The structure and electronic properties of the (111)-surface of alkali metal oxides

Abstract

Under normal conditions, М 2 О oxides (M = Li, Na, K) crystallize into a structure of the antifluorite type with one formula unit (N = 1) [1]. Each (111) surface layer of the М 2 О oxide consists of three charged atomic layers M-O-M. Such surfaces often prove relatively stable [2]. Oxides of alkali metals were investigated both by experimental and theoretical methods, however, the electronic structure and geometry relaxation of the (111) surface were studied only for Li 2 O [3,4]. In so doing, in the framework of the Hartree-Fock method, a model of a 2-layer thick plate was applied [3], which is insufficient for studying surfaces. Thus, the surface of oxides is poorly understood.In this work, calculations for М 2 О are performed using the CRYSTAL06 package [5]. In the calculations, use was made of the basis of linear combinations of atomic orbitals (LCAO) and the hybrid functional B3PW [6]. The bases 6-411G, 6-1G, 8-511G, and 86-511G were used for an oxygen atom and Li-, Na-, and K cations, respectively [7]. In so doing, the calculated lattice constants in the volume deviate from the experimental values [1] only by 0.18, 0.46 and 1.8% for Li 2 O, Na 2 O, and K 2 O, respectively. In addition, the experimental oxygen atom charge (-1.90 |e|) for Li 2 O is very well reproduced (-1.897 |e|) [8]). The surface was modeled by a two-dimensional plate (2D), which was "cut out" from a three-dimensional crystal (3D). The Cartesian axes were arranged in such a way that the Z axis was perpendicular to the examined (hkl) surface. The surface is defined as a set of atomic layers, whose properties are different from the bulk ones, and each layer is determined by the atoms with fixed Z. A two-dimensional unit cell was chosen in such a way that the properties of the middle layer reproduced the volume properties with an accuracy of 0.2%. The surface energy was calculated using the formula Е surf (n) = (Е 2D -nE 3D )/2∆S surf , where Е 2D is the energy of a 2D-unit cell of the plate, Е 3D is the energy of a 3D-unit cell of the crystal, n = N 2D /N 3D , and ∆S surf is the area of the basis of a 2D unit cell. Table 1 shows the atomic charges and displacements of atomic layers of the M 2 O (111) surface caused by surface relaxation. The numbering of atomic layers starts from the topmost layer. Fourteen atomic layers were sufficient to consider surface properties. Displacements of oxygen atoms in Li 2 O reach a maximum value of 1.08% (layer 2) and those for lithium atoms are 0.96% (layer 3). Then, the displacements decrease into the depth, so that for the twelfth layer, they are only 0.16%, which is an order lower than the relaxation for layer 3. The displacements of layers of sodium and potassium oxides decrease in a similar way, while with increasing atomic number of the cation, the surface atoms undergo greater relaxation. The changes of atomic charges in the upper layers of oxide surface relative to the volume increase with the atomic number of the cation, but they do not exceed 0.005, 0.010, and 0.012 |e|, respectively. Starting fro...