Homework 3F.5 Part C

Moderators: Chem_Mod, Chem_Admin

Ellis Song 4I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

Homework 3F.5 Part C

Postby Ellis Song 4I » Mon Nov 11, 2019 6:56 pm

Suggest, giving reasons, which substance in each of the following pairs is likely to have the higher normal melting point: c) CHI3 or CHF3

The solutions manual says CHI3 would have a higher melting point because it has stronger London Dispersion forces but how is the strength of London Dispersion determined?

Andrew Pfeiffer 2E
Posts: 101
Joined: Sat Sep 28, 2019 12:16 am

Re: Homework 3F.5 Part C

Postby Andrew Pfeiffer 2E » Mon Nov 11, 2019 7:49 pm

It is determined that triiodomethane will have stronger London dispersion forces on the basis of polarizability. Triiodomethane is bigger than trifluoromethane, and its electrons are held more loosely (and can be distorted more easily). Thus, triiodomethane undergoes stronger London dispersion forces. Hope this helped!

Shail Avasthi 2C
Posts: 101
Joined: Fri Aug 30, 2019 12:17 am

Re: Homework 3F.5 Part C

Postby Shail Avasthi 2C » Mon Nov 11, 2019 10:53 pm

CHI3 is a much larger molecule than CHF3, because the size of Iodine's electron cloud is much larger than the size of Fluorine's. The larger the molecule (by molar mass and by size), the greater its polarizability and therefore the greater its LDFs (induced dipole/induced dipole)

Return to “Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)”

Who is online

Users browsing this forum: No registered users and 1 guest