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A Comment on Molecular Geometry Frank J. Gomba Department of Chemistry, U. S. Naval Academy, Annapolis, MD 21402-5026;
[email protected] In browsing through several currently published general chemistry texts (1) one will find that the “directions” for determining the proper molecular geometry for a simple molecule or ion always start with a statement such as “first draw the Lewis structure”. In this commentary, a proposal is made that eliminates this usual “first step”. The Lewis structure, in fact, can be left for the very last step, after the correct molecular geometry has been determined. The proposed method does assume that one has familiarity with Lewis (2, 3) structures and the valence shell electron pair repulsion (VSEPR) approach to bonding as given by Gillespie (4, 5). With the proposed sequence given in this paper, one may determine the correct geometry for just about any molecule or ion with one central atom. This includes any of the traditional types having integral numbers of electron pairs usually identified by “generic” formulas such as AX2, AX3, AX2E, AX4, AX3E, AX2E2, AX5, AX4E, AX3E2, AX3E2, AX6, AX5E, and AX4E2 (6 ), where A indicates the central atom, X stands for the ligand atoms (atoms bonded to the central atom), and E means an unshared pair of electrons. The subscripts have the usual meaning. Discussion of the Method If one draws the Lewis structure for monochloromethane (methyl chloride), CH3Cl, we note that there are (not counting the bonding electron pairs) no electrons “about each of the hydrogen atoms” and six electrons “about the chlorine atom”. Thus, we may make a generalization that not counting the bonding electrons (to the central atom), there are six electrons about the ligand atom. If the ligand atom is hydrogen, then, not counting the bonding electrons to the central atom, there are no electrons about the hydrogen atoms: Cl H
C
H
H
Now, we must first determine the central atom (CA) in our simple molecule or ion. We usually pick the atom written just once in the formula, or if we have molecules such as carbon monoxide, CO (or others with more than one atom written just once), we will go with the more electropositive of the two atoms; here, it is carbon. From the group position of the CA in the periodic table, the number of valence electrons is easily assigned to the CA. Now, if hydrogen is a ligand atom (LA), we will always “donate” the one electron to the CA. Other LAs will “adjust” their usual valence number of electrons to the total of six electrons as required. For example, halogen atoms acting as LAs will also donate one electron to the CA. The atoms from group 6, having six electrons, will neither donate electrons to nor accept them from the CA. Group 5 atoms will have to accept one electron; group 4 atoms will take two, and group 3 atoms must come up with three electrons, all of these at the “expense” of the CA. The total 1732
number of electrons is now determined as the algebraic sum. If the atomic species is a cation, one electron must be subtracted for each positive charge. For anions, one electron must be added for each negative charge. The sum is then divided by two to get the number of electron pairs about the CA. The number of electron pairs about the CA will now give us the necessary electron pair arrangement (EPA) as determined by VSEPR. That is, two electron pairs equates to a linear EPA; three electron pairs, to a trigonal planar arrangement; four electrons pairs means a tetrahedral EPA; five, a trigonal bipyramidal EPA; and finally, a six-electron-pair arrangement takes on an octahedral EPA. (As an additional note, the hybridization, if required, can now be assigned by the guidelines given by that approach to bonding: linear (two electron pairs) means sp hybridization; three electron pairs dictates sp2 hybridization; etc.). The “molecular geometry” (MG) will be the same as the EPA provided that the number of electron pairs (EP) about the CA is the same as the number of bonding pairs (BP). The molecular geometry will, of course, be different when the number of EP is not equal to the number of BP. The resultant geometry is determined by the relative repulsion between the pairs of electrons involved in the bonding along with the so-called lone pairs (LP) of electrons as given by Gillespie’s approach (4, 5). This, in summary, is that the repulsion is greatest between lone pairs; then, next with lone pairs–bonding pairs, and, finally, the least repulsion between two bonding pairs of electrons. The molecular geometry is usually fairly easy to picture, especially with the many types of flexible model kits available or by making your own with balloons. A more permanent set could be made using the plastic soda bottle approach, as given by V. V. Samoshin in this Journal (7 ). The distorted structure involving five electron pairs where one, two, or three may be lone pairs does give the most problem to determining the proper molecular geometry. In this trigonal bipyramidal EPA the fact that there are 90° and 120° angles does introduce a bit of discussion as to the repulsion between the lone pairs and the bonding pairs. After a thorough discussion (with the help of models) one is led to the easy enough conclusion. It will then be quickly seen and understood that the lone pair will be best placed in the plane rather than along the axial positions to minimize the repulsion between electron pairs (8). The Lewis structure may now be “drawn”, if desired. Examples Some worked out examples are shown in the box. If the Lewis structure is desired, that may be done after all of the above have been completed. For example, let us do that for CO2. It had been established that the carbon atom has two pairs of electrons about it and the molecule is linear, since the number of BP equals the number of EP. So, we may write O C O .
