MODULE 10

1. MODULE 10 THE CHEMICAL BOND Atoms associate into specific groupings that we know as molecules. The inter-atomic interactions that allow stable groupings to exist make up what we call chemical bonds. The nature of bonding has always fascinated the community of chemists. My opinion is that there are two types of bonding-ionic and covalent. Ionic bonding results from the classical attraction between oppositely charged ions and it can be completely understood through the application of the Coulomb law. Covalent bonding is not understandable from a classical approach and we shall see that the covalent interaction is a purely quantum mechanical phenomenon.

2. MODULE 10 One of the major outcomes of our study of atomic structure is the concept of the atomic orbital. This is the physical representation of a solution to a Schrödinger equation and is designated by a set of quantum numbers. We also found that two electrons can occupy an atomic orbital if, and only if, their spins are opposed (antisymmetric wavefunction/SD). In this and ensuing modules we develop the concept of molecular orbitals, which also can be occupied by a pair of electrons with opposed spins. We start our development with (as usual) the simplest of molecules, the hydrogen molecule ion

3. MODULE 10 The Hydrogen Molecule Ion This is a hydrogen molecule that has one electron missing, or a hydrogen atom with an additional proton. It has two nucleons and a single electron, which as we shall see, provides the bonding force between the protons. It is a three-particle problem (as He atom). It is a real molecule that is stable (but reactive). It has a bond length of 2.00 a.u. (106 pm), and a binding energy of 268 kJ mol-1 (a weak covalent bond). As with all problems in quantum chemistry, our first task is to write down the Hamiltonian operator in atomic units:

4. e rB rA HA HB MODULE 10 The first term on the RHS contains the kinetic energy operators for the two nuclei, labeled as A and B (figure). The next term is the Laplacian for the single electron. The third and fourth terms are the Coulomb potential terms for the interactions between the electron and each of the nuclei. The last term is the inter-nuclear repulsion. We cannot solve this full Hamiltonian as it stands for the same reason that we encountered when dealing with the He atom-the inter-particle repulsion term. However, for molecules there is a very useful approximation we can use.

5. MODULE 10 This approximation depends on the fact that the protons have much greater mass than the electron and for a given kinetic energy they have a much smaller velocity. This means that we can imagine that the electrons “instantaneously” adjust their coordinates to the positions of the nucleons (like flies on a horse). Therefore we can separate the electronic and nuclear contributions to the Hamiltonian. This is the Born-Oppenheimer Approximation.

6. MODULE 10 In the calculation we • omit the nuclear motion terms in the hamiltonian, • fix the inter-nuclear distance R at an arbitrary value, • solve the resulting Schrödinger equation for the electron energy, • select another value of R, • solve the Schrödinger equation again, and so on. In this way we generate a potential energy vs. R curve. Thus R is used as a parameter in the calculation of the electronic energy.

7. MODULE 10 Within the Born-Oppenheimer approximation the hamiltonian is where the sole Laplacian is that for the single electron. R is treated as a parameter so that at any R value the inter-nuclear potential is a constant. where the set of functions yj are the eigenfunctions of the hamiltonian having eigenvalues Ej. Each member of the series is a MOLECULAR ORBITAL that extends over both nuclei.

8. MODULE 10 THE HYDROGEN MOLECULE ION is a one electron species The B-O approximation keeps the inter-nuclear repulsion from gumming things up and so the Schrödinger equation can be solved exactly. However the solutions are not easy to use and the mathematical form gives little insight into the whys and wherefores of the nature of the bonding interaction. It is more useful to go for an approximate solution to build molecular wavefunctions. This provides physical insight into the nature of chemical bonds. It also gives good agreement with experimental observations and this can always be improved and corrected using perturbation theory.

9. MODULE 10 To make a start on solving the Schrödinger equation we use a variation theory approach and employ a trial function. The obvious function to use is a linear combination of hydrogen 1s orbitals, each centered on one of the H atoms in the molecule. The linear combination can be a sum or difference This combination describes a (pair of) molecular orbital(s) formed by a linear combination of atomic orbitals (LCAO method). The figure shows a sketch of y+2 with the coefficients equal.

10. MODULE 10 Note that the MO spreads over both nuclei, as required. The nuclei are identical so we expect the coefficients to be equal, and we can set them equal to unity, for convenience. At some point we shall need to normalize the molecular wavefunction in order to obtain probability densities. We now do a variational calculation on our trial function The usual way to proceed with a variational calculation is to write down the secular determinant, obtain the secular equation, and minimize the energy with respect to chosen parameters

11. MODULE 10 The basis set is composed of two AOs The determinant will be 2x2 and the secular equation will be a quadratic yielding two values of E of which we select the lower. The other root is the energy of the second MO that is formed out of the LCAO procedure. Remember that there is a MO formed for every AO that contributes to the basis set The combination of two atomic orbitals leads to the formation of two molecular orbitals. For the sum combination we have a Schrödinger equn

12. MODULE 10 1s is a real orbital (no imaginary components) hence 1s* = 1s. Also the atomic wave functions are normalized. Thus the first two integrals are equal to unity, and the last two are identical. Hence the denominator becomes

