Question

1. rationalize the difference in boiling points for each of the following pairs of substances:

a. n - pentane ch3ch2ch2ch2ch3 36.2°c
neopentane h3c - c(ch3)3 9.5°c
b. hf 20°c
hcl - 85°c
c. hcl - 85°c
licl 1360°c
d. n - pentane ch3ch2ch2ch2ch3 36.2°c
n - hexane ch3ch2ch2ch2ch2ch3 69°c

Explanation:

Step1: Analyze intermolecular forces in a

n - pentane has a linear structure and neopentane has a more branched structure. Linear molecules have greater surface - area contact, leading to stronger London dispersion forces. So n - pentane has a higher boiling point due to stronger London dispersion forces.

Step2: Analyze intermolecular forces in b

HF has hydrogen bonding due to the highly electronegative F atom bonded to H. HCl has only dipole - dipole forces and London dispersion forces. Hydrogen bonding is stronger than dipole - dipole and London dispersion forces, so HF has a higher boiling point.

Step3: Analyze intermolecular forces in c

HCl is a covalent compound with relatively weak intermolecular forces (dipole - dipole and London dispersion forces). LiCl is an ionic compound. Ionic compounds have strong electrostatic forces between ions. So LiCl has a much higher boiling point.

Step4: Analyze intermolecular forces in d

Both n - pentane and n - hexane are non - polar hydrocarbons with only London dispersion forces. n - hexane has more electrons and a larger molar mass, so it has stronger London dispersion forces and a higher boiling point.

Answer:

a. n - pentane has stronger London dispersion forces due to its linear shape.
b. HF has hydrogen bonding while HCl has weaker dipole - dipole and London dispersion forces.
c. LiCl is ionic with strong electrostatic forces, HCl is covalent with weak intermolecular forces.
d. n - hexane has stronger London dispersion forces due to more electrons and larger molar mass.