Predicting the Stability of Hypervalent Molecules

In the Classroom edited by

Molecular Modeling Exercises and Experiments

Alan J. Shusterman Reed College Portland, OR 97202-8199

Predicting the Stability of Hypervalent Molecules

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Tracy A. Mitchell, Debbie Finocchio, and Jeremy Kua* Department of Chemistry, University of San Diego, San Diego, CA 92110; *[email protected]

Hypervalent molecules are defined as those containing atoms that seem to exceed the octet rule. The octet rule was first introduced by Lewis (1) to establish the general observation that in most stable molecules, atoms have eight electrons in their valence shell. These four pairs of electrons can be categorized as either covalent bond pairs within a covalent bond or lone pairs that do not participate in bonds. Many introductory college chemistry texts explain the existence of stable hypervalent molecules by invoking the mixing of empty d orbitals even though the contribution of d orbitals to bonding is much smaller than expected from idealized hybrid dsp3 or d2sp3 orbitals (2). There is often no explanation as to why hypervalent molecules are observed only when the central atom forms bonds with more electronegative outer atoms, but not vice versa. It is well known that the fluorides of P, S, and Cl form stable hypervalent molecules, while the corresponding hypervalent hydrides are not observed in nature. In fact, calculations have shown that when the outer atoms are much more electronegative than the central atom, the octet rule is not actually exceeded (2–4) in a molecule that has more than four bonds around a central atom. This is because the electronegative atoms pull sufficient electron density away from the central atom such that there is a net of eight or fewer electrons around the central atom. Gillespie and Silvi have provided an excellent review on the history, terminology, and use of different models to explain bonding in hypervalent molecules (5). Bonding descriptions of hypervalent molecules have also been previously discussed in this Journal by Curnow (6). In the standard textbook model, the Lewis structure of PF5 is represented by five covalent P⫺F bonds. Since there are formally five electron pairs surrounding the central P atom, it exceeds the octet rule; therefore PF5 is considered hypervalent. An alternative model that represents PF5 without the central P atom exceeding the octet rule is shown in Scheme I. In this model, PF5 has five equivalent resonance structures, each having four covalent bonds and one ionic bond. Since there are only four formal covalent bonds, these resonance structures of PF5 conform to the octet rule. The advantage of this model over the standard textbook model is that it can be used to explain why PF5 is stable, but not PH5. This is because F, with of its high electronegativity, supports the negative charges in Scheme I. The same argument can be applied to SF4 and ClF3, which are both observed in nature, while the corresponding hydrides are not. The goal of this exercise is to introduce students to using concepts in thermochemistry (bond energies, ionization potentials, electron affinities) to predict the relative stability of a hypervalent molecule relative to its non-hypervalent counterpart. Via calculations, students will compare the energies of formation for both the fluoride and the hydride. A www.JCE.DivCHED.org

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novel part of this exercise is the ability to predict a priori whether a hypervalent molecule should be stable based only on information gathered from stable non-hypervalent molecules. Students will explore the issue of partial ionic character in polar covalent bonds and use a variety of sources including their textbook and the Internet. They will also gain experience using computational chemistry as a resource by running their own calculations. Improved speed and userfriendly interfaces of molecular modeling software have enhanced the use of computational chemistry in chemical education. In fact, molecular modeling of PF5 has been used to study the mechanism of Berry pseudorotation (7). As part of our general chemistry and organic chemistry curriculum, we teach students how molecular modeling can be used to enhance comprehension of the subject material by visualizing what is occurring at the molecular level. In their first semester of general chemistry, students build, calculate, and analyze computer models of simple molecules incorporating some of the exercises by Shusterman et al. (8), as well as others designed by our faculty. This exercise described here is aimed at students in general chemistry, although it can be easily adapted for students in inorganic or physical chemistry. We have also included an alternate exercise in the Supplemental Material,W numerically equivalent to this exercise, that might make it easier for students to understand the differences between F and H without being potentially distracted by a formal thermochemical cycle containing species with non-integral net charge.

