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Which force acts between the molecules

Dispersion, dipole-dipole and hydrogen bonding as consequences of the bonds a molecule has and the shape it is in.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to decide, for a small molecule, whether its bonds are polar, whether its shape leaves it with a dipole, and which of dispersion, dipole-dipole and hydrogen bonding therefore act between one molecule and the next. You will be able to say why a molecule can have strongly polar bonds and no dipole at all, and why a hydrogen bond needs a hydrogen attached to nitrogen, oxygen or fluorine rather than just any hydrogen.

2. What you already have

You can read a formula as a count of atoms, you know that a covalent bond is a shared pair of electrons, and from the last unit you can work out the shape of a small molecule from the number of bonding pairs and lone pairs around its central atom. You also know that electronegativity rises across a period and falls down a group. All four of those are used in this lesson, and the shape is used hardest.

3. Words for this lesson

4. Two questions, and the list falls out of them

Do not start from three forces to be remembered. Start from two questions about the molecule in front of you.

Question one: are the bonds polar? Two atoms of different elements pull a shared pair by different amounts, and the difference in electronegativity is how much. Two atoms of the same element pull equally, so $\mathrm{Cl_2}$, $\mathrm{O_2}$ and $\mathrm{N_2}$ have no polar bond however electronegative the element is. Carbon and hydrogen differ by only $0.4$, and this course treats a carbon-to-hydrogen bond as non-polar.

Question two: does the shape cancel them? Polar bonds are little arrows. Arrange them symmetrically and they add to nothing. $\mathrm{CO_2}$ has two of the most polar bonds in ordinary chemistry, pointing in exactly opposite directions, and the molecule has no dipole at all. Bend it, as water is bent, and they no longer cancel.

Now the three forces are just the answers:

• No dipole — dispersion forces, and nothing else.
• A dipole — dipole-dipole forces, as well as dispersion.
• A dipole, with hydrogen on nitrogen, oxygen or fluorine — hydrogen bonding, as well as dispersion.

The phrase as well as is doing real work. Dispersion forces act between every pair of molecules in the universe; they are never switched off. What the other two do is add to them.

5. The method, step by step, and how to check it

Check each bond's polarity. Different elements with a real electronegativity difference make a polar bond; carbon to hydrogen is treated as non-polar.

Work out the shape. From the bonding pairs and lone pairs on the central atom.

Ask whether the dipoles cancel. A symmetrical shape with identical outer atoms cancels them.

Name the forces. Dispersion always; dipole-dipole if the molecule is polar; hydrogen bonding if a hydrogen sits on nitrogen, oxygen or fluorine.

Compare substances. Where the same force acts in each, compare electron counts; where different forces act, the stronger force usually wins.

Check the work. Have you named dispersion as present in every substance? Did a molecule with polar bonds and a symmetrical shape come out non-polar? Is every claimed hydrogen bond a hydrogen on nitrogen, oxygen or fluorine, not on carbon? And does any boiling point you know agree with the ranking you reached?

6. Why each step is allowed

Starting from bond polarity is allowed because a dipole can only come from charge that is unevenly shared. With no polar bonds, there is nothing for the shape to add up.

Checking the shape next is allowed because dipoles are directional. Two equal pulls in opposite directions cancel exactly, the same way two equal ropes pulled from opposite sides leave the knot where it was.

Treating dispersion as always present is allowed because every molecule has electrons, and every electron cloud flickers. The flicker in one molecule induces a matching flicker in the next, and the two attract.

Singling out hydrogen on nitrogen, oxygen or fluorine is allowed because those three are small and highly electronegative. They strip the hydrogen's electron so far that a nearly bare proton is left, and a lone pair on a neighbor can approach it unusually closely. Hydrogen on carbon is not stripped enough, and hydrogen on sulfur sits beside an atom too large and too weakly pulling.

7. Dispersion forces are not the weak case

It is easy to read the list as a ladder with dispersion at the bottom, and easy to conclude that dispersion forces are the ones that hardly matter. They are the ones that always matter.

A dispersion force grows with the number of electrons in the molecule, because a larger, looser electron cloud is easier to distort. Compare four substances with no dipole at all:

Iodine boils eighty-four degrees above water, on dispersion forces and nothing else. Any sentence of the form dispersion forces are weak, so it must boil low is wrong as often as it is right.

Molar mass is the number most people reach for, and it works most of the time because heavier molecules usually have more electrons. Where the two disagree, the electron count is the one that is right.

8. The pair the whole unit turns on

$\mathrm{C_2H_6O}$ is the formula of two different substances.

Ethanol, $\mathrm{CH_3CH_2OH}$: the oxygen is on the end of the chain and carries a hydrogen. Boils at $78$ °C.

Methoxymethane, $\mathrm{CH_3OCH_3}$: the oxygen is in the middle, between the two carbons. Boils at $-25$ °C.

