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van der Waals radius

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Size of an atom's imaginary sphere representing how close other atoms can get

van der Waals radii
Element	radius (Å)
Hydrogen	1.2 (1.09)^[1]
Carbon	1.7
Nitrogen	1.55
Oxygen	1.52
Fluorine	1.47
Phosphorus	1.8
Sulfur	1.8
Chlorine	1.75
Copper	1.4
van der Waals radii taken from Bondi's compilation (1964).^[2] Values from other sources may differ significantly (see text)

Types of radii
Atomic radius Ionic radius Covalent radius Metallic radius Van der Waals radius
v t e

Thevan der Waals radius,r_w, of anatom is theradius of an imaginary hardsphere representing thedistance of closest approach for another atom. It is named afterJohannes Diderik van der Waals, winner of the 1910Nobel Prize in Physics, as he was the first to recognise that atoms were not simplypoints and to demonstrate the physical consequences of their size through thevan der Waals equation of state.

van der Waals volume

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Thevan der Waals volume,V_w, also called theatomic volume ormolecular volume, is the atomic property most directly related to the van der Waals radius.^[3] 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 a single atom, it is the volume of a sphere whose radius is the van der Waals radius of the atom: $V_{\rm {w}}={4 \over 3}\pi r_{\rm {w}}^{3}.$

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 bonds.

The van der Waals volume of an atom or molecule may also be determined by experimental measurements on gases, notably from thevan der Waals constantb, thepolarizabilityα, or themolar refractivityA. In all three cases, measurements are made on macroscopic samples and it is normal to express the results asmolar quantities. To find the van der Waals volume of a single atom or molecule, it is necessary to divide by theAvogadro constantN_A.

The molar van der Waals volume should not be confused with themolar volume of the substance. In general, at normal laboratory temperatures and pressures, the atoms or molecules of gas only occupy about1⁄1000 of the volume of the gas, the rest is empty space. Hence the molar van der Waals volume, which only counts the volume occupied by the atoms or molecules, is usually about1000 times smaller than the molar volume for a gas atstandard temperature and pressure.

Table of van der Waals radii

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Van der Waals radius of the elements in theperiodic table
Group →	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18
↓ Period
1	H110^[1] or 120																	He140
2	Li182	Be153^[4]											B192^[4]	C170	N155	O152	F147	Ne154
3	Na227	Mg173											Al184^[4]	Si210	P180	S180	Cl175	Ar188
4	K275	Ca231^[4]	Sc211^[4]	Ti	V	Cr	Mn	Fe	Co	Ni163	Cu140	Zn139	Ga187	Ge211^[4]	As185	Se190	Br185	Kr202
5	Rb303^[4]	Sr249^[4]	Y	Zr	Nb	Mo	Tc	Ru	Rh	Pd163	Ag172	Cd158	In193	Sn217	Sb206^[4]	Te206	I198	Xe216
6	Cs343^[4]	Ba268^[4]	Lu	Hf	Ta	W	Re	Os	Ir	Pt175	Au166	Hg155	Tl196	Pb202	Bi207^[4]	Po197^[4]	At202^[4]	Rn220^[4]
7	Fr348^[4]	Ra283^[4]	Lr	Rf	Db	Sg	Bh	Hs	Mt	Ds	Rg	Cn	Nh	Fl	Mc	Lv	Ts	Og

			La	Ce	Pr	Nd	Pm	Sm	Eu	Gd	Tb	Dy	Ho	Er	Tm	Yb
			Ac	Th	Pa	U186	Np	Pu	Am	Cm	Bk	Cf	Es	Fm	Md	No
Legend
Values for the van der Waals radii are inpicometers (pm or1×10⁻¹² m)
The shade of the box ranges from red to yellow as the radius increases; Gray indicate a lack of data.
Unless indicated otherwise, the data is fromMathematica'sElementData function fromWolfram Research, Inc.^[5]
Primordial From decay Synthetic Border shows natural occurrence of the element

Methods of determination

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Van der Waals radii may be determined from themechanical properties of gases (the original method), from thecritical point, from measurements of atomic spacing between pairs of unbonded atoms incrystals 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.^[2]

