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boiling point

Posted: Tue Nov 19, 2019 7:18 pm
by Christineg1G
Why would HI have a higher boiling point than HBr if HBr has a stronger dipole-dipole force? Would we go based off of the fact that HI is larger than HBr?

Re: boiling point

Posted: Tue Nov 19, 2019 7:38 pm
by AVerma_F19
I believe that since Iodine is larger than the Bromine, the induced-dipole/induced-dipole forces would cause the IMFs in the HI to be stronger than those in the HBr molecules.

Re: boiling point

Posted: Tue Nov 19, 2019 7:38 pm
by Daniel Honeychurch1C
HI has a higher boiling point than HBr because HI has more electrons and thus stronger induced dipole-induced dipole forces.

Re: boiling point

Posted: Wed Nov 20, 2019 2:51 am
by Sion Hwang 4D
As you go down a group, the molar mass of each element, as well as the number of electrons within in each atom, increases. Hence, stronger dispersion forces can be created.

Although HBr may have a greater difference in electronegativity than HI, iodine has a greater polarizability. This means that HI's electron cloud will be diffused more effectively, creating stronger intermolecular forces.

Re: boiling point

Posted: Fri Nov 22, 2019 7:29 pm
by 005321227
The number of lone pairs (more electrons) increases polarizability thus a higher melting point.

Re: boiling point

Posted: Fri Nov 29, 2019 4:17 pm
by Chetas Holagunda 3H
HI would have a higher boiling point because it has a greater molar mass and more electrons, causing it to have more London dispersion forces. This means it requires more heat to separate the atoms.

Re: boiling point

Posted: Sat Nov 30, 2019 6:22 pm
by CMaduno_1L
To add to this, among molecules that contain only London dispersion forces, the way to tell whether one is stronger than another is by the size of the entire molecule. The larger the molecule is, the greater the strength of LDFs.