To melt diamond, we have to break the covalent bonds, which we can consider 'intermolecular' because it is one giant molecule.

To melt Methane, we have to break the van der Waals (intermolecular) forces.

For $\ce{NaCl}$, ionic bonds, which are intermolecular as well in a sense.

But I've read that melting graphite also requires breaking the covalent $\ce{C-C}$ bonds rather than just breaking the van der Waals forces between the layers.

And for cross-linked polymers, I've read that melting involves breaking only the bonds linking the two polymer chains.

So I guess what I want is a precise definition of melting. Does it involve breaking the weakest bonds? The intermolecular forces? The intramolecular bonds?

When we melt a molecule of, say, $\ce{H2O}$, we imagine molecules of $\ce{H2O}$ roaming around with little intermolecular forces, as most hydrogen bonds have been broken.

What do we 'imagine' when we melt diamond, or a cross-linked polymer, or graphite, or ionic compounds like $\ce{NaCl}$?

  • $\begingroup$ Have a look first for Lindemann theory (or Lindemann Criterion) for melting. To find out if diamond /graphite melts before decomposing you will have to find a phase diagram. Some types of polymers melt e.g. polythene, many types do not, e.g. bakelite It all depends on their molecular structure. $\endgroup$
    – porphyrin
    Apr 26, 2017 at 17:05
  • $\begingroup$ I'm only a highschool student. I looked at it, but all I can make out is that melting occurs when the average thermal vibrations of atoms is greater than the inter-atomic distances. I cannot relate this to the question. $\endgroup$
    – John
    Apr 26, 2017 at 17:28
  • 2
    $\begingroup$ Ignoring your (not necessarily helpful) distinction between bond types, the main point is that a liquid is a distinct thermodynamic phase. Above the melt temperature, it is preferred, having a lower Gibbs free energy. There is still 'bonding', with inter-atomic forces holding the liquid together. $\endgroup$
    – Jon Custer
    Apr 26, 2017 at 17:47
  • $\begingroup$ So melting involves breaking the intermolecular forces? What if we have a polymer with partial cross-linking? $\endgroup$
    – John
    Apr 26, 2017 at 18:13
  • $\begingroup$ Liquids and solids both have intermolecular forces, a liquid must as it has a meniscus, i.e it is not a gas. The difference liquid to solid is one of the amount of energy available. Liquids have enough energy that molecules can become mobile and rapidly move past one another whereas in a solid this does not normally happen. Chemical bonds do not need to break to cause melting or to vaporise a substance. $\endgroup$
    – porphyrin
    Apr 27, 2017 at 7:57

1 Answer 1


This is a good question. Unfortunately there are several criteria by which chemists indentify whether a process is melting or not. One of them is called the Lindemann Criteria which says:

"Crystals are considered to melt, when the vibrational amplitude becomes half of the interatomic spacing in the crystal lattice."

What does it mean? Normally, at temperatures greater than 0K, atoms possess kinetic energy. Infact, temperature itself is a measure of the kinetic energy of the constituent atoms. The kinetic energy of an atom is related to the temperature as:

$$\text{Kinetic Energy}=\frac{3}{2}k\text{T}$$

where $k$ is a constant called Boltzmann Constant, with a value of $1.38×10^{-23}\text{J/mol}$. Atoms in a solid are characterized by their fixed mean positions. They only vibrate within a certain boundary around them. Since they have kinetic energy (and velocity as well), they tend to leave their current location, but repulsive forces from the other atoms push it back to its original position. In this way, you could consider atomic bonds like "little springs".


The extent to which the atom displaces itself from the mean location is called it's vibrational amplitude. As we increase temperature, the atoms will have more velocity, and can consequently displace itself further from the mean position. Lindemann defines the melting temperature as the temperature at which the amplitude becomes half of the space between two adjacent atoms of the crystal.

