Question

question 1 (1 point)
why does the boiling point of an alkane increase as its chain length increases?
a) there is more hydrogen bonding.
b) there are hydrogen atoms available for hydrogen bonds to form.
c) the molecules are highly branched.
d) the strength of the dispersion forces increases with increased molecular size.
e) none of the above

Explanation:

Alkanes are non - polar molecules. They only have London dispersion forces (a type of van der Waals force). London dispersion forces are weak intermolecular forces that arise due to temporary dipoles in molecules. As the chain length (molecular size) of alkanes increases:

• Option a and b: Alkanes do not have hydrogen bonding. Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom (such as N, O, or F). In alkanes (general formula \(C_nH_{2n + 2}\)), the C - H bonds are not polar enough to form hydrogen bonds.
• Option c: The question is about the effect of chain length (not branching). Branching (for isomeric alkanes) actually decreases the boiling point (by reducing the surface area of contact between molecules, thus reducing the strength of London dispersion forces).
• Option d: The strength of London dispersion forces is related to the number of electrons in the molecule (which is related to molecular size). Larger molecules (longer - chain alkanes) have more electrons. More electrons mean more temporary dipoles can be formed, and the strength of the London dispersion forces increases. As the strength of intermolecular forces (London dispersion forces in the case of alkanes) increases, more energy is required to separate the molecules (to change the state from liquid to gas), so the boiling point increases.

Answer:

d) The strength of the dispersion forces increases with increased molecular size.