Journal of Chemical Education • Vol. 76 No. 12 December 1999 • JChemEd.chem.wisc.edu
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NOTE: We have put the six electrons around each oxygen atom as “required”. But, since each atom must have eight electrons around it, we must rewrite the Lewis structure as O C O. The Central Atom Revisited If you have read this far, you may be interested in doing an “exercise left to the reader”. Suppose you follow the method given in determining the molecular geometry of the water molecule. Now, let us suppose you chose to use the hydrogen atom as the CA. Then you have one hydrogen atom and the oxygen atom each acting as LA. Then the hydrogen atom that acts as the CA has just two electrons (one EP). Try this with methane, CH4. That is, use one of the hydrogens as a CA. The other three hydrogens and the carbon atom now act as ligand atoms. Using the “recipe” given for this method, there is, again, just one EP about the hydrogen atom acting as a CA. It becomes clear that we cannot use hydrogen as the central atom because it keeps turning up with just one electron pair—truly insufficient to have more than one
CA LA
CA LA
CA LA
CA LA
Methane, CH4 C 4e{ 4H +4e{ Total 8e{/2 = 4 e{ pairs (EP), tetrahedral EPA (sp3 hybridization); there are 4 BP; number of BP = number of EP; so molecule is also tetrahedral (MG is tetrahedral) Ammonia, NH3 N 5e{ 3H +3e{ Total 8e{/2 = 4 e{ pairs (EP), tetrahedral EPA (sp3 hybridization); there are 3 BP ≠ 4 EP, so MG is a distorted tetrahedron/trigonal pyramidal with one LP Water, H2 O O 6e{ 2H +2e{ Total 8e{/2 = 4 EP, tetrahedral EPA (sp3 hybridization); there are 2 BP ≠ 4 EP, so MG is a distorted tetrahedron/bent molecule with 2 LP Sulfur Trioxide, SO3 S 6e{ 3O 0e{ Total 6e{/2 = 3 EP, trigonal planar EPA (sp2 hybridization); 3 BP = 3 EP, so MG is trigonal planar, also
CA I 7e{ LA 2Cl +2e{ Charge is {1 +1e{
Iodine Dichloride Ion, ICl2 {
Some Final Thoughts
Total 10e{/2 = 5 EP, EPA is trigonal bipyramidal (sp3 d hybridization); 2 BP ≠ 5 EP, so MG is a distorted trigonal bipyramidal/linear with 3 LP CA LA
CA LA LA
CA LA
bonded atom around the hydrogen. Of course, you say, we knew that already—hydrogen atoms can have only one bond (one pair of electrons). This comes out “naturally” by using the method. We quickly note that if you have a diatomic molecule in which one atom is hydrogen, the hydrogen may be used as the CA, such as in HCl, and the correct electron pair arrangement is obtained. Using the chlorine as the central atom would also work. And, while we are at it, suppose we work on the carbon tetrachloride molecule, CCl4. Try one of the chlorine atoms as the CA. Then three of the chlorine atoms and the carbon atom are acting as LAs. You will wind up with eight electrons about the chlorine atom designated as the CA; that is, 4 EP. The electron pair arrangement is tetrahedral and, since there are also four bonding pairs of electrons, we have a tetrahedral molecule—the same geometry as when we picked the carbon atom as the CA. But, let us go back and check the formal charges on each atom…some of you are way ahead of our discussion, I am sure. The formal charge on our chlorine as the CA is +3 and that of carbon is {3. While this may not seem outrageous by itself, it becomes so if we examine the formal charges on these atoms when the carbon atom is chosen as the CA. Then carbon has a formal charge of zero and each chlorine has a formal charge (all are LAs) zero, also. A much better arrangement. Just one more “exercise”. Go back to the example XeF4. Suppose we choose one of the fluorine atoms to act as the CA. Then we would have 7e{ from the fluorine plus 1e{ from each of the three other fluorine atoms (acting as a LA) plus 2e{ from the xenon atom (also acting as a LA), which gives us a total of 12e{ or 6 EP. The EPA is octahedral and, since there are only four BP, the MG is square planar. We get the same result as when we chose the xenon atom to be the CA. If you check the formal charges on the atoms, the fluorine CA comes out to be {1; each of the fluorine atoms acting as LA has the same formal charge of zero; and the xenon atom’s formal charge is +1. Please note that the formal charge on the central atom (in this case fluorine) turns out to be {1. It is not a positive number as you might wish to guess, since it serves as the CA. If we use the “correct” atom for the CA (as in the original example), all the formal charges come out to be zero—a much more desirable situation.
