Textbook 3f 5

[Sameer Chowdhury 3C]

Part c of this problem asked between CHF3 and CHI3 which is likely to have the higher normal melting point. I chose CHF3 because F has the highest electronegativity and as result it would create a stronger dipole moment than I. According to the textbook the answer is CHI3. Could someone explain why?

[Ethan Laureano 3H]

You are right that CH₃F has a higher dipole moment. However, size and weight trump this. Since iodine is larger and heavier than fluorine (not to mention that there are three of them), the molecule will have stronger LDF and thus a higher melting point.

[Silvi_Lybbert_3A]

Hi! I was wondering why size trumps difference in electronegativity. Is a difference in electronegativity less impactful than a difference in size, even if dipole-dipole interactions are stronger than LDF interactions? I don't know if that makes sense; I am a little confused. Thanks!

[Edward Tang 1k]

Hi! I was wondering why size trumps difference in electronegativity. Is a difference in electronegativity less impactful than a difference in size, even if dipole-dipole interactions are stronger than LDF interactions? I don't know if that makes sense; I am a little confused. Thanks!

This is because the equation for intermolecular energy is proportional to -(a1*a2)/(r^6), meaning that each increment in distance is raised to the power of 6. So even if the charge in the numerator, a1 and a2, increases by a bit, the increase in the distance in the denominator would far outweigh that and decrease overall energy.

[Silvi_Lybbert_3A]

Hi! I was wondering why size trumps difference in electronegativity. Is a difference in electronegativity less impactful than a difference in size, even if dipole-dipole interactions are stronger than LDF interactions? I don't know if that makes sense; I am a little confused. Thanks!

This is because the equation for intermolecular energy is proportional to -(a1*a2)/(r^6), meaning that each increment in distance is raised to the power of 6. So even if the charge in the numerator, a1 and a2, increases by a bit, the increase in the distance in the denominator would far outweigh that and decrease overall energy.

Thank you! As a follow up, one thing I was confused about with this equation is what exactly the 'r' means in relation to a bigger molecule. Because if 'r' is the distance between molecules, how does that relate to the size of a molecule itself. And if 'r' is the radius of an atom, wouldn't an increase in an atoms size lead to a smaller negative number (a potential energy closer to zero) and doesn't this mean a less attractive (more unfavorable) force rather than a more attractive one?

[Edward Tang 1k]

Hi! I was wondering why size trumps difference in electronegativity. Is a difference in electronegativity less impactful than a difference in size, even if dipole-dipole interactions are stronger than LDF interactions? I don't know if that makes sense; I am a little confused. Thanks!

This is because the equation for intermolecular energy is proportional to -(a1*a2)/(r^6), meaning that each increment in distance is raised to the power of 6. So even if the charge in the numerator, a1 and a2, increases by a bit, the increase in the distance in the denominator would far outweigh that and decrease overall energy.

Thank you! As a follow up, one thing I was confused about with this equation is what exactly the 'r' means in relation to a bigger molecule. Because if 'r' is the distance between molecules, how does that relate to the size of a molecule itself. And if 'r' is the radius of an atom, wouldn't an increase in an atoms size lead to a smaller negative number (a potential energy closer to zero) and doesn't this mean a less attractive (more unfavorable) force rather than a more attractive one?

r refers to the internuclear distance, or the distance between the nuclei in a bond, which in a single bond is basically the sum of the radii of the two atoms in that bond. If the size of the molecule increases, r would increase, which, as you said, leads to a smaller negative value in the energy and makes the bond more stable. In here, however, melting point has nothing to do with intramolecular forces(bonds), it's only the intermolecular forces that are overcome. CHI3 is stronger than CHF3 because it's large THEREFORE would have trouble holding on to it's electrons, making it more polarizable, which then means stronger LDF and higher melting point. Sorry for bringing bond into this and making the question more confusing.

[Silvi_Lybbert_3A]

This is because the equation for intermolecular energy is proportional to -(a1*a2)/(r^6), meaning that each increment in distance is raised to the power of 6. So even if the charge in the numerator, a1 and a2, increases by a bit, the increase in the distance in the denominator would far outweigh that and decrease overall energy.

Thank you! As a follow up, one thing I was confused about with this equation is what exactly the 'r' means in relation to a bigger molecule. Because if 'r' is the distance between molecules, how does that relate to the size of a molecule itself. And if 'r' is the radius of an atom, wouldn't an increase in an atoms size lead to a smaller negative number (a potential energy closer to zero) and doesn't this mean a less attractive (more unfavorable) force rather than a more attractive one?

r refers to the internuclear distance, or the distance between the nuclei in a bond, which in a single bond is basically the sum of the radii of the two atoms in that bond. If the size of the molecule increases, r would increase, which, as you said, leads to a smaller negative value in the energy and makes the bond more stable. In here, however, melting point has nothing to do with intramolecular forces(bonds), it's only the intermolecular forces that are overcome. CHI3 is stronger than CHF3 because it's large THEREFORE would have trouble holding on to it's electrons, making it more polarizable, which then means stronger LDF and higher melting point. Sorry for bringing bond into this and making the question more confusing.

No you didn't make it more confusing at all! Thank you so much; that cleared everything up; I really appreciate it!!