Ice I_h (hexagonal ice crystal) (pronounced: ice one h, also known as ice-phase-one) is the hexagonal crystal form of ordinary ice, or frozen water.^[1] Virtually all ice in the biosphere is ice I_h, with the exception only of a small amount of ice I_c that is occasionally present in the upper atmosphere. Ice I_h exhibits many peculiar properties that are relevant to the existence of life and regulation of global climate. For a description of these properties, see Ice, which deals primarily with ice I_h.

The crystal structure is characterized by the oxygen atoms forming hexagonal symmetry with near tetrahedral bonding angles. Ice I_h is stable down to −268 °C (5 K; −450 °F), as evidenced by x-ray diffraction^[2] and extremely high resolution thermal expansion measurements.^[3] Ice I_h is also stable under applied pressures of up to about 210 megapascals (2,100 atm) where it transitions into ice III or ice II.^[4]

Physical properties

The density of ice I_h is 0.917 g/cm³ which is less than that of liquid water. This is attributed to the presence of hydrogen bonds which causes atoms to become closer in the liquid phase.^[5] Because of this, ice I_h floats on water, which is highly unusual when compared to other materials. The solid phase of materials is usually more closely and neatly packed and has a higher density than the liquid phase. When lakes freeze, they do so only at the surface while the bottom of the lake remains near 4 °C (277 K; 39 °F) because water is densest at this temperature. No matter how cold the surface becomes, there is always a layer at the bottom of the lake that is 4 °C (277 K; 39 °F). This anomalous behavior of water and ice is what allows fish to survive harsh winters. The density of ice I_h increases when cooled, down to about −211 °C (62 K; −348 °F); below that temperature, the ice expands again (negative thermal expansion).^[2][3]

The latent heat of melting is 5987 J/mol, and its latent heat of sublimation is 50911 J/mol. The high latent heat of sublimation is principally indicative of the strength of the hydrogen bonds in the crystal lattice. The latent heat of melting is much smaller, partly because liquid water near 0 °C also contains a significant number of hydrogen bonds. The refractive index of ice I_h is 1.31.

Crystal structure

The accepted crystal structure of ordinary ice was first proposed by Linus Pauling in 1935. The structure of ice I_h is the wurtzite lattice, roughly one of crinkled planes composed of tessellating hexagonal rings, with an oxygen atom on each vertex, and the edges of the rings formed by hydrogen bonds. The planes alternate in an ABAB pattern, with B planes being reflections of the A planes along the same axes as the planes themselves.^[6] The distance between oxygen atoms along each bond is about 275 pm and is the same between any two bonded oxygen atoms in the lattice. The angle between bonds in the crystal lattice is very close to the tetrahedral angle of 109.5°, which is also quite close to the angle between hydrogen atoms in the water molecule (in the gas phase), which is 105°. This tetrahedral bonding angle of the water molecule essentially accounts for the unusually low density of the crystal lattice – it is beneficial for the lattice to be arranged with tetrahedral angles even though there is an energy penalty in the increased volume of the crystal lattice. As a result, the large hexagonal rings leave almost enough room for another water molecule to exist inside. This gives naturally occurring ice its rare property of being less dense than its liquid form. The tetrahedral-angled hydrogen-bonded hexagonal rings are also the mechanism that causes liquid water to be densest at 4 °C. Close to 0 °C, tiny hexagonal ice I_h-like lattices form in liquid water, with greater frequency closer to 0 °C. This effect decreases the density of the water, causing it to be densest at 4 °C when the structures form infrequently.

Hydrogen disorder

The hydrogen atoms in the crystal lattice lie very nearly along the hydrogen bonds, and in such a way that each water molecule is preserved. This means that each oxygen atom in the lattice has two hydrogens adjacent to it, at about 101 pm along the 275 pm length of the bond. The crystal lattice allows a substantial amount of disorder in the positions of the hydrogen atoms frozen into the structure as it cools to absolute zero. As a result, the crystal structure contains some residual entropy inherent to the lattice and determined by the number of possible configurations of hydrogen positions that can be formed while still maintaining the requirement for each oxygen atom to have only two hydrogens in closest proximity, and each H-bond joining two oxygen atoms having only one hydrogen atom.^[7] This residual entropy S₀ is equal to 3.4±0.1 J mol⁻¹ K⁻¹ ${\displaystyle =R\ln(1.50\pm 0.02)}$.^[8]

By contrast, the structure of ice II is hydrogen-ordered, which helps to explain the entropy change of 3.22 J/mol when the crystal structure changes to that of ice I. Also, ice XI, an orthorhombic, hydrogen-ordered form of ice I_h, is considered the most stable form at low temperatures.

Theoretical calculation

There are various ways of approximating this number from first principles. The following is the one used by Linus Pauling.^[9][10]

Suppose there are a given number N of water molecules in an ice lattice. To compute its residual entropy, we need to count the number of configurations that the lattice can assume. The oxygen atoms are fixed at the lattice points, but the hydrogen atoms are located on the lattice edges. The problem is to pick one end of each lattice edge for the hydrogen to bond to, in a way that still makes sure each oxygen atom is bond to two hydrogen atoms.

The oxygen atoms can be divided into two sets in a checkerboard pattern, shown in the picture as black and white balls. Focus attention on the oxygen atoms in one set: there are N/2 of them. Each has four hydrogen bonds, with two hydrogens close to it and two far away. This means there are ${\textstyle {\tbinom {4}{2))=6}$ allowed configurations of hydrogens for this oxygen atom (see Binomial coefficient). Thus, there are 6^N/2 configurations that satisfy these N/2 atoms. But now, consider the remaining N/2 oxygen atoms: in general they won't be satisfied (i.e., they will not have precisely two hydrogen atoms near them). For each of those, there are 2⁴ = 16 possible placements of the hydrogen atoms along their hydrogen bonds, of which 6 are allowed. So, naively, we would expect the total number of configurations to be

$${\displaystyle 6^{N/2}(6/16)^{N/2}=(3/2)^{N}.}$$

Using Boltzmann's entropy formula, we conclude that

$${\displaystyle S_{0}=k\ln(3/2)^{N}=nR\ln(3/2),}$$

${\displaystyle R\ln(3/2)=3.37\mathrm {J} \cdot \mathrm {mol} ^{-1}\mathrm {K} ^{-1))$

The same answer can be found in another way. First orient each water molecule randomly in each of the 6 possible configurations, then check that each lattice edge contains exactly one hydrogen atom. Assuming that the lattice edges are independent, then the probability that a single edge contains exactly one hydrogen atom is 1/2, and since there are 2N edges in total, we obtain a total configuration count ${\displaystyle 6^{N}\times (1/2)^{2N}=(3/2)^{N))$, as before.

Refinements

This estimate is 'naive', as it assumes the six out of 16 hydrogen configurations for oxygen atoms in the second set can be independently chosen, which is false. More complex methods can be employed to better approximate the exact number of possible configurations, and achieve results closer to measured values. Nagle (1966) used a series summation to obtain ${\displaystyle R\ln(1.50685\pm 0.00015)}$.^[11]

As an illustrative example of refinement, consider the following way to refine the second estimation method given above. According to it, six water molecules in a hexagonal ring would allow ${\displaystyle 6^{6}\times (1/2)^{6}=729}$ configurations. However, by explicit enumeration, there are actually 730 configurations. Now in the lattice, each oxygen atom participates in 12 hexagonal rings, so there are 2N rings in total for N oxygen atoms, or 2 rings for each oxygen atom, giving a refined result of ${\displaystyle R\ln(1.5\times (730/729)^{2})=R\ln(1.504)}$.^[12]