(a) CH3OH boils at 65 degrees Cel., CH3SH boils at 6 degrees Cel. (b) Xe is liquid at atmospheric pressure and 120 K, whereas Ar is gas. (c) Kr, atomic weight 84, boils at 120.9 K, whereas Cl2, molecular weight about 71, boils at 120.9 K. (d) Acetone boils at 56 degrees Cel., whereas 2-methylpropane boils at -12 degrees Cel.
a) Due to intermolecular hydrogen bonding in CH3OH, it boils at high temperature. No intermolecular hydrogen bonds in CH3SH.
b) The atomic size and weight of Xe is higher than that of Ar. Hence the vander waal’s forces of attractions are stronger in Xe.
c) It is due to almost same size of Kr and Cl2
d)Acetone has polar C=O bond. As a result, there are attractions like dipole dipole attractions and hydrogen bonds especially in the enolic form of acetone
a. CH3OH contains strong hydrogen bonds between its molecules due to the presence of a hydrogen atom attached to an electronegative atom (oxygen) and a lone pair in the oxygen atom. CH3SH contains only weak Van der Waals forces of attraction between its molecules. Since hydrogen bonds are stronger than Van der Waals forces, more energy is needed to break the stronger hydrogen bond than the weaker Van der Waals forces. Consequently, CH3OH has a higher b.p. than CH3SH
b. Xe has a larger molecular mass than Ar and hence Xe atoms are bigger than Ar atoms. The extent of distortion of electron cloud in Xe is greater than in Ar due to its larger atomic size. This resutls in more extensive and stronger Van der Waals intermolecular forces in Xe than in Ar. More energy is needed to break stronger van der Waals intermolecular forces. Hence, Ar exists as a gas while Xe exists as a liquid.
c. There exists induced dipole – induced dipole (or instantaneous dipole) attraction between Cl2 molecules in addition to Van der Waals forces of attraction.
d. same reasoning as in (a)
A) Hydrogen Bonding
B) London dispersion forces
C) London dispersion forces
D) Dipole-dipole bonding
Finally, that’s what I was searching for! Thanks to author of this question.