6.1 and 6.19 question

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Nicole Shak 1L
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6.1 and 6.19 question

Postby Nicole Shak 1L » Wed May 30, 2018 5:22 pm

For question 6.1 in the textbook, it asks to identify what kinds of intermolecular forces arise between the molecules given. For SO2 the solutions say it could be a dipole-dipole interaction, but I don't understand why this could occur.

Also in question 6.19, I'm confused on how London forces/Hydrogen bonding affects melting and boiling points. Why does CH3(CH2)3CH3 have a higher boiling point than C(CH3)4?

Chem_Mod
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Re: 6.1 and 6.19 question

Postby Chem_Mod » Wed May 30, 2018 6:09 pm

Draw the Lewis structure for SO2 and determine the shape. If done correctly, you will see that the two bond dipole moments do not cancel.

Stronger London forces means molecules will stick to each other more. In order to melt or especially to boil, molecules must (partially) detach from each other. Therefore, stronger London forces means the temperature must be raised higher to melt/boil a substance.

CH3(CH2)3CH3 has more surface area than C(CH3)4, which means more dispersion forces. There is also a mechanical aspect to it: it is harder to separate strands of spaghetti from each other compared to pieces of macaroni, even if they have the same amount of pasta per piece.

Maya Khoury
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Re: 6.1 and 6.19 question

Postby Maya Khoury » Sat Jun 02, 2018 2:09 pm

so how can you tell which has a greater london force? Is it solely based on the mass of the molecule? could someone explain the logic behind the answer for 6.19 (b)?

Tarek Abushamma
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Re: 6.1 and 6.19 question

Postby Tarek Abushamma » Sun Jun 03, 2018 2:07 pm

No it has to do with the shape. Since CH3(CH2)3CH3 is linear, it has a greater surface area, and thus more contact points for london forces to occur. The more interactions there are the stronger they are as a unit, which translates to the higher the boiling point.


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