A new uniformist model of water

Abstract

A simple uniformist model is proposed for the structure of liquid water which consists of a puckered hexagonal ring of six oxygen atoms with six hydrogen-bonded hydrogen atoms within the ring and six non-bonded noncyclic hydrogen atoms. This model differs from previous ones in that it postulates that of the two hydrogen bonds one oxygen can form with two hydrogen atoms, one is stronger than the other. Both exist in ice. The weaker hydrogen bond (5.9 kJmol^-1: at 0°C.) breaks when ice melts while the stronger one (37.7 kJmol^-1 at 100°C.) persists in water and breaks only when water boils. The stronger bond also breaks during sublimation of ice. This explains why the heat of melting, due to breaking of the weak H-bond, is only 13% of the sublimation energy of the solid. The puckered six-membered ring of the H₂0 molecule is enclosed in a hexagonal cell, the volume of which is controlled by the oxygen-oxygen distance and the bond angles. The hexagonal cell in water is thinner than that in ice. This uniformist model is in good agreement with the distances of nearest neighbours as obtained from X-ray diffraction data for both water and ice-I. It also offers an explanation of not only why the number of nearest neighbours increases from 4.0 to 4.4 when ice melts but also why there is a neighbour at 0.35 nm in water but not in ice. Due to breaking of the weak hydrogen bonds on melting, the hexagonal cells can approach one another closer than they can in ice which explains the density increases on melting. On the basis of an empirical equation for the thickness of the hexagonal cells in water the predicted bond angle is 112°. Based on the distances of the nearest neighbors in water, it is calculated that the O - O - O bond angle is 106° at O°C.