Research: Science & Education
A Simple Molecular Orbital Treatment of the Barrier to Internal Rotation in the Ethane Molecule Derek W. Smith Department of Chemistry, University of Waikato, Hamilton, New Zealand
Recognition of the fact that rotation about the carbon– carbon bond in the ethane molecule is not “free”, but involves an energy barrier of 12.1 kJ mol᎑1, is a necessary preliminary to the study of conformational analysis (1). The origin of this energy barrier has aroused much discussion (1–3) and there appears to be a consensus that it is not caused simply by greater steric repulsion between vicinal hydrogen atoms in the eclipsed form relative to the staggered form. As summarized in ref 3, calculations performed by Pitzer and coworkers in the 1960s, which were substantially corroborated by other groups in the 1970s, attribute the barrier to overlap repulsion. The Pauli principle requires that electrons of the same spin occupy orthogonal (i.e., nonoverlapping) orbitals. Thinking in terms of localized bond orbitals, the need to maintain orthogonality of vicinal C–H bond orbitals as the molecule is subjected to internal rotation alters the wave function, and the calculated difference in energy between the eclipsed and staggered forms is in good agreement with the experimental value. Teachers and students who lack a firm grounding in quantum chemistry may find this explanation rather difficult to grasp and may remain unconvinced that the staggered form should be more stable. The purpose of this paper is to formulate the overlap repulsion effect in the ethane molecule in terms of simple molecular orbital theory, with a minimum of mathematics, in a manner that should be comprehensible to most undergraduate students. The explanations given in detail here are implicit in the qualitative arguments— supported by semiempirical MO calculations—previously set out by Lowe (4–6 ). The Antibonding Effect The simplest problem in molecular orbital (MO) theory involves the case of two atomic orbitals (AOs) φA and φ B centered on different nuclei A and B. If these are nonorthogonal—that is, if the overlap integral SAB defined by eq 1 S AB = ∫ φ A φ B d τ
(1)
is non-zero, we must construct a pair of linear combinations ψ 1 and ψ 2, such that ∫ ψ 12 d τ = ∫ ψ 22d τ = 1
(2)
∫ ψ1ψ2d τ = 0
(3)
H AA – E
H AB – ES AB =0
H AB – ES AB
(6)
H BB – E
where H AA = ∫ ψ A H ψ A d τ HBB = ∫ ψ B H ψ B d τ HAB = ∫ ψ A H ψ B d τ The Hamiltonian operator H need not be specified in the simplest treatments. Expanding eq 6 as a quadratic, normalized to the form E 2 – aE + b = 0 the sum of the roots (E 1 + E2) is equal to a. In the case where φ A and φ B are equivalent, so that HAA = H BB = E 0, it is easy to show that (E2 – E 0) – (E0 – E 1) = ᎑2HABSAB/(1 – SAB2)
(7)
The right-hand side of eq 7 must be positive, because (i) HAB is a negative energy quantity (representing, in simplistic terms, the enhanced electron–nucleus attraction in the overlap region); (ii) the phases of φA and φB have been chosen so that SAB is positive (otherwise, ψ 2 would be bonding and ψ 1 antibonding); and (iii) (1 – SAB2) must be positive because SAB cannot exceed unity. It follows then from eq 7 that the destabilization of an antibonding MO is always greater than the stabilization of the corresponding bonding MO, relative to the uncombined AOs. This antibonding effect (shown in Fig. 1) disappears if the overlap integral is zero in eqs 4–6, an approximation often invoked in elementary texts. The result of filling the antibonding MO is not merely to cancel the bonding arising from occupancy of the bonding MO: the net effect is destabilization. Thus in the He2 molecule, quantitative calculations lead to net repulsion between the atoms at an internuclear distance as great as 2.5 Å (7 ) (cf the equilibrium distance of 0.74 Å in the H2 molecule). This is usually attributed to repulsive interactions attending the enforcement
where eqs 2 and 3 represent the requirements of normalization and orthogonality, respectively. In the simplest case where φ A and φB are equivalent, so that the coefficients of φ1 and φ2 must be equal in magnitude, we obtain: ψ1 = [2(1 + SAB)]᎑1/2(φA + φB )
(4)
ψ 2 = [2(1 – S AB)]᎑1/2(φA – φB)
(5)
The energies E 1 and E 2 of ψ1 and ψ 2, respectively, are equal to the roots of a determinantal equation (eq 6):
Figure 1. Relative energies of two equivalent AOs and the resultant MOs, illustrating the antibonding effect.
