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64 Journal of Chemical Education Vol. 79 No. 1 January 2002 JChemEd.chem.wisc.edu

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 relationto that of a surrounding cube (1); Kawa considered a similarapproach but he used the law of cosines (2). Brittin (3),Snatzke (4), Duffey(5) and Sutcliffe(6) employed vectoralgebra. Gombert (7), Cockburn (8) and Apak (9) usedanalytical geometry. Glaister(10) used spherical polar coor-dinates and the dot product, and later on he employed trigo-nometry(11). Woolf used trigonometry and a close-packingapproach (12). Some of the demonstrations are rather complexand not appropriate for students who attend the first courses.Some students cannot understand these concepts and theycan only learn a qualitative estimation. Our approach is similar

to McCulloughs

(13)more complete, and we have madeit very suitable.

The following presentation is based on the knowledge ofthe permanent dipole moment of molecules. We usually teachthe tetrahedral and pyramidal geometry of simple moleculesfollowing a short derivation. We use geometry, projections,trigonometry, and the bond and dipole moments.

The method of calculating the dipole moment of a mol-ecule is based on an idea suggested by Thomson in 1923 (14).The dipole moment of a molecule is the vector sum of theindividual bond moments. Measurements of dipole momentsat one time served as a principal source of information aboutmolecular structure; the dielectric constant and the refractive

index were determined and , dipole moment was calculated.Later on, bond lengths and angles were measured by variousspectroscopic and diffraction methods. Several computer pro-grams have been published for conversion of those bondlengths and bond angles to atomic coordinates (15). We thinkthat the approach presented below, with simple molecules,is most appropriate to extract bond angles and moments,before working out bigger molecules. Moreover, students mayinvestigate dipole moment measurement (16).

Classroom Methodology in Our Organic ChemistryCourse

Part I: Calculation of the Tetrahedral Angle

We begin drawing an A4C molecule, (methane, carbontetrachloride, neopentane, etc.). Atoms or groups, A, arearranged symmetrically; each A is equidistant from the carbonatom and any pair of As forms an equal angle with the vertex atC (Fig. 1a). We can appreciate that ACA is the tetrahedralangle. Methane, carbon tetrachloride, and neopentane havezero dipole moments, = 0 D. We would expect that theindividual bond dipole moments are equal. For the purpose ofthe calculations, the molecule may be placed on a Cartesiancoordinate system and the orientation of the molecule is that

shown in Figure 1b. The zaxis is directed along one CA bond,

and the xy plane is defined by the other three As with theorigin at O; m represents the bond dipole moment or bondmoment, in debye units, that belongs to CA bond, and isthe angle between CA bond and CO segment.

We have avoided pointing out the bond moments di-rection, in order to make a general treatment. The directiondepends on atoms electronegativity and teachers may assigndirection depending on their textbooks convention or theymay follow our suggestion.

We can observe in Figure 1c that the bond moments areresolved into mz and mxy, which are the projections in the zaxis and thexyplane, respectively.

First, lets work with the projection in the xyplane, mxy.

We are going to show that a resultant component in thexyplanedoes not exist. Figure 1d illustrates three projections mxy thatare equidistant in the xyplane, with AOA = 120. Alongthe yaxis we have the contribution from one CA, mxy, andthe contribution from the other two CA projections, 2my.These components are added and the resultant is calculated:

mxy= m sin

my= mxycos 60

my(resultant)= mxy 2my

my(resultant)=msin2msincos60=m sin(12cos 60)

my(resultant) = 0

Along thexaxis we only have the contributions from two CAprojections, which are equal in magnitude and in oppositedirection.

mx= mxy sin 60; mx(resultant) = mx mx= 0

Therefore we have verified that there is not any contributionto the dipole moment in the planexy.

Now, consider the bond dipole moment projection in thezaxis, mz. It may be obtained from the cosine of the angle be-tween the component desired and the direction of the bondmoment. COA is the angle between the zaxis and thexyplane; it is 90 (Figs. 1b and 1c).

mz= m cos

Along the zaxis we have m from one CA bond and three mzfrom the other CA bonds, which are in the opposite direction,so we subtract 3mz from m in eq 1. As = 0 the resultantmust be zero because there is no resultant in thexyplane.