Scheme I. Resonance structures of PF5 where the octet rule is not exceeded.

Vol. 84 No. 4 April 2007

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Journal of Chemical Education

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In the Classroom

Table 1. Selected Average Bond Energies Bond

Energy/(kcal/mol)

F–F

037

P–F

117

S–F

068

Cl–F

—

H–H

104

P–H

077

S–H

087

Cl–H

103

PH3 + H2

It is important to note that this exercise makes substantial approximations that have been chosen to keep the exercise both procedurally and conceptually practical for students at the general chemistry level. Computational Methods All computations were performed with Gaussian98 (9) using the Unix Spartan02 (10) interface at the standard B3LYP//6-311+G(d) level of theory and basis set.1 Similar results can be obtained with later versions of these programs (Gaussian03 and Spartan04) or other computational chemistry programs (e.g., Hyperchem, QChem, Jaguar, GAMESS). Electrostatic potential (ESP) charges were calculated using the Merz–Kollman approach (11). We also computed zero point energies (ZPEs) and thermal corrections to 298 K; their inclusion changes the numerical results only by a small factor and does not alter the conclusions. Since this exercise is aimed at general chemistry students, ZPEs and thermal corrections are omitted in the exercise given to students. Thus, students will be calculating values for ∆E in place of ∆H. Experimental ionization potentials (IP) and electron affinities (EA) were obtained from the NIST Chemistry WebBook (12). Given the methods used and our approximations, we have rounded all final calculated energy values in kcal兾mol to integer values. Procedure and Results A handout for this exercise, provided in the Supplemental Material,W is available for students to download before they come into lab. As a preparation exercise students use average bond energies from their textbook to calculate the relative stability (given by ∆E ) of PF5 compared to PF3 and F2. The relevant bond energies from our current textbook (13) are shown in Table 1. Assuming that the P⫺F bonds have the same bond energy in both PF5 and PF3, the calculated ∆E value is PF5

∆E = −197 kcal/mol

This is because the net reaction involves breaking one F⫺F bond (cost of 37 kcal兾mol) and forming two P⫺F bonds (gain of 2 × 117 = 234 kcal兾mol). For comparison, the ex-

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PH5

∆E = −50 kcal/mol

This calculated ∆E value suggests that PH5 should be stable relative to PH3 and H2. In reality, PH5, is not observed in nature. The assumptions and approximations of this preparation exercise are then discussed. We highlight the following three points:

Note: Bond Energies are from ref 13.

PF3 + F2

perimental enthalpy change for the formation of PF5 from PF3 and F2 is ᎑152 kcal兾mol as calculated from ∆fH ⬚ values (11). Using the same assumption for the hydrides, breaking one H⫺H bond (104 kcal兾mol) and forming two P⫺H bonds (2 × 77 = 154 kcal兾mol) leads to

• There is a difference between average bond energies and actual bond energies of specific molecules. For example, the average P⫺F bond energy (117 kcal兾mol) is not necessarily the P⫺F bond energy in PF3, PF5, or any particular molecule that contains P⫺F bonds; rather it is an average value of P⫺F bond energies from a variety of molecules. • The P⫺F bond energy in PF3 is not necessarily the same as in PF5. In addition, the P⫺F bonds within trigonal bipyramidal PF5 may also have different bond energies depending on whether they are axial or equatorial bonds. The same is true for the P⫺H bonds in PH3 and PH5. The approximation that the P⫺F bond energy is the same in PF3 and PF5 assumes the bonds are purely covalent in character with the same bond energies and bond lengths. • Pauli repulsion was not taken into account. The repulsion is larger in hypervalent molecules (five bond pairs surrounding the central atom) compared to their non-hypervalent counterparts (three bond pairs and one lone pair surrounding the central atom). Pauli repulsion is also larger in the hydrides compared to fluorides because (i) the P⫺H bonds are shorter and (ii) the bond pairs are on average closer to the central P atom in the hydride since H is less electronegative than P (while F is more electronegative than P).