Van der Waals equation of state

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Main article:Van der Waals equation

The van der Waals equation of state is the simplest and best-known modification of theideal gas law to account for the behaviour ofreal gases: $\left(p+a\left({\frac {n}{\tilde {V}}}\right)^{2}\right)({\tilde {V}}-nb)=nRT,$ wherep is pressure,n is the number of moles of the gas in question anda andb depend on the particular gas, ${\tilde {V}}$ is the volume,R is the specific gas constant on a unit mole basis andT the absolute temperature;a is a correction for intermolecular forces andb corrects for finite atomic or molecular sizes; the value ofb 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 thevan der Waals force), which increases whenn increases orV decreases and molecules become more crowded together.

Gas	d (Å)	b (cm³mol^–1)	V_w (Å³)	r_w (Å)
Hydrogen	0.74611	26.61	34.53	2.02
Nitrogen	1.0975	39.13	47.71	2.25
Oxygen	1.208	31.83	36.62	2.06
Chlorine	1.988	56.22	57.19	2.39
van der Waals radiir_w in Å (or in 100 picometers) calculated from thevan der Waals constants of some diatomic gases. Values ofd andb from Weast (1981).

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

Forhelium,^[6]b = 23.7 cm³/mol. Helium is amonatomic gas, and each mole of helium contains6.022×10²³ atoms (theAvogadro constant,N_A): $V_{\rm {w}}={b \over {N_{\rm {A}}}}$ Therefore, the van der Waals volume of a single atomV_w = 39.36 Å³, which corresponds tor_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 is2r_w and the internuclear distance isd. The algebra is more complicated, but the relation $V_{\rm {w}}={4 \over 3}\pi r_{\rm {w}}^{3}+\pi r_{\rm {w}}^{2}d$ can be solved by the normal methods forcubic functions.

Crystallographic measurements

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The molecules in amolecular crystal are held together byvan der Waals forces rather thanchemical bonds. In principle, the closest that two atoms belonging todifferent 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 workThe 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 ofhydrogen as opposed to Bondi's 1.20 Å.^[1] A more recent analysis of theCambridge Structural Database, carried out by Santiago Alvarez, provided a new set of values for 93 naturally occurring elements.^[8] The values of different authors are sometimes very different, so that one has to chose the ones which are closest in their physical meaning to those one wants to compare with. Here is a table with entries of four different authors. The valuse of Bondi from 1966 are those mostly used in crystallography:

