (and tetrahedral) they are nonpolar and dipole dipole interactions will not be that important. However, they still have London dispersion forces despite being nonpolar.

[Abigail Sanders 1E]

I agree with the previous responses, it is very helpful to draw ou thte structures for these molecules before determining their interactions. Overall, structures that are nonpolar will only have very weak london forces. Structures that are polar will have dipole-dipole and structures that are polar and have a Hydrogen attatched to a Nitrogen, Oxygen, or Fluorine will also have hydrogen bonds. So, A and E have only London Forces and thus dipole will not be important. For B, C, and D, the molecules are polar and dipole-dipole will be important.