13. MODULE 10 The integral on the RHS is over the product of two identical H atom orbitals, one centered on A and one on B (Figure) [the integral on the RHS is not zero because the two 1s functions are not orthogonal since they are on different atoms] Both wavefunctions decline exponentially from maximum amplitude at the nucleus and so the spatial overlap will be greatest when the nuclei are near one another, i.e. when R is small. The integral is known as the overlap integral,S. S can take values between 0 and 1. After a difficult evaluation, S, a function of R, is found to be

14. MODULE 10 Now for the numerator The 1s functions are normalized and they are equivalent eigenfunctions of the respective operator with identical energy eigenvalues

15. MODULE 10 then Using these in the above brackets leads to

16. MODULE 10 The first two integrals on the RHS of the last equation are identical except for interchange of the indices. We call these Coulomb integrals and represent them by J (or a). Since R is fixed during the integration over r we can express J as The electronic charge in dr is 1s*A1sAdr and the Coulomb interaction energy with a proton at distance rB is 1s*A1sAdr/rB. If we integrate this over all the volume elements in the allotted space and we add the Coulomb repulsion between the two nuclei (= 1/R au), then we have J.

17. MODULE 10 A physical picture for J is that it is the composite of the electron density around nucleus A interacting with the nucleus B, together with the inter-nuclear repulsion. The other Coulomb integral is identical, except with the atoms interchanged. The second and third integrals are also identical since they differ only by the interchange of electron labels. These are the exchange integrals and are represented by K (or b). Thus, using the definition of S from above and the constancy of R during the integration leads to

18. MODULE 10 K has no classical analog, but one physical representation might be that it represents the interaction of one of the nucleons with the overlap electron density existing between the two nuclei. It is intriguing to realize that K arises directly from the fact that our MO is formed from a linear combination of AOs of a pair of atoms. with an equivalent term for the other atom.

19. MODULE 10 the RHS of the last equation represents the energy of the hydrogen molecule ion with respect to a H(1s) atom /proton pair at R = infinity

20. MODULE 10 The J, K, and S integrals can be evaluated and their R-dependence is as follows:

21. MODULE 10 Plots of the first and second terms on the RHS of equation and their sum are shown on Figure. We can see that the Coulomb term is always positive, but the exchange term, which is the difference between two positive quantities, can be positive or negative. The sum of the two (DE+) has negative values at R/a0 >1.5 a.u. And in this region the energy of the coupled system of three particles is lower than the energy of the separated proton and H1s CHEMICAL BOND

22. MODULE 10 Thus, y+is the wavefunction of a bonding molecular orbital that encompasses both the nuclei. From the Figure it is apparent that the minimum in energy (vs.R) occurs at R = 2.5 au (the experimental value is 2.00 au). The dissociation energy at Req is 170 kJ mol-1 (experimental value is 268 kJ mol-1). So we see that our simple calculation (no variational parameters) does a poor job in reproducing the experimental values. Though we lose in accuracy we discover the nature of the covalent bond. The bonding effect arises only because the K integral can take larger negative values then J can positive. The K integral is a purely quantum effect, and so is bonding.

23. MODULE 10 Turning our attention to the second possible combination of a pair of 1s atomic orbitals we use -Plots of DE+ and DE- vs. R0 are shown in Figure. The difference function never gives negative energy solutions, and y-is the wavefunction of an antibonding MO.

24. MODULE 10 In our evaluation of the two energy quantities, although variation theory has been used no variational parameters have been employed, nor have we used the secular determinant approach. Had we done so we would have arrived at the same result. Normalization gives Node Zero probability

25. MODULE 10 The two orbitals we have calculated describe the ground and excited states of the hydrogen molecule-ion. Some terminology for MOs. The bonding MO that we formed above (y+) was builtby thelinear combination of a pair of spherically symmetrical (l = 0) 1s orbitals. As a result the electronic charge accumulates between the atoms in a cylindrical symmetry. The resulting orbital is termed 1s, where the s tells about the OAM around the inter-nuclear axis, zero for s orbitals. The antibonding orbital is designated 2s. Many times you will also find antibonding orbitals with an asterisk, e.g. 2s*.

26. MODULE 10 Note that the 2s orbital is more antibonding than the 1s is bonding. This can also be seen in a previous Figure by comparing the two curves in the region of the minimum of R. This asymmetry is caused by the repulsion between the nuclei. Another item is the classification of orbitals by their parity. This is the behavior of the symmetry properties under the parity operator (inversion of the electron coordinates).

27. MODULE 10 The symmetric combination of the two AOs generates the egg-shaped 1s where the whole of the orbital is positive. Starting  from the molecular center and drawing two lines of equal length in exactly opposite directions will lead to exactly identical intensities and signs. This corresponds to a symmetrical inversion and the orbital is given the gerade (g) classification. This symbol is written as a subscript to the s. If the sign of the molecular wave function changes under this symmetry operation, the orbital is ungerade (u)

28. MODULE 10 In our evaluation of the structure of atoms we saw that for the H atom, for example, an infinite set of allowed orbitals and associated energy states exists, and yet for the simplest molecule we have found only two. The reason is in our choice of basis set. In the procedure we employed only two AOs (the H1s on the two atoms) in our linear combination; this forced the two MO result. This was solely in the interests of simplification. We could just as well used a LC of six atomic orbitals such as This would lead to six MOs and six associated energies. This could be extended and would result in improvement in the energy and the wavefunction estimates. All this is within the Born-Oppenheimer Approximation.

29. MODULE 10