		r_vdW / Å	r_vdW / Å	r_vdW / Å	r_vdW / Å
Element	Atomic number	Bondi^[2] 1966	Batsanov^[9] 2001	Hu^[10] 2009	Alvarez^[8] 2014
H	1	1.2	1.0	1.08	1.2
Li	3	1.82	2.15	2.14	2.12
Be	4		1.85	1.69	1.98
B	5		1.75	1.68	1.91
C	6	1.7	1.7	1.53	1.77
N	7	1.55	1.6	1.51	1.66
O	8	1.52	1.55	1.49	1.5
F	9	1.47	1.45	1.48	1.46
Na	11	2.27	2.45	2.38	2.5
Mg	12	1.73	2.15	2.00	2.51
Al	13		2.05	1.92	2.25
Si	14	2.1	2.05	1.93	2.19
P	15	1.8	1.95	1.88	1.9
S	16	1.8	1.8	1.81	1.89
Cl	17	1.75	1.8	1.75	1.82
Se	17	1.90	1.9	1.92	1.82
K	19	2.75	2.85	2.52	2.73
Ca	20		2.45	2.27	2.62
Sc	21		2.25	2.15	2.58
Ti	22		2.10	2.11	2.45
V	23		2.05	2.07	2.42
Cr	24		2.0	2.06	2.45
Mn	25		2.0	2.05	2.45
Fe	26		2.0	2.04	2.44
Ni	28	1.63	1.95	1.97	2.4
Cu	29	1.40	1.9	1.96	2.38
Zn	30	1.39	2.0	2.01	2.39
Ga	31	1.87	2.05	2.03	2.32
Ge	32		2.05	2.05	2.29
As	33	1.85	2.05	2.08	1.88
Br	35	1.85	1.9	1.9	1.86
Rb	37		3.0	2.61	3.21
Sr	38		2.6	2.42	2.84
Y	39		2.4	2.32	2.75
Zr	39		2.3	2.23	2.52
Nb	41		2.15	2.18	2.56
Mo	42		2.1	2.17	2.45
Tc	43		2.1	2.16	2.44
Ru	44		2.05	2.13	2.46
Rh	45		2.0	2.1	2.44
Pd	46	1.63	2.05	2.1	2.15
Ag	47	1.72	2.05	2.11	2.53
Co	47		1.95	2	2.4
Cd	48	1.62	2.2	2.18	2.43
In	49	1.93	2.25	2.21	2.43
Sn	50	2.17	2.2	2.23	2.42
Sb	51		2.25	2.24	2.47
Te	52	2.06	2.15	2.11	1.99
I	53	1.98	2.1	2.09	2.04
Cs	55		3.15	2.75	3.48
Ba	56		2.7	2.59	3.03
La	57		2.5	2.43	2.98
Hf	72		2.25	2.23	2.63
Ta	73		2.2	2.22	2.53
W	74		2.15	2.18	2.57
Re	75		0.21	2.16	2.49
Os	76		2.0	2.16	2.48
Ir	77		2.0	2.13	2.41
Pt	78	1.72	2.05	2.13	2.32
Au	79	1.66	2.0	2.14	2.32
Hg	80	1.70	2.1	2.23	2.45
Tl	81	1.96	2.25	2.27	2.47
Pb	82	2.02	2.3	2.37	2.6
Bi	83		2.35	2.38	2.54
Th	90		2.45	2.45	2.93
U	91	1.86	2.4	2.41	2.71

A simple example of the use of crystallographic data (hereneutron diffraction) is to consider the case of solid helium, where the atoms are held together only by van der Waals forces (rather than bycovalent ormetallic 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 is0.214(6) g/cm³,^[11] corresponding to amolar volumeV_m =18.7×10⁻⁶ m³/mol. The van der Waals volume is given by $V_{\rm {w}}={\frac {\pi V_{\rm {m}}}{N_{\rm {A}}{\sqrt {18}}}}$ where the factor of π/√18 arises from thepacking of spheres:V_w =2.30×10⁻²⁹ m³ = 23.0 Å³, corresponding to a van der Waals radiusr_w = 1.76 Å.

Molar refractivity

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Themolar refractivityA of a gas is related to itsrefractive indexn by theLorentz–Lorenz equation: $A={\frac {RT(n^{2}-1)}{3p}}$ The refractive index of heliumn =1.0000350 at 0 °C and 101.325 kPa,^[12] which corresponds to a molar refractivityA =5.23×10⁻⁷ m³/mol. Dividing by the Avogadro constant givesV_w =8.685×10⁻³¹ m³ = 0.8685 Å³, corresponding tor_w = 0.59 Å.

Polarizability

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Thepolarizabilityα of a gas is related to itselectric susceptibilityχ_e by the relation $\alpha ={\varepsilon _{0}k_{\rm {B}}T \over p}\chi _{\rm {e}}$ and the electric susceptibility may be calculated from tabulated values of therelative permittivityε_r using the relationχ_e =ε_r − 1. The electric susceptibility of heliumχ_e =7×10⁻⁵ at 0 °C and 101.325 kPa,^[13] which corresponds to a polarizabilityα =2.307×10⁻⁴¹ C⋅m²/V. The polarizability is related the van der Waals volume by the relation $V_{\rm {w}}={1 \over {4\pi \varepsilon _{0}}}\alpha ,$ so the van der Waals volume of heliumV_w =2.073×10⁻³¹ m³ = 0.2073 Å³ by this method, corresponding tor_w = 0.37 Å.

When the atomic polarizability is quoted in units of volume such as Å³, 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.