In the Classroom
Tetrahedral Geometry and the Dipole Moment of Molecules Sara N. Mendiara* and Luis J. Perissinotti† Departamento de Química, Facultad de Cs. Exactas y Naturales, Universidad Nacional de Mar del Plata, Funes 3350, Mar del Plata 7600, Argentina; *
[email protected] Many publications that report the determination of the tetrahedral bond angle have been recorded in this Journal. Dore considered the geometry of the tetrahedron in relation to that of a surrounding cube (1); Kawa considered a similar approach but he used the law of cosines (2). Brittin (3), Snatzke (4), Duffey (5) and Sutcliffe (6 ) employed vector algebra. Gombert (7), Cockburn (8) and Apak (9) used analytical geometry. Glaister (10) used spherical polar coordinates and the dot product, and later on he employed trigonometry (11). Woolf used trigonometry and a close-packing approach (12). Some of the demonstrations are rather complex and not appropriate for students who attend the first courses. Some students cannot understand these concepts and they can only learn a qualitative estimation. Our approach is similar to McCullough’s (13)—more complete, and we have made it very suitable. The following presentation is based on the knowledge of the permanent dipole moment of molecules. We usually teach the tetrahedral and pyramidal geometry of simple molecules following a short derivation. We use geometry, projections, trigonometry, and the bond and dipole moments. The method of calculating the dipole moment of a molecule is based on an idea suggested by Thomson in 1923 (14). The dipole moment of a molecule is the vector sum of the individual bond moments. Measurements of dipole moments at one time served as a principal source of information about molecular structure; the dielectric constant and the refractive index were determined and µ, dipole moment was calculated. Later on, bond lengths and angles were measured by various spectroscopic and diffraction methods. Several computer programs have been published for conversion of those bond lengths and bond angles to atomic coordinates (15). We think that the approach presented below, with simple molecules, is most appropriate to extract bond angles and moments, before working out bigger molecules. Moreover, students may investigate dipole moment measurement (16 ). Classroom Methodology in Our Organic Chemistry Course
Part I: Calculation of the Tetrahedral Angle We begin drawing an A4C molecule, (methane, carbon tetrachloride, neopentane, etc.). Atoms or groups, A, are arranged symmetrically; each A is equidistant from the carbon atom and any pair of A’s forms an equal angle with the vertex at C (Fig. 1a). We can appreciate that ∠ACA is the tetrahedral angle. Methane, carbon tetrachloride, and neopentane have zero dipole moments, µ = 0 D. We would expect that the individual bond dipole moments are equal. For the purpose of the calculations, the molecule may be placed on a Cartesian coordinate system and the orientation of the molecule is that †
64
Also CIC/Provincia de Buenos Aires, Argentina.
shown in Figure 1b. The z axis is directed along one CA bond, and the xy plane is defined by the other three A’s with the origin at O; m represents the bond dipole moment or bond moment, in debye units, that belongs to CA bond, and α is the angle between CA bond and CO segment. We have avoided pointing out the bond moment’s direction, in order to make a general treatment. The direction depends on atom’s electronegativity and teachers may assign direction depending on their textbook’s convention or they may follow our suggestion. We can observe in Figure 1c that the bond moments are resolved into mz and mxy, which are the projections in the z axis and the xy plane, respectively. First, let’s work with the projection in the xy plane, mxy. We are going to show that a resultant component in the xy plane does not exist. Figure 1d illustrates three projections mxy that are equidistant in the xy plane, with ∠AOA = 120°. Along the y axis we have the contribution from one CA, mxy, and the contribution from the other two CA projections, 2my. These components are added and the resultant is calculated: mxy = m sin α my = mxy cos 60° my(resultant) = mxy – 2my my(resultant) = m sin α – 2m sin α cos 60° = m sin α(1 – 2 cos 60°) ⇒ my(resultant) = 0 Along the x axis we only have the contributions from two CA projections, which are equal in magnitude and in opposite direction. mx = mxy sin 60°; mx(resultant) = mx – mx = 0 Therefore we have verified that there is not any contribution to the dipole moment in the plane xy. Now, consider the bond dipole moment projection in the z axis, mz. It may be obtained from the cosine of the angle between the component desired and the direction of the bond moment. ∠COA is the angle between the z axis and the xy plane; it is 90° (Figs. 1b and 1c). mz = m cos α Along the z axis we have m from one CA bond and three mz from the other CA bonds, which are in the opposite direction, so we subtract 3mz from m in eq 1. As µ = 0 the resultant must be zero because there is no resultant in the xy plane. µ = mz(resultant) = m – 3mz
(1)
µ = mz(resultant) = m – 3m cos α = m(1 – 3 cos α)
As µ = 0 D, ⇒ mz(resultant) = 0 ⇒ 1 – 3 cos α = 0 ⇒ cos α = 1⁄3; α = 70.53° and tetrahedral angle = ∠ACA = 180° – 70.53° = 109.47°.