We then proceed to the “meat” of the exercise: the recalculation of ∆E using an alternative model for hypervalent molecules that does not exceed the octet rule. As shown in Scheme I, if the central P in PF5 has a formal +1 charge, and hence has formally only four bond pairs surrounding it, the octet rule will not be exceeded. Formation of PF5 requires the following steps: 1. Breaking the F⫺F bond pair in F2 2. Moving charge away from P so that it acquires a formal +1 charge 3. The excess charge is picked up by F atoms that will form axial P⫺F bonds 4. Forming the axial P⫺F bonds, of which only one is a formal covalent bond pair

Table 2 lists the raw data from the computer calculations for this exercise and also includes data for the analogous fluorides and hydrides of S and Cl. The results for S and Cl are discussed later. Table 3 lists the experimental ionization potentials and electron affinities from ref 12. Using


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In the Classroom Table 2. Calculated Electronic Energies, Atomic Charges, and Bond Distances Moiety

Spin Mult

Energy/ hartree

Bond Length/Å

PF 3

1

᎑641.08753

1.605

P(+0.443)

SF2

1

᎑597.89946

1.639

S(+0.314) F(᎑0.157)

ClF

1

᎑560.00975

1.679

Cl(+0.150) F(᎑0.150)

PH3

1

᎑343.15681

1.423

P(᎑0.230) H(+0.077)

SH2

1

᎑399.41332

1.351

S(᎑0.354) H(+0.178)

ClH

1

᎑460.82006

1.292

Cl(᎑0.265) H(+0.265)

F2

1

᎑199.57176

1.409

H2

1

᎑1.17663

0.742

P

4

᎑341.28173

S

3

᎑398.13307

Cl

2

᎑460.16688

F

2

᎑99.76059

H

2

᎑0.50216

ESP Atomic Charges (a.u.) F(᎑0.147)

Moiety

Ionization Potential/eV

PF3

11.38

SF2

10.08

ClF

12.66

PH3

09.87

SH 2

10.46

ClH

12.74 Electron Affinity/eV

F

3.401

H

0.754

Note: Data from ref 12.

data from these two tables, we illustrate in detail the entire analysis for PF5 in a series of bite-sized steps corresponding to the steps outlined in the exercise handout given to students. We also present the numerical results for the same process for PH5. Energy conversion values are also provided in the exercise handout.

Step 1: Calculating Bond Energy of F2 The total bond energy of an F⫺F bond in F2 can be obtained by comparing the difference in computed energy of the molecule with respect to its atomic fragments. This can be determined by first computing the total electronic energies of the molecule and its fragments using molecular modeling software. The bond energies are then calculated from the difference in electronic energies of the molecule and its fragments. Thus, the F⫺F bond energy is given by ∆E of the reaction F 2 → 2F, where ∆E = [(2)(᎑99.76059) + 199.57176] × 627.51 = 32 kcal兾mol. (Conversion: 1 hartree = 627.51 kcal兾mol.) Steps 2a–2c: Calculating the Ionic and Covalent Contributions to the P⫺F Bond in PF3 The change in energy for the reaction PF3 → P + 3F corresponds to the bond energy of three P⫺F bonds in PF3. For this reaction, ∆E = [᎑341.28173 + (3)(᎑99.76059) + 641.08753] × 627.51 = 329 kcal兾mol. Hence, one P⫺F bond in PF3 is worth 110 kcal兾mol. Since the P⫺F bond is polar, if we assume that the ionic contribution to the bond energy can be estimated using Coulomb’s law, the ionic contribution can be calculated using E ionic = 332.0

Table 3. Experimental Ionization Potentials and Electron Affinities

q1 q 2 r

where q1 and q2 are the charges on the P and F atoms in atomic units, r is the P⫺F bond length in Å, and 332.0 is the conversion factor to express energy in kcal兾mol. Note that the use of Coulomb’s law for the ionic contribution is www.JCE.DivCHED.org