Yet another criteria for defining the melting point is the Born criteria that says:

"Crystals are considered to melt, when the shear modulus approaches zero"

You might wonder, what is this shear modulus? It's infact, a measure of how much tangential stress a solid object can handle, and the tangential strain caused by it. Tangential forces on an object are forces that act parallel to the surface of the object, hence the name. See the picture below that shows a tangential force applied on an originally cuboidal object:

enter image description here

Tangential (or shearing) stress is defined as:

$$\sigma _{shear}=\frac{F_{shear}}{A}$$

where $F_{shear}$ is the tangential (shearing) force, and $A$ denotes area of surface in question.

Considering the picture above, shearing strain would be defined as:

$$\epsilon _{shear}=\frac{\Delta x}{h}$$

There exists a unique proportionality between the stress applied and strain produced, and is given by:

$$\frac{\sigma _{shear}}{\epsilon _{shear}}=S$$

where $S$ is called the shear modulus. Its a constant for a given material and a given temperature.

In a physical sense, liquids are considered as substances that cannot withstand a tangential stress. When a tangential stress is applied, liquids simply keep on increasing the strain, even under small stress. This itself is a useful way to consider when a solid melts, when its shear modulus becomes zero!

Now, let's consider how we consider melting for the different examples that you have asked for:

Diamond: Diamond has a structure of covalent bonds arranged tetrahedrally to each carbon. Each bond has it's characteristic bond length. Applying Lindemann condition over here would be considered advisable , so that melting point is where vibrations of carbon atoms is half of the $\ce{C-C}$ bond distance. Ultimately some of the $\ce{C-C}$ would break and the molten mixture would largely consist of radicals of varying sizes, that permit free movement between themselves.

Graphite: While each layer of graphite is only weakly bonded to the other layer by van der Waals' forces, each layer (called graphene) itself is a quite a large molecule. You can consider a liquid as a large number of small molecules, so clearly graphene doesn't look like this. To melt it, you would have to break some of the $\ce{C-C}$ bonds so that we can produce small molecules that can move freely. We apply Lindemann condition in this case. The molten mixture would be similar to that of diamond.

Branched polymers: Remember how I told to consider liquids as small molecules that can move freely sliding among themselves? We consider the same thing over here. The branch bonds are comparatively weaker than the rest of the bonds in the polymer, and these are the ones that will break when heated. As far as bond-breakings are concerned, we use the Lindemann condition. While the unbranched polymers are big molecules, they are still small enough to exhibit liquid character at the high temperature required to break the branched bonds. The molten mixture would consist of the straight-chain polymers in radical forms.

  • 2
    $\begingroup$ Your answer tells me what melting is from a very broad perspective, but I can't apply it to actually find out what I want to know. $\endgroup$
    – John
    Apr 30, 2017 at 14:31
  • $\begingroup$ Oh, that's a rather unique way to give an answer. I shall be waiting! $\endgroup$
    – John
    Apr 30, 2017 at 17:03
  • $\begingroup$ Thank you for the answer! The biggest follow-up question I have is what does it mean for a molecule to be 'small'? We say graphene isn't small, but polymer chains are. How do we draw the distinction? If we have a sample of a branched polymer consisting of 100 chains and every 2 chains are joined to each other (so there are 50 pieces), why do we need to break the branch bonds rather than the Van Der Waals forces between the pieces for melting? That would allow the pieces to move freely. Lastly, aren't Van Der Waals forces weaker than the branch bonds? If so, why aren't these broken first? $\endgroup$
    – John
    May 1, 2017 at 7:41
  • $\begingroup$ @Saad there is a lot that goes on. Youre right, my explanation using 'big' is quite vague, while what you must consider are the van der Waals' attractions. If the pieces are large enough for the attractions to make it solid, it will be solid. This is not for network solids. In case of graphite, van der Waals' bonds are broken first, but the graphene sheets still remain in solid state. You must break a few $\ce{C-C}$ bonds to get it to liquid state. $\endgroup$ May 1, 2017 at 9:26
  • $\begingroup$ For network solids, we have small molecules bonded to each other, so breaking the Van Der Waals forces would cause no difference as the molecules would stay the same distance apart, and these forces would soon reform. Is this correct? Also, could you answer my 'pieces' question? $\endgroup$
    – John
    May 1, 2017 at 9:42

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