= mz(resultant) = m 3mz (1)

= mz(resultant) = m 3m cos = m(1 3 cos )

As =0D,mz(resultant)=013cos=0cos=13;

= 70.53 and tetrahedral angle = ACA = 180 70.53 = 109.47.

Tetrahedral Geometry and the Dipole Moment of Molecules

Sara N. Mendiara* and Luis J. Perissinotti

Departamento de Qumica, Facultad de Cs. Exactas y Naturales, Universidad Nacional de Mar del Plata, Funes 3350,Mar del Plata 7600, Argentina; *792947@copetel.com.ar

Also CIC/Provincia de Buenos Aires, Argentina.

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JChemEd.chem.wisc.edu Vol. 79 No. 1 January 2002 Journal of Chemical Education 65

Part II: Calculation of Bond Moments

Case 1

When we observe other molecules we find other charac-teristic angles. For example, in the ammonia molecule (NH3),there is only one type of bond moment, a characteristic angle and the bond angle HNH. Figure 2a shows an ammoniamolecule scheme. We observe that along the zaxis there are

only three mzcontributions, all in the same direction.

mz= m cos

So, instead of eq 1 for mz(resultant), in this case we have eq 2.

= mz(resultant) = 3mz (2)

The value for the dipole moment, NH3 = 1.47 D, wasobtained from the literature cited shown in Table 1. Now, ineq 3, the bond moment m and the angle are unknownquantities. In order to knowm we must first calculate (eq 4).

NH3 = 3m cos = 1.47 D (3)

m = mHN = 1.47/(3 cos ) (4)

From spectroscopic techniques we know that HNH is equalto 107.3; therefore we can calculate ONH, , as shownin Figure 2. We see that NOH = 90 and it is the anglebetween the zaxis and thexyplane.

We are going to solve for HH(segment), OH(segment),and . First, we solve for HH(segment) (see Fig. 2 b):

HH(segment) = 2NH(segment) sin (HNH/2)

HH(segment) = 2NH(segment) sin (107.3/2) (5)

Second, we solve for OH(segment) (see Fig. 2c):

OH(segment) =HH(segment)/2

cos 30(6)

OH(segment) = NH(segment) sin 53.65/cos 30

Last, we solve for (see Fig. 2d):

sin = OH(segment)/NH(segment) (7)

sin = sin 53.65/cos 30 = 68.44

We may now calculate the HN bond moment replac-ing in eq 4:

m = mHN = 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 inFigure 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 bondmoment. We have tabulated some values from handbooksand textbooks in Tables 1 and 2 (17). We shall make use, inour calculations, of the textbook quantities for dipole andbond moments. For example, for chloromethane textbooks

elopiDcirtcelE.1elbaTehtnistnemoM saG esahP

eluceloM D/ feR

H3N 174.174.174.164.1

81 44.9p,81 85-Ep,71 821.5p,91

H3 lCC 298.178.178.168.1

81 54.9p,81 95-Ep,71 79.5p,91

elopiDdnoB.2elbaTstnemoM

dnoBdnoB

D/tnemoMfeR

NH 13.113.1

71

91

CH 3.04.0

71

91

lCC 78.1 a

64.165.1

5.15.1

71

91

02

12

22

aAliphatic.

Figure 1. A4C type molecule with = 0.

xaxis

yaxis

zaxis

O

A

A

A

C

A

xaxis

xaxis

yaxis

yaxis

zaxis

O

60

A

A

AA

C

A

A

A

C

(a) (b)

(c) (d)

A

mz

mx

my

m

mxy

mxy

mxy

Figure 2. H3N, ammonia molecule, HNH =107.3.

z axis

O

H

H

N

N

H

H

O

H

H

H

N

H

mz

m

m

60

107.3

HH/2

HH/2

O

H

(a) (b)

(c) (d)

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give a dipole moment of 1.86 D and an HC bond dipole of0.4 D; CRC tables give an HCH of 110.

H3CCl = 1.86 D; HCH = 110; mHC = 0.4 D

As in ammonia calculations, from eqs 57, we now have =71.06.

