Answer:
a. Van der Waals forces, C8H18 has higher boiling point.
b. Van der Waals forces in both compounds
Dipole -Dipole in CH3OCH3
c. Van der Waals and dipole-dipole in both, higher boiling point for HOOH.
Additionally H bonding in HOOH
d. Van der Waals for both compounds, additionally dipole- dipole and hydrogen bonding for NH2NH2.
Higher boling point NH2NH2
Explanation:
To answer this question we have to know the hybridization of the atoms of interest and/or the presence of resultant dipole moments which make the molecule polar, and the van der Waals interactions. Also we have to look for the presence of hydrogen bonding.
Remember van der Waals forces will be present in all of the molecules since they are of the instantaneous type produced by the momentaneous distortions of the electron clouds.
a. These two compounds are hydrocarbons, the electronegativity difference between carbon and hydrogen is very small, so the only factor we need to consider are the van der Waals interactions.
The van der Waals interactions increase with molecular weight since they are interactions which are momentaneous between the atoms in the molecule, and the heavier the molecule the more these distortions are possible.
Since C8H18 has a higher molecular than C6H14, it follows it has a greater boiling point.
b. CH3OCH3 differs only on the added O atom to C3H8. They have similar molecular weights, 46 vs 44 g/mol, but the presence of the tehedral sp³ hybridized oxygen atom with its high electronegativity produces dipole moments and imparts a polarity to the molecule which make us to predict CH3OCH3 will have a higher boiling point.
c. In HOOH we have dipole-dipole moments since the oxygen atom is sp³ hybridized and the O-O-H is tetrahedral, the molecule is polar. Additionally we have the presence of H bond in this compound which is not possible in HSSH.
While the molecule HSSH is also polar since it has two dipole moments in the sp³ hybridized ( tetrahedral ) S, it is not as big as the dipole moment in O-H.
Therefore Hydrogen peroxide, HOOH, has a higher boiling point than HSSH
d. The N-H bond has a dipole moment, and since the N is tetrahedral there is a resultant dipole moment. As in case d., we have hydrogen bonding between H and N.
NH2NH2 then will have a higher boiling point than CH3CH3 where only van der Waals interactions are possible.
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