van der Waals radius

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:

$${\displaystyle \left(p+a\left({\frac {n}{\tilde {V}}}\right)^{2}\right)({\tilde {V}}-nb)=nRT,}$$

where p is pressure, n is the number of moles of the gas in question and a and b depend on the particular gas, ${\displaystyle {\tilde {V}}}$ 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. Their values vary from gas to gas.

The van der Waals equation also has a microscopic interpretation: molecules interact with one another. The interaction is strongly repulsive at a very short distance, becomes mildly attractive at the intermediate range, and vanishes at a long distance. The ideal gas law must be corrected when attractive and repulsive forces are considered. For example, the mutual repulsion between molecules has the effect of excluding neighbors from a certain amount of space around each molecule. Thus, a fraction of the total space becomes unavailable to each molecule as it executes random motion. In the equation of state, this volume of exclusion (nb) should be subtracted from the volume of the container (V), thus: (V - nb). The other term that is introduced in the van der Waals equation, ${\textstyle a\left({\frac {n}{\tilde {V}}}\right)^{2}}$, describes a weak attractive force among molecules (known as the van der Waals force), which increases when n increases or V decreases and molecules become more crowded together.

The van der Waals constant b volume can be used to calculate the van der Waals volume of an atom or molecule with experimental data derived from measurements on gases.

For helium,[5] b = 23.7 cm³/mol. Helium is a monatomic gas, and each mole of helium contains 6.022×10²³ atoms (the Avogadro constant, N_A):

$${\displaystyle V_{\rm {w}}={b \over {N_{\rm {A}}}}}$$

Therefore, the van der Waals volume of a single atom V_w = 39.36 Å³, which corresponds to r_w = 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 2r_w and the internuclear distance is d. The algebra is more complicated, but the relation

$${\displaystyle V_{\rm {w}}={4 \over 3}\pi r_{\rm {w}}^{3}+\pi r_{\rm {w}}^{2}d}$$

can be solved by the normal methods for cubic functions.

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. By examining a large number of structures of molecular crystals, it is possible to find a minimum radius for each type of atom such that other non-bonded atoms do not encroach any closer. This approach was first used by Linus Pauling in his seminal work The Nature of the Chemical Bond.[6] 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 Alvarez, provided a new set of values for 93 naturally occurring elements.[7]

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,[8] corresponding to a molar volume V_m = 18.7×10⁻⁶ m³/mol. The van der Waals volume is given by

$${\displaystyle V_{\rm {w}}={\frac {\pi V_{\rm {m}}}{N_{\rm {A}}{\sqrt {18}}}}}$$

where the factor of π/√18 arises from the packing of spheres: V_w = 2.30×10⁻²⁹ m³ = 23.0 Å³, corresponding to a van der Waals radius r_w = 1.76 Å.

The molar refractivity A of a gas is related to its refractive index n by the Lorentz–Lorenz equation:

$${\displaystyle A={\frac {RT(n^{2}-1)}{3p}}}$$

The refractive index of helium n = 1.0000350 at 0 °C and 101.325 kPa,[9] which corresponds to a molar refractivity A = 5.23×10⁻⁷ m³/mol. Dividing by the Avogadro constant gives V_w = 8.685×10⁻³¹ m³ = 0.8685 Å³, corresponding to r_w = 0.59 Å.

The polarizability α of a gas is related to its electric susceptibility χ_e by the relation

$${\displaystyle \alpha ={\varepsilon _{0}k_{\rm {B}}T \over p}\chi _{\rm {e}}}$$

and the electric susceptibility may be calculated from tabulated values of the relative permittivity ε_r using the relation χ_e = ε_r − 1. The electric susceptibility of helium χ_e = 7×10⁻⁵ at 0 °C and 101.325 kPa,[10] which corresponds to a polarizability α = 2.307×10⁻⁴¹ C⋅m²/V. The polarizability is related the van der Waals volume by the relation

$${\displaystyle V_{\rm {w}}={1 \over {4\pi \varepsilon _{0}}}\alpha ,}$$

so the van der Waals volume of helium V_w = 2.073×10⁻³¹ m³ = 0.2073 Å³ by this method, corresponding to r_w = 0.37 Å.

When the atomic polarizability is quoted in units of volume such as ų, as is often the case, it is equal to the van der Waals volume. However, the term "atomic polarizability" is preferred as polarizability is a precisely defined (and measurable) physical quantity, whereas "van der Waals volume" can have any number of definitions depending on the method of measurement.