- Van der Waals radius
The

**van der Waals radius**, "r"sub|w, of anatom is the radius of an imaginary hardsphere which can be used to model the atom for many purposes. It is named afterJohannes Diderik van der Waals , winner of the 1910Nobel Prize in Physics , as he was the first to recognise that atoms had a finite size (ie, that atoms were not simply points) and to demonstrate the physical consequences of their size through the van der Waals equation of state.**Van der Waals Volume**The

**van der Waals volume**, "Vsub|w", also called the**atomic volume**or**molecular volume**, is the atomic property most directly related to the van der Waals radius. It is the volume "occupied" by an individual atom (or molecule). The van der Waals volume may be calculated if the van der Waals radii (and, for molecules, the inter-atomic distances and angles) are known. For spherical single atoms, it is the volume of a sphere whose radius is the van der Waals radius of the atoms ::$V\_\{\; m\; w\}\; =\; \{4over\; 3\}pi\; r\_\{\; m\; w\}^3$ and $r\_\{\; m\; w\}\; =\; sqrt\; [3]$3over {4piV_{ m w, respectively.For a molecule, it is the volume enclosed by thevan der Waals surface . 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 formchemical bond s.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", thepolarizability "α" or themolar 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 theAvogadro constant "N"sub|A.The molar van der Waals volume should not be confused with the

molar volume of the substance. In general, at normal laboratory temperatures and pressures, the atoms or molecules of a gas only occupy about 1/1000 of the volume of the gas, the rest being empty space. Hence the molar van der Waals volume, which only counts the volume occupied by the atoms or molecules, is usually about 1000-times smaller than the molar volume for a gas atstandard temperature and pressure .**Methods of Determination**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

crystal s or from measurements of electrical or optical properties (thepolarizability and themolar refractivity ). These various methods give values for the van der Waals radius which are similar (1–2 Å, 100–200 pm) but not identical. Tabulated values of van der Waals radii are obtained by taking aweighted mean of a number of different experimental values, and, for this reason, different tables will often have different values for the van der Waals radius of the same atom. Indeed, there is no reason to assume that the van der Waals radius is a fixed property of the atom in all circumstances: rather, it tends to vary with the particular chemical environment of the atom in any given case.**Van der Waals Equation of State**The van der Waals equation of state is the simplest and best-known modification of the

ideal gas law to account for the behaviour ofreal gas es::$left\; (p\; +\; aleft\; (frac\{n\}\{\; ilde\{V\; ight\; )^2\; ight\; )\; (\; ilde\{V\}\; -\; nb)\; =\; nRT$,where "n" is theamount of substance of the gas in question and "a" and "b" are adjustable parameters; "a" is a correction for intermolecular forces and "b" corrects for finite atomic or molecular sizes; the value of "b" equals the volume of one mole of the atoms or molecules.

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 very short distance, becomes mildly attractive at intermediate range, and vanishes at 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, $aleft\; (frac\{n\}\{\; ilde\{V\; ight\; )^2$, 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.**Crystallographic Measurements**The molecules in a

molecular crystal are held together byvan der Waals force s rather thanchemical bond s. 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 byLinus Pauling in his seminal work "The Nature of the Chemical Bond". [*Pauling (1945).*] Bondi also conducted a study of this type, published in 1964, 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. Rowland and 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 ofhydrogen as opposed to Bondi's 1.20 Å. [*Rowland & Taylor (1996).*]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 ormetallic bond s) 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/cmsup|3, [*Henshaw (1958).*] corresponding to amolar volume "V"sub|m = 18.7×10sup|–6 msup|3/mol. The van der Waals volume is given by:$V\_\{\; m\; w\}\; =\; frac\{pi\; V\_\{\; m\; m\{N\_\{\; m\; A\}sqrt\{18$where the factor of π/√18 arises from the packing of spheres: "V"sub|w = 2.30×10sup|–29 msup|3 = 23.0 Åsup|3, corresponding to a van der Waals radius "r"sub|w = 1.76 Å.**Molar Refractivity**The

molar refractivity "A" of a gas is related to itsrefractive index "n" by theLorentz–Lorenz equation ::$A\; =\; frac\{R\; T\; (n^2\; -\; 1)\}\{3p\}$The refractive index of helium "n" = 1.000 0350 at 0 ºC and 101.325 kPa, [*Kaye & Laby Tables, [*] which corresponds to a molar refractivity "A" = 5.23×10sup|–7 msup|3/mol. Dividing by the Avogadro constant gives "V"sub|w = 8.685×10sup|–31 msup|3 = 0.8685 Åsup|3, corresponding to "r"sub|w = 0.59 Å.*http://www.kayelaby.npl.co.uk/general_physics/2_5/2_5_7.html Refractive index of gases*] .**Polarizability**The

polarizability "α" of a gas is related to itselectric susceptibility "χ"sub|e by the relation:$alpha\; =\; \{epsilon\_0\; k\_\{\; m\; B\}Tover\; p\}chi\_\{\; m\; e\}$and the electric susceptibility may be calculated from tabulated values of the relative permittivity "ε"sub|r using the relation "χ"sub|e = "ε"sub|r–1. The electric susceptibility of helium "χ"sub|e = 7×10sup|–5 at 0 ºC and 101.325 kPa, [*Kaye & Laby Tables, [*] which corresponds to a polarizability "α" = 2.307×10sup|–41 Cmsup|2/V. The polarizability is related the van der Waals volume by the relation:$V\_\{\; m\; w\}\; =\; \{1over\{4piepsilon\_0alpha\; ,$so the van der Waals volume of helium "V"sub|w = 2.073×10sup|–31 msup|3 = 0.2073 Åsup|3 by this method, corresponding to "r"sub|w = 0.37 Å.*http://www.kayelaby.npl.co.uk/general_physics/2_6/2_6_5.html Dielectric Properties of Materials*] .When the atomic polarizability is quoted in units of volume such as Åsup|3, 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.**References*** [

*http://pubs.acs.org/cgi-bin/abstract.cgi/jpchax/1964/68/i03/f-pdf/f_j100785a001.pdf First page*] .

* [*http://prola.aps.org/abstract/PR/v109/i2/p328_1 Abstract*] .

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***External links*** [

*http://periodictable.com/Properties/A/VanDerWaalsRadius.v.html Van Der Waals Radius of the elements*] at PeriodicTable.com

* [*http://www.webelements.com/periodicity/van_der_waals_radius/ Van der Waals Radius – Periodicity*] at WebElements.com

* [*http://www.ccdc.cam.ac.uk/products/csd/radii/ Elemental Radii*] from theCambridge Structural Database

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