Journal of Chemical Education • Vol. 79 No. 1 January 2002 • JChemEd.chem.wisc.edu
In the Classroom
A
Part II: Calculation of Bond Moments
A
(a)
(b)
Case 1 When we observe other molecules we find other characteristic angles. For example, in the ammonia molecule (NH3), there is only one type of bond moment, a characteristic angle α and the bond angle ∠HNH. Figure 2a shows an ammonia molecule scheme. We observe that along the z axis there are only three mz contributions, all in the same direction.
z axis
y axis C A
C α O
x axis
A
A
A A
mz = m cos α
A
So, instead of eq 1 for mz(resultant), in this case we have eq 2. (c)
µ = mz(resultant) = 3mz
(d) A
z axis
mz α
mxy x axis
m x axis
O
The value for the dipole moment, µNH = 1.47 D, was obtained from the literature cited shown in Table 1. Now, in eq 3, the bond moment m and the angle α are unknown quantities. In order to know m we must first calculate α (eq 4). 3
y axis
C
60° A
mxy
mxy
my A
mx
A
y axis
(b)
z axis
N
N
m
H
α O
µNH 3 = 3 m cos α = 1.47 D
(3)
m = mH–N = 1.47/(3 cos α)
(4)
From spectroscopic techniques we know that ∠HNH is equal to 107.3°; therefore we can calculate ∠ONH, α, as shown in Figure 2. We see that ∠NOH = 90° and it is the angle between the z axis and the xy plane. We are going to solve for HH(segment), OH(segment), and α. First, we solve for HH(segment) (see Fig. 2 b):
Figure 1. A4C type molecule with µ = 0.
(a)
(2)
HH(segment) = 2NH(segment) sin (∠HNH/2)
107.3°
H
HH(segment) = 2NH(segment) sin (107.3/2)
H
(5)
H
HH/2
Second, we solve for OH(segment) (see Fig. 2c):
H
OH(segment) = (c)
H
(d)
H
60°
HH/2
Last, we solve for α (see Fig. 2d):
m
sin α = OH(segment)/NH(segment)
O
H
H
Molecule H 3N
H3CCl
µ/D
Ref
1.471 1.47 1.47 1.46
18, p 9.44 18, p E-58 17, p 5.128 19
1.892 1.87 1.87 1.86
18, p 9.45 18, p E-59 17, p 5.97 19
We may now calculate the H–N bond moment replacing α in eq 4:
Table 2. Bond Dipole Moments Bond
Bond Ref Moment/D
H–N
1.31 1.31
17 19
H–C
0.3 0.4
17 19
C–Cl
1.87a 1.46 1.56 1.5 1.5
17 19 20 21 22
aAliphatic.
(7)
sin α = sin 53.65°/cos 30° ⇒ α = 68.44°
Figure 2. H3N, ammonia molecule, ∠HNH =107.3°.
Table 1. Electric Dipole Moments in the Gas Phase
(6)
OH(segment) = NH(segment) sin 53.65°/cos 30°
N
mz α
O
HH(segment)/2 cos 30°
m = mH–N = 1.47/(3 cos 68.44°) = 1.33 D The result is quite near the tabulated one (Table 2). Students may complete the Cartesian coordinates as in Figure 1. Case 2 We suggest that our students look for the chloromethane (H3CCl) dipole moment, HC bond moment, and ∠HCH. Then they may first calculate ∠ClCH and later the CCl bond moment. We have tabulated some values from handbooks and textbooks in Tables 1 and 2 (17 ). We shall make use, in our calculations, of the textbook quantities for dipole and bond moments. For example, for chloromethane textbooks
JChemEd.chem.wisc.edu • Vol. 79 No. 1 January 2002 • Journal of Chemical Education
65
In the Classroom
give a dipole moment of 1.86 D and an HC bond dipole of 0.4 D; CRC tables give an ∠HCH of 110°. As in ammonia calculations, from eqs 5–7, we now have α = 71.06°. We may also calculate ∠ClCH = 180° – α = 108.94°. In this case, we have along the z axis mCCl from the CCl bond and three mz from the HC bonds. The chlorine is more electronegative than carbon and the carbon is more electronegative than hydrogen, and now mCCl and the three mz are all in the same direction. So, the mz(resultant) equation is similar to eq 1 but in this case we have to add 3mz.