We may also calculate ClCH = 180 = 108.94.In this case, we have along the zaxis mCCl from the CCl bond

and three mz from the HC bonds. The chlorine is moreelectronegative than carbon and the carbon is more electro-negative than hydrogen, and nowmCCl and the three mzareall in the same direction. So, the mz(resultant) equation issimilar to eq 1 but in this case we have to add 3mz.

mz= mHC cos

From eq 1, = mz(resultant) = mCCl + 3mHC cos = 1.86 D:

mCCl = 1.5 D

The CCl bond moment mCCl calculated is quite similar tothe values shown in Table 2. Langes handbook gives 1.87 D,but this value corresponds to an aliphatic CCl bond mo-

ment, where a CC bond moment group environment andnot an HC one has been considered. Moreover, we have totake into account that the bond moment assigned to HCbond is a supposed value. So, the tabulated bond momentsmay be different in the future (23).

In the development of the present work students are activelearners. They follow an interactive approach between experi-mental values and calculations. They also gain an appreciationfor the importance of geometrical structure. We hope thatthe methodology described here is didactic, amusing, andcapable of training the students for the analysis of moredifficult situations.

Literature Cited1. Dore, W. H.J. Chem. Educ. 1942,19, 2930.2. Kawa, C. J.J. Chem. Educ.1988,65, 884885.

3. Brittin, W. E.J. Chem. Educ. 1945,22, 145.4. Snatzke, G.J. Chem. Educ. 1963,40, 94.5. Duffey, G. H.J. Chem. Educ.1990,67, 3536.6. Sutcliffe, B. T.; Smith, S. J.J. Chem. Educ.1992,69, 171.7. Gombert, G. L.J. Chem. Educ. 1941,18, 336337.8. Cockburn, B. L.J. Chem. Educ. 1963,40, 94.9. Apak, R.; Tor, I.J. Chem. Educ. 1991,68, 970.

10. Glaister, P.J. Chem. Educ. 1993,70, 546547.

11. Glaister, P.J. Chem. Educ. 1997,74, 1086.12. Woolf, A. A.J. Chem. Educ. 1995, 72, 1920.13. McCullough, T.J. Chem. Educ. 1962,39, 476.14. Thomson, J. J. Philos. Mag. 1923, 46, 513.15. McEachern, D. M.; Lehmann, P. A.J. Chem. Educ.1970,47,

389390 and references there in.16. Rapp, R. D.; Sturm, J. E.J. Chem. Educ. 1969, 45, 851853.

Coe, D. A.; Nibler, J. W. J. Chem. Educ. 1973,50, 8284.Baron. M.; Arevalo, E. S.J. Chem. Educ. 1988, 65, 644645.

17. Langes Handbook of Chemistry, 4th ed.; Dean, J. A., Ed.;McGraw-Hill: New York, 1992.

18. CRC Handbook of Chemistry and Physics, 73rd ed.; Lide, D. R.,Ed.; CRC Press: London, 19921993; CRC Handbook of

Chemistry and Physics, 67thed.; Lide, D. R., Ed.; CRC Press:

London, 19861987.19. Organic Chemistry Textbooks: Fessenden, R.; Fessenden, J.

Organic Chemistry, 2nd ed.; PWS: Boston, 1982; p 23.Morrison, R.; Boyd, R. Organic Chemistry, 6th ed.; Prentice-Hall: Englewood Cliffs, NJ, 1992; p 24. Carey, F. A.;Sundberg, R. J.Advanced Organic Chemistry. Part A: Structureand Mechanisms, 3rd ed.; Plenum: New York, 1990; p 16.

20. Adamson, A. W.A Textbook of Physical Chemistry; Academic:New York, 1979; Chapter 3.

21. Levine, I. N. Physical Chemistry, 4th ed.; McGraw-Hill: NewYork, 1995; Chapter 20.

22. Glasstone, S. Textbook of Physical Chemistry; Van Nostrand:

New York, 1949; Chapter 8.23. Wiberg,K.B.; Wendoloski,J.J.J.Phys.Chem.1984,88,586593.Wiberg, K. B.; Hadad, C. M.; Breneman, C. M.; Laidig, K. E.;Murcko, M. A.; LePage T. J. Science1991,252, 12661272.

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