Water basis vectors

If one looks up the term component in practically any text on physical chemistry or thermodynamics, one finds it is defined as the minimum number of chemical formula units needed to describe the composition of all parts of the system. We say formulas rather than substances because the chemical formulas need not correspond to any actual compounds. For example, a solution of salt in water has two components, NaCl and H2O, even if there is a vapor phase and/or a solid phase (ice or halite), because some combination of those two formulas can describe the composition of every phase. Similarly, a mixture of nitrogen and hydrogen needs only two components, such as N2 and H2, despite the fact that much of the gas may exist as species NH3. Note that although there is always a wide choice of components for a given system (we could equally well choose N and H as our components, or N10 and H10), the number of components for a given system is fixed. The components are simply building blocks , or mathematical entities, with which we are able to describe the bulk composition of any phase in the system. The list of components chosen to represent a system is, in mathematical terms, a basis vector, or simply the basis . 

To show how the matrix approach works, we will go over the (Zp c, 2py) example more formally. We have used an equivalent set of basis vectors for the water problem before. Figure 2.2 shows a set of vectors labelled x,y and z on the O atom of H2O along with the transformation that occurs after each of the symmetry operations in Cav is applied. These vectors have exactly the same symmetry properties as the p-orbital set on the O atom, since they are their functional forms. The paper models from Appendix 1 can also be used to follow the transformations discussed with this basis. If we consider the x and y vectors together, the C2 transformation can be written as. 

Let us illustrate the method by finding SALCs equivalent to Equations (6.25) and (6.26), which were simply stated in Section 6.4, using the two-vector basis for the O—H stretching modes of water shown in Figure 6.18a. If we take the basis vector, which is along the O—Hi bond, and apply each of the 2 operations, the results obtained are as given in the first row of Table 6.10. The E and CTv (TZ) operations (the plane of the molecule) do not affect the positions of the basis vectors, and so b is simply transformed into itself. In contrast, the C2 and a XZ) operations swap over and bj, these two basis functions are linked by the operations of the group. 

The actual program used at NPL was written by N.P. Barry on the basis of the methods described previously. It is written in FORTRAN and has been implemented on IBM 370 and UNIVAC 1100 computers operated by computer bureaux. Vector algebra is employed. The reason why the graphs have double boundaries is that the calculation can be performed for boundaries of any convex polygon of up to 30 sides. This permits calculations to be restricted to the stability range of particular components, for example, that of water or chloride. 

Consider next the water molecule. As we have seen, it has a dipole moment, so we expect at least one IR-active mode. We have also seen that it has CIt, symmetry, and we may use this fact to help sort out the vibrational modes. Each normal mode of iibratbn wiff form a basis for an irreducible representation of the point group of the molecule.13 A vibration will be infrared active if its normal mode belongs to one of the irreducible representation corresponding to the x, y and z vectors. The C2 character table lists four irreducible representations A, Ait Bx, and B2. If we examine the three normal vibrational modes for HzO, we see that both the symmetrical stretch and the bending mode are symmetrical not only with respect to tbe C2 axis, but also with respect to the mirror planes (Fig. 3.21). They therefore have A, symmetry and since z transforms as A, they are fR active. The third mode is not symmetrical with respect to the C2 axis, nor is it symmetrical with respect to the ojxz) plane, so it has B2 symmetry. Because y transforms as Bt, this mode is also (R active. The three vibrations absorb at 3652 cm-1, 1545 cm-1, and 3756 cm-, respectively. 

On a vector basis (Fig. 7.2b), the major route of release was residues , responsible for 82.4% of the total, followed by air (12.9%) and water (4.2%). The land and products together contributed to only 0.5% of the total annual release. It was observed that for the land and products vectors, a blank release value was assigned to many classes of potential emission sources due to a general lack of data on EFs. 

First, an appropriate basis set has to be found. Considering that a molecule has 37V degrees of motional freedom, a system of 37V so-called Cartesian displacement vectors is a convenient choice. A set of such vectors is shown in Figure 5-5 for the water molecule. A separate Cartesian coordinate system is attached to each atom of the molecule, with the atoms at the origin. The orientation of the axes is the same in each system. Any displacement of the atoms can be expressed by a vector, and in turn this vector can be expressed as the vector sum of the Cartesian displacement vectors. 

Continuing with the water molecule as an example, the basis of the Cartesian displacement vectors will consist of nine vectors... 

Figure 5-5. Cartesian displacement vectors as basis for representation of the water molecule.

Here V is the molal volume of the liquid, N is Avagadro s number, K the. Boltzmann constant, T the absolute temperature, and is the angle between the vectors jj, and Ji. ag is the polarizability of the molecules in the liquid it can be determined from the refractive index. For highly polar substances is small compared to the second term in the parenthesis. The high dielectric constant of water is well accounted for by equation (47) on the basis of the known dipole moment of water, and the orientation of the water molecules relative to one another. The molar polarization of the liquid is defined in terms of the polarizability and the dipole moment of the molecules by equation (47). In a solution containing several components equation (47) becomes... 

Continuing with the water molecule as an example, the basis of the Cartesian displacement vectors will consist of nine vectors (see Figure 5-5). Operation E brings all of them into themselves, and the character is 9. Operation Cj changes the position of the two hydrogen atoms, so only the three coordinates of the oxygen atom have to be considered. The corresponding block of the matrix representation is... 

---