Xenon Tetrafluoride, XeF4 Xe 8e{ 4F +4e{ Total 12e{/2 = 6 EP, EPA is octahedral (sp3 d2 hybridization); 4 BP ≠ 6 EP, so MG is a distorted octahedral/square planar molecule with 2 LP Dinitrogen Oxide, N2 O N 5e{ N {1e{ O 0e{ Total 4e{/2 = 2 EP, EPA is linear (sp hybridization); 2 BP = 2 EP, so MG is linear, also Carbon Dioxide, CO2 C 4e{ 2O 0e{ Total 4e{/2 = 2 EP, linear EPA (sp hybridization); 2 BP = 2 EP, so MG is linear, also
The method described does focus on the many simple molecules and ions covered in most general chemistry texts, having one central atom. One may choose any atom as the central atom and the method does show that when hydrogen is chosen, it can only deal with one bond. If we choose other than the “correct” central atom, the formal charges may deviate too far from the much-sought-after zero for each atom. These outcomes are a natural consequence of the procedure outlined in this paper. This may be argued from the fact that we are dealing with the valence electrons and electrons are indistinguishable. These electron pairs will take on certain preferred arrangements about the much slower moving atomic kernel of a so-called central atom. These preferred arrangements are argued from the VSEPR approach. The strategy of getting the needed six electrons about each ligand atom (excepting hydrogen) has its roots in the Lewis approach, of course. The “magic number” of eight is not dis-
JChemEd.chem.wisc.edu • Vol. 76 No. 12 December 1999 • Journal of Chemical Education
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carded, but the focus of attention is on six electrons only, because two must be in place for the usual bond to the central atom. Odd numbers of electron pairs may still be handled. For example, NO will give 2.5 EP about the N and the MG is not linear, which would be true for 2 EP, and not trigonal planar, which would be the case for 3 EP, but some angle in between. The method works quite well for the usual set of examples (as given in ref 6 ). It is quick and using it makes it almost “fun”. As one uses it more and more, Lewis structures can also be written with greater ease. Acknowledgment I would like to thank my colleague Robert F. Ferrante for his encouragement and his computer tutorials.
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Literature Cited 1. See, for example, Chang, R.; Chemistry, 6th ed.; McGraw-Hill: New York, 1998; p 374. 2. Lewis, G. N. J. Am. Chem. Soc. 1916, 38, 762–785. 3. Lewis, G. N. Valence and the Structure of Atoms and Molecules; Reinhold: New York, 1923. 4. Gillespie, R. J. J. Chem. Educ. 1963, 40, 295–301. 5. Gillespie, R. J. J. Chem. Educ. 1970, 47, 18–23. 6. See, for example, Umland, J. B. General Chemistry; West: St Paul, MN, 1993; pp 340–341. 7. Samoshin, V. V. J. Chem. Educ. 1998, 75, 985. 8. For a brief discussion on assigning lone pairs to the planar (equatorial) position, several general chemistry texts may be consulted; for example, McMurry, J.; Fay, R. C.; Chemistry, 2nd ed.; Prentice Hall: Upper Saddle River, NJ, 1998; p 262.
Journal of Chemical Education • Vol. 76 No. 12 December 1999 • JChemEd.chem.wisc.edu
