Question 6.19 (Sixth Edition)

[Steve Magana 2I]

Question: Account for the following observations in terms of the type and strength of intermolecular forces. (a) The melting point of solid xenon is -112 Celsius and that of solid argon is -189 Celsius. (b) The vapor pressure of diethyl ether
(C2H5OC2H5) is greater than that of water. (c) The boiling point of pentane, CH3(CH2)3CH3, is 36.1 Celsius, whereas that of 2,2-dimethylpropane (also known as neopentane), C(CH3)4, is 9.5 Celsius.

Can someone help in by explaining this? Thank you!

[Mona Lee 4L]

a. Xenon has more electrons, so it is more polarizable and has stronger LDF than Ar. Because of the stronger LDF, the melting point will be much higher because it requires more energy to break the stronger bonds.
b. H2O has hydrogen bonding while C2H5OC2H5 is nonpolar and only has LDF. Because LDF are weaker than hydrogen bonds, C2H5OC2H5 will be held less tightly together, increasing the vapor pressure.
c. pentane is an unbranched molecule while neopentane is branched. Both are nonpolar so will only experience LDF. Because pentane is unbranched / straight, pentane molecules "stack" better, reducing the space between molecules, increasing the LDF. Because of the increased LDF, it will require more energy to break the bonds from liquid to gas, leading to a higher boiling point.

[Andrea Zheng 1H]

a) Xenon is larger and has more electrons. This means it will have increased attractive forces, making the melting point higher.
b) There will be a higher vapor pressure when the attractive forces are not as strong. Because water has hydrogen bonding, the diethyl ether will have the higher vapor pressure.
c) pentane is a linear structure, unlike 2,2-dimethylpropane, which is spherical. Therefore, the more linear structure (pentane) will have increased attractive forces and thus a higher boiling point.