Same atoms. Same molar mass, $46$ g/mol. Same twenty-six electrons, so the same dispersion forces. Both polar, because in both the oxygen pulls harder than what it is attached to and the shape does not cancel it. A hundred and three degrees apart.

The only difference is that ethanol has a hydrogen attached directly to oxygen and methoxymethane does not. An oxygen-to-hydrogen bond leaves that hydrogen nearly stripped of its electron — a small, bare, positively charged nucleus — and a lone pair on a neighboring oxygen can get very close to it. That closeness is what makes a hydrogen bond so much stronger than an ordinary dipole-dipole attraction, and it is available to ethanol and not to its isomer.

If you can explain this pair, you can explain almost every surprising boiling point in chemistry.

9. In the world: why the propane tank and the lighter differ

An American backyard grill runs on propane and a pocket lighter on butane, and dispersion forces explain the choice. Both are made of carbon and hydrogen only, so neither has a dipole and neither can hydrogen bond; dispersion forces are all that hold their molecules together. Propane, with 26 electrons, boils at $-42$ °C. Butane, with 34, boils at $-1$ °C.

That 41-degree difference decides where each one works. A propane tank stays useful through a Minnesota winter, because even at $-20$ °F (about $-29$ °C) propane is still above its boiling point and keeps turning to vapor to feed the grill. A butane lighter left in a car on a January morning may barely light, because at those temperatures butane is close to or below its boiling point and gives off little vapor.

Gas plants near Houston separate these fuels by exactly this property. Natural gas liquids are cooled and pressurized, and the components are drawn off at different temperatures in a distillation column: methane at $-162$ °C, ethane at $-89$, propane at $-42$, butane at $-1$. Each extra carbon adds electrons, strengthens the dispersion forces, and raises the boiling point, so the column sorts the fuels by the size of their electron clouds.

10. In the world: why water beads on a waxed car

Car wax is a long-chain hydrocarbon with dispersion forces only. Water molecules hydrogen bond to each other far more strongly than they are attracted to the wax, so they pull together into beads rather than spreading out.

11. Where this goes wrong

Polar bonds read as a polar molecule. They are different questions and the shape is what connects them. $\mathrm{CO_2}$, $\mathrm{BF_3}$ and $\mathrm{CCl_4}$ all have strongly polar bonds and no dipole whatever.

Bonds confused with forces. A bond and an intermolecular force are not the same size of thing. Boiling water pulls whole molecules apart from each other and leaves every $\mathrm{O-H}$ bond exactly where it was, which is why water boils at a hundred degrees and does not decompose there. A substance with strong bonds inside its molecules can still boil low, and the question which forces act between the molecules is a different question from what holds each molecule together.

Dispersion treated as negligible. It is the only force iodine has, and iodine boils above water. Dispersion is the floor, not the exception.

Hydrogen bonding claimed wherever there is a hydrogen. Methane is full of hydrogens and has no hydrogen bonding at all. The hydrogen has to be attached to nitrogen, oxygen or fluorine, and there has to be a lone pair on another molecule for it to reach.

Ranking by molar mass without checking. It works whenever the same force acts in every substance being compared, and it fails badly the moment it does not — hydrogen sulfide is twice the mass of water and boils a hundred and sixty degrees lower.

12. Carbon dioxide against water

1. Check carbon dioxide's bonds.

   $3.5 - 2.5 = 1.0$

   Oxygen against carbon: strongly polar.

2. Work out its shape.

   $\text{linear}$

   Two bonding regions, no lone pairs.

3. Ask whether the dipoles cancel.

   $\text{yes: opposite directions}$

   So no dipole.

4. Name carbon dioxide's forces.

   $\text{dispersion only: a gas}$

   Only 22 electrons.

5. Do the same for water.

   $\text{bent; hydrogen on oxygen: hydrogen bonding}$

   Same kind of polar bonds, opposite answer: a liquid.

13. Why hydrogen sulfide boils below water

1. Compare the masses.

   $\mathrm{H_2S}: 34; \ \mathrm{H_2O}: 18 \text{ g/mol}$

   Mass says hydrogen sulfide should boil higher.

2. Compare the shapes.

   $\text{both bent}$

   Both are polar molecules.

3. Check for hydrogen on N, O or F.

   $\text{water yes; hydrogen sulfide no}$

   Sulfur is not on the list.

4. Name each one's strongest force.

   $\text{hydrogen bonding; dipole-dipole}$

   Water's is far stronger.

5. Compare the boiling points.

   $100 - (-60) = 160 \ ^\circ\mathrm{C}$

   Water boils 160 degrees higher.

6. Draw the conclusion.

   $\text{the force beat the mass}$

   Which force acts beats how heavy it is.

14. Ranking three halogens by boiling point

1. Check the bonds.

   $\mathrm{Cl_2}, \mathrm{Br_2}, \mathrm{I_2}: \text{same element}$

   No polar bonds in any of them.