JChemEd.chem.wisc.edu • Vol. 75 No. 7 July 1998 • Journal of Chemical Education
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Research: Science and Education
of the Pauli principle as the 1s orbitals begin to overlap appreciably, and Sovers et al. assert that “…the dominant term in the ethane barrier can be considered analogous to the closed-shell repulsion between a pair of helium atoms” (2). Thus at the level of simplification being attempted in the present paper, the ethane problem can be addressed in terms of the antibonding effect. Lowe (4–6 ) takes an alternative approach to the antibonding effect. Following Coulson (8), he notes that, in enforcement of the requirements of normalization and orthogonality, the nodal surface in an antibonding MO requires that the coefficients of the AOs be greater than those in the corresponding bonding MO. This leads to a negative bond order where both MOs are filled. The present paper focuses attention on orbital energies rather than on coefficients and bond orders, but the two approaches are ultimately equivalent. MO Treatment of the Ethane Molecule (4–6, 9) The simplest approach is to deal first with a pyramidal (C3v) CH3 radical, having the same geometry as in ethane, and then to combine the MOs of two methyls to give a description of the bonding in ethane. We use only the valence AOs 1s(H), 2s(C) and 2p(C). In CH3, the hydrogens are numbered according to Figure 2; the z axis coincides with the threefold rotation axis. The group orbitals (or symmetry orbitals), that is, normalized and orthogonal linear combinations of 1s(H) orbitals belonging to irreducible representations of the C3v point group, are given by eqs 8–10. Φ(a1) = [3(1 + 2S g)]᎑1/2( φ1 + φ 2 + φ 3)
(8)
Φ(e y) = [6(1 – S g)]᎑1/2(2φ1 – φ 2 – φ 3)
(9)
Φ(ex) = [2(1 – Sg)]
( φ2 – φ3)
-1/2
make only a small contribution. 3a1 is strongly antibonding, while 1e and 2e are respectively bonding and antibonding pairs. The ground state of the radical arises from the configuration (1a1)2(1e)4(2a1). We now consider how these methyl MOs are combined in an ethane molecule having the staggered form (D3d). The methyl a1 MOs can be combined in-phase or out-of-phase, to give, respectively, ethane MOs a1g and a2u , while the methyl 1e MOs lead to ethane MOs 1e g and 1eu, as shown in Figure 3. According to photoelectron spectra (9), the eg (C–C πantibonding) orbitals are higher in energy than the C–C bonding a1g, but the diagram is a little clearer with the ordering reversed. Thus the ground state of staggered ethane arises from the configuration (1a1g)2(1a 2u) 2(1eu) 4(1eg) 4(2a1g)2. A similar diagram is obtained for the eclipsed form (D3h), subject to translation of the symmetry labels: a1g → a1′, a2u → a2′′, eg → e′′ and eu → e′. In either conformation, it is apparent that the overlap of the methyl 2a1 MOs is largely responsible for the stability of the ethane molecule. It should also be apparent that the only orbital overlaps that are affected by rotation about the C–C bond are those between vicinal 1s(H) orbitals, and it is in these, then, that we must seek the reasons for the energy barrier. This focuses our attention on the e MOs, because of all the occupied MOs these are expected to have the largest contribution from 1s(H). Here we expect a significant antibonding effect: how will it be affected by rotation about the C–C bond? The extent of the antibonding effect (eq 7) is proportional to H ABSAB/(1 – SAB2), where SAB is the overlap integral between the methyl e MOs, labeling the carbon atoms A and B. In semiempirical treatments, for
(10)
where the 1s(H) AOs are labeled φ i and Sg is the overlap integral between the 1s orbitals for a pair of geminal H atoms. Φ(a1) can overlap with both 2s(C) and 2pz(C), while the degenerate group orbitals Φ(ex) and Φ(ey) overlap with 2px(C) and 2py(C), respectively. We thus obtain three MOs of a1 symmetry, labeled 1a1, 2a1 and 3a1 in increasing order of energy, and two pairs of degenerate e MOs, 1e and 2e. 1a1 is a bonding MO, mostly 2s(C) in character. 2a1 is essentially nonbonding and can be approximated as a hybrid of 2s(C) and 2pz(C) with most of its electron density directed along the z axis, away from the H atoms whose 1s orbitals
Figure 2. Numbering of hydrogen atoms and coordinate axes in methyl.
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Figure 3. MOs of ethane, constructed from MOs of two methyl groups.
Journal of Chemical Education • Vol. 75 No. 7 July 1998 • JChemEd.chem.wisc.edu
Research: Science & Education
example in extended Hückel calculations (10), H AB is often deemed to be proportional to SAB, so that if SAB2
