Place the following compounds in order of increasing strength of intermolecular forces.

a. ch3ch3 ch3(ch2)8ch3 ch3ch2ch2ch3
b. ch3ch3 < ch3(ch2)8ch3 < ch3ch2ch2ch3
c. ch3ch2ch2ch3 < ch3ch3 < ch3(ch2)8ch3
d. ch3(ch2)8ch3 < ch3ch3 < ch3ch2ch2ch3
e. ch3ch3 < ch3ch2ch 2ch3 < ch3(ch2)8ch3
f. ch3(ch2)8ch3 < ch3ch2ch2ch3 < ch3ch3

CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)

Explanation:

Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are, on average, farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles.

CH3CH3 has a molar mass of 30.07 g/mol

CH3(CH2)8CH3 has a molar mass of 142.28 g/mol

CH3CH2CH2CH3 has a molar mass of 58.12 g/mol

CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.