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an approximation because the atoms do not “see” each other as point charges. For the P⫺F bond in PF 3 , E ionic = (332.0)(0.443)(᎑0.147)兾(1.605) = ᎑13 kcal兾mol. The negative sign simply indicates that the ionic contribution stabilizes the bond since P and F have opposite charges. To obtain the covalent bond energy we make the following assumption: The total bond energy is separable into just two quantities, an ionic contribution and a covalent contribution. Assuming that the sum of these two contributions add up to the total bond energy, the covalent contribution is 110 − 13 = 97 kcal兾mol for the P⫺F bond in PF3. It should be emphasized that this method for separating the ionic and covalent contributions to the total bond energy is not rigorous, but is easily performed and understood by students. A further discussion of this approximation can be found in the Supplemental Material.W

Step 2d: Moving Charge Away from P in PF3 So That P Has a +1 Formal Charge Since a novel aspect of this exercise is the prediction of stability of the hypervalent molecule without calculating any of its properties, we chose as a reference point the lower limit of polarity sufficient to maintain the octet rule (3), corresponding to a charge of +1 on P. In our model of PF5, the P atom has a charge of +1. Since the ESP charge of P in PF3 is +0.443, an additional 0.557 units of charge must be moved off the P in PF3. We make the assumption that the cost of moving the charge off P can be calculated as a fraction of the full ionization potential of PF3, that is, the charge removed from PF3 primarily comes from the “lone pair” on the central P atom (where the highest occupied molecular orbital is localized). This approximation and the use of different charge schemes are further discussed in the Supplemental Material.W From Table 3, the full ionization potential of PF3 is 11.38 × 23.06 = 262 kcal兾mol (where 1 eV = 23.06 kcal兾mol). Thus, the energy cost of moving 0.557 units of charge away from P is 55.7% of 262 kcal兾mol, or 146 kcal兾mol.


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Scheme II. Energetics of formation of PF5 from PF3 and F2. (All energies are in kcal/mol.)

Scheme III. Energetics of formation of PH5 from PH3 and H2. (All energies are in kcal/mol.)

Step 2e: Enhancement of the Ionic Bond Now That P Has a +1 Charge Before moving charge away from P, the ESP charge of P was +0.447, leading to an ionic contribution of 13 kcal兾mol. However, when the charge on P is +1, the ionic contribution increases. Using the Coulomb law approximation, Eionic = (332.0)(1)(᎑0.147)兾(1.605) = ᎑30 kcal兾mol, that is, an enhancement of 17 kcal兾mol per bond or 51 kcal兾mol for the three P⫺F bonds.

For the sake of simplicity, we assume that the covalent bond energy for the new P⫺F bond is equal to the covalent bond energy previously estimated for PF3, 97 kcal兾mol. We also assume that the covalent bond energies for the three existing P⫺F bonds do not change upon formation of PF5. These assumptions are not likely to be valid since changes in P⫺F bond length, the position of F in PF5, and changes in coordination number and charge at P, will influence covalent interactions. The ionic contribution for forming the two new axial P⫺F bonds is estimated using Coulomb’s law as

Step 3: Moving the Charge to F Results in an Energy Gain There is a small gain in energy as 0.557 units of charge is moved to the two F atoms that will be forming the two new axial P⫺F bonds. This is approximated as 55.7% of the electron affinity of F (3.401 × 23.06 = 78 kcal兾mol), or 43 kcal兾mol. Note that this is numerically equivalent to moving 0.5 × 0.557 units of charge to two F atoms in our approximation. Steps 4a-4b: Calculating the Bond Energies of the Two New P⫺F Bonds in PF5 The P in PF3+0.55 carries a formal charge of +1 and is bonded to three fluorines. To form PF5, we must combine PF3+0.55 with two F᎑0.275, which involves the formation of one more covalent P⫺F bond (Scheme I) and the creation of two new P