3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
mz = mHC cos α
16.
µH3CCl = 1.86 D; ∠HCH = 110°; mHC = 0.4 D
From eq 1, µ = mz(resultant) = mCCl + 3mHC cos α = 1.86 D: mCCl = 1.5 D The C–Cl bond moment mCCl calculated is quite similar to the values shown in Table 2. Lange’s handbook gives 1.87 D, but this value corresponds to an aliphatic C–Cl bond moment, where a C–C bond moment group environment and not an H–C one has been considered. Moreover, we have to take into account that the bond moment assigned to H–C bond is a supposed value. So, the tabulated bond moments may be different in the future (23). In the development of the present work students are active learners. They follow an interactive approach between experimental values and calculations. They also gain an appreciation for the importance of geometrical structure. We hope that the methodology described here is didactic, amusing, and capable of training the students for the analysis of more difficult situations.
17. 18.
19.
20. 21. 22.
Literature Cited 1. Dore, W. H. J. Chem. Educ. 1942, 19, 29–30. 2. Kawa, C. J. J. Chem. Educ. 1988, 65, 884–885.
66
23.
Brittin, W. E. J. Chem. Educ. 1945, 22, 145. Snatzke, G. J. Chem. Educ. 1963, 40, 94. Duffey, G. H. J. Chem. Educ. 1990, 67, 35–36. Sutcliffe, B. T.; Smith, S. J. J. Chem. Educ. 1992, 69, 171. Gombert, G. L. J. Chem. Educ. 1941, 18, 336–337. Cockburn, B. L. J. Chem. Educ. 1963, 40, 94. Apak, R.; Tor, I. J. Chem. Educ. 1991, 68, 970. Glaister, P. J. Chem. Educ. 1993, 70, 546–547. Glaister, P. J. Chem. Educ. 1997, 74, 1086. Woolf, A. A. J. Chem. Educ. 1995, 72, 19–20. McCullough, T. J. Chem. Educ. 1962, 39, 476. Thomson, J. J. Philos. Mag. 1923, 46, 513. McEachern, D. M.; Lehmann, P. A. J. Chem. Educ. 1970, 47, 389–390 and references there in. Rapp, R. D.; Sturm, J. E. J. Chem. Educ. 1969, 45, 851–853. Coe, D. A.; Nibler, J. W. J. Chem. Educ. 1973, 50, 82–84. Baron. M.; Arevalo, E. S. J. Chem. Educ. 1988, 65, 644–645. Lange’s Handbook of Chemistry, 4th ed.; Dean, J. A., Ed.; McGraw-Hill: New York, 1992. CRC Handbook of Chemistry and Physics, 73rd ed.; Lide, D. R., Ed.; CRC Press: London, 1992–1993; CRC Handbook of Chemistry and Physics, 67th ed.; Lide, D. R., Ed.; CRC Press: London, 1986–1987. Organic Chemistry Textbooks: Fessenden, R.; Fessenden, J. Organic Chemistry, 2nd ed.; PWS: Boston, 1982; p 23. Morrison, R.; Boyd, R. Organic Chemistry, 6th ed.; PrenticeHall: Englewood Cliffs, NJ, 1992; p 24. Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry. Part A: Structure and Mechanisms, 3rd ed.; Plenum: New York, 1990; p 16. Adamson, A. W. A Textbook of Physical Chemistry; Academic: New York, 1979; Chapter 3. Levine, I. N. Physical Chemistry, 4th ed.; McGraw-Hill: New York, 1995; Chapter 20. Glasstone, S. Textbook of Physical Chemistry; Van Nostrand: New York, 1949; Chapter 8. Wiberg, K. B.; Wendoloski, J. J. J. Phys. Chem. 1984, 88, 586–593. Wiberg, K. B.; Hadad, C. M.; Breneman, C. M.; Laidig, K. E.; Murcko, M. A.; LePage T. J. Science 1991, 252, 1266–1272.
Journal of Chemical Education • Vol. 79 No. 1 January 2002 • JChemEd.chem.wisc.edu
