[big]

If the CO₂ molecule were bent rather than linear, you would expect:
A) Both the boiling and freezing points would be higher
B) Both the boiling and freezing points would be lower.
C) The boiling point would be higher but the freezing point would be lower.
D) The freezing point would be higher but the boiling point would be lower.

I feel like the answer should be B since if it were bent, the London dispersion forces would not be as strong, and so both boiling and freezing points should be lower, but is this correct? Is there anything else that I am overlooking?

[UG]

I feel like the answer should be B since if it were bent, the London dispersion forces would not be as strong, and so both boiling and freezing points should be lower

I agree with this answer

[fledarmus]

1) What is the difference in electronegativity between carbon and oxygen?

2) Which direction are the bonds polarized?

3) What is the vector sum of the bond polarization?

4) If the molecule were bent, what would be the vector sum of the bond polarization?

[big]

1) What is the difference in electronegativity between carbon and oxygen?

2) Which direction are the bonds polarized?

3) What is the vector sum of the bond polarization?

4) If the molecule were bent, what would be the vector sum of the bond polarization?

1. The difference in electronegativity is 0.89, slightly less than that of H and O.

2. The bonds are polarized towards the O atoms.

3. The vector sum, then, would be 0, right? Since they cancel each other out?

4. If the molecule were bent, the vector sum would be something other than 0 because the molecule would become polar.

Ohh. Does that mean then that if it's bent, there'll be dipole-dipole forces, which would be why both the boiling and freezing point would be higher?

[UG]

The stronger dipole-dipole attractions may well outweigh the decrease in dispersion forces, which I had not considered

[fledarmus]

Bingo!