It is named after Johannes Diderik van der Waals, winner of the 1910 Nobel Prize in Physics, as he was the first to recognise that atoms were not simply points and to demonstrate the physical consequences of their size through the van der Waals equation of state.
The van der Waals volume of a molecule is always smaller than the sum of the van der Waals volumes of the constituent atoms: the atoms can be said to "overlap" when they form chemical bonds.
The van der Waals volume of an atom or molecule may also be determined by experimental measurements on gases, notably from the van der Waals constant b, the polarizability α, or the molar refractivity A.
In all three cases, measurements are made on macroscopic samples and it is normal to express the results as molar quantities.
To find the van der Waals volume of a single atom or molecule, it is necessary to divide by the Avogadro constant NA.
Primordial From decay Synthetic Border shows natural occurrence of the element Van der Waals radii may be determined from the mechanical properties of gases (the original method), from the critical point, from measurements of atomic spacing between pairs of unbonded atoms in crystals or from measurements of electrical or optical properties (the polarizability and the molar refractivity).
These various methods give values for the van der Waals radius which are similar (1–2 Å, 100–200 pm) but not identical.
[2] The van der Waals equation of state is the simplest and best-known modification of the ideal gas law to account for the behaviour of real gases:
is the volume, R is the specific gas constant on a unit mole basis and T the absolute temperature; a is a correction for intermolecular forces and b corrects for finite atomic or molecular sizes; the value of b equals the van der Waals volume per mole of the gas.
The van der Waals equation also has a microscopic interpretation: molecules interact with one another.
The ideal gas law must be corrected when attractive and repulsive forces are considered.
Thus, a fraction of the total space becomes unavailable to each molecule as it executes random motion.
Therefore, the van der Waals volume of a single atom Vw = 39.36 Å3, which corresponds to rw = 2.11 Å (≈ 200 picometers).
This method may be extended to diatomic gases by approximating the molecule as a rod with rounded ends where the diameter is 2rw and the internuclear distance is d. The algebra is more complicated, but the relation
The molecules in a molecular crystal are held together by van der Waals forces rather than chemical bonds.
In principle, the closest that two atoms belonging to different molecules can approach one another is given by the sum of their van der Waals radii.
This approach was first used by Linus Pauling in his seminal work The Nature of the Chemical Bond.
[7] Arnold Bondi also conducted a study of this type, published in 1964,[2] although he also considered other methods of determining the van der Waals radius in coming to his final estimates.
Some of Bondi's figures are given in the table at the top of this article, and they remain the most widely used "consensus" values for the van der Waals radii of the elements.
Scott Rowland and Robin Taylor re-examined these 1964 figures in the light of more recent crystallographic data: on the whole, the agreement was very good, although they recommend a value of 1.09 Å for the van der Waals radius of hydrogen as opposed to Bondi's 1.20 Å.
[1] A more recent analysis of the Cambridge Structural Database, carried out by Santiago Alvareza, provided a new set of values for 93 naturally occurring elements.
[8] A simple example of the use of crystallographic data (here neutron diffraction) is to consider the case of solid helium, where the atoms are held together only by van der Waals forces (rather than by covalent or metallic bonds) and so the distance between the nuclei can be considered to be equal to twice the van der Waals radius.
The density of solid helium at 1.1 K and 66 atm is 0.214(6) g/cm3,[9] corresponding to a molar volume Vm = 18.7×10−6 m3/mol.
where the factor of π/√18 arises from the packing of spheres: Vw = 2.30×10−29 m3 = 23.0 Å3, corresponding to a van der Waals radius rw = 1.76 Å.
Dividing by the Avogadro constant gives Vw = 8.685×10−31 m3 = 0.8685 Å3, corresponding to rw = 0.59 Å.
The electric susceptibility of helium χe = 7×10−5 at 0 °C and 101.325 kPa,[11] which corresponds to a polarizability α = 2.307×10−41 C⋅m2/V.
so the van der Waals volume of helium Vw = 2.073×10−31 m3 = 0.2073 Å3 by this method, corresponding to rw = 0.37 Å.