2. Name the force in each.

   $\text{dispersion only}$

   With no dipole, that is all there is.

3. Count the electrons.

   $34, \ 70, \ 106$

   Dispersion grows with the cloud.

4. Rank by electron count.

   $\mathrm{Cl_2} < \mathrm{Br_2} < \mathrm{I_2}$

   More electrons, stronger forces.

5. Check with boiling points.

   $-34, \ 59, \ 184 \ ^\circ\mathrm{C}$

   The order agrees.

6. Find the biggest step.

   $184 - 59 = 125$

   Bromine to iodine.

7. Compare iodine with water.

   $184 - 100 = 84$

   Dispersion alone beats water's hydrogen bonds here.

15. Your turn: propanone, $\mathrm{C_3H_6O}$, is $58$ g/mol and boils at $56$ °C. Butane, $\mathrm{C_4H_{10}}$, is also $58$ g/mol and boils at $-1$ °C. What acts in each, and why the difference?

1. Name butane's forces.

   $\text{dispersion only}$

   Carbon and hydrogen bonds are treated as non-polar.

2. Your turn: work this step out. Its working is at the end of the packet.

   Name propanone's forces.

3. Your turn: work this step out. Its working is at the end of the packet.

   Check for hydrogen bonding.

16. Guided practice

Match each substance to the strongest force acting between its molecules. More than one substance may take the same answer.

17. Guided practice

Complete the worked solution: four substances with dispersion forces only boil at methane $-162$ °C ($10$ electrons), chlorine $-34$ °C ($34$), bromine $59$ °C ($70$) and iodine $184$ °C ($106$). Find the rise in boiling point at each step.

1. Compare chlorine with methane.

   $(\text{chlorine}) - (\text{methane}) =$ _a

   More electrons, stronger dispersion forces.

2. Compare bromine with chlorine.

   $(\text{bromine}) - (\text{chlorine}) =$ _b

   The electron cloud keeps growing.

3. Compare iodine with bromine.

   $(\text{iodine}) - (\text{bromine}) =$ _c

   Iodine boils far above water on dispersion alone.

18. Guided practice

Ethanol and methoxymethane both have the molecular formula $\mathrm{C_2H_6O}$, a molar mass of $46$ g/mol and $26$ electrons in a molecule. Ethanol boils at $78$ °C and methoxymethane at $-25$ °C. Which statement explains the difference?

• In ethanol a hydrogen is attached directly to the oxygen, so ethanol molecules can hydrogen bond to each other; in methoxymethane every hydrogen is attached to carbon, so they cannot.
• Ethanol's oxygen-to-hydrogen bond is stronger than methoxymethane's carbon-to-oxygen bond, so more energy is needed to boil it.
• Ethanol molecules are heavier, so the dispersion forces between them are stronger.
• Methoxymethane is non-polar, so there are no forces between its molecules at all.

19. Practice

For each molecule, say whether its bonds are polar, whether the molecule as a whole is polar, and what the strongest force between its molecules is. The shape of each is given with its name.

20. Practice

A gas plant near Houston separates fuel gases by boiling point. pentane, $\mathrm{C_5H_{12}}$, boils at $36$ °C and propane, $\mathrm{C_3H_8}$, at $-42$ °C. Both have dispersion forces only. How many degrees higher does pentane boil?

The answer: _a.

21. Somewhere new

A paint works keeps three solvents in a store. A safety officer wants them ranked by the strength of the forces acting between their molecules, weakest first, because that is what decides which drum vents first on a hot day. Put them in order. Their molar masses are close enough that mass will not settle it.

Number the steps in order (write the number in the box):

• water
• hydrogen sulfide
• ammonia

22. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

23. Test question

For each of these three substances, say whether the molecule as a whole is polar, and name the strongest force acting between one molecule and the next. The number of electrons in each molecule is given, because that is what dispersion forces scale with.

24. What you can do now

You can get from a structure to the forces acting between its molecules in two questions: are the bonds polar, and does the shape cancel them. Say out loud why carbon dioxide is not a polar molecule although its bonds are, and why ethanol boils a hundred degrees above methoxymethane although the two are made of exactly the same atoms. Next: what those forces do to boiling point, viscosity and how fast a puddle dries.

Working for the steps left to you

15. Your turn: propanone, $\mathrm{C_3H_6O}$, is $58$ g/mol and boils at $56$ °C. Butane, $\mathrm{C_4H_{10}}$, is also $58$ g/mol and boils at $-1$ °C. What acts in each, and why the difference?, step 2

15. Your turn: propanone, $\mathrm{C_3H_6O}$, is $58$ g/mol and boils at $56$ °C. Butane, $\mathrm{C_4H_{10}}$, is also $58$ g/mol and boils at $-1$ °C. What acts in each, and why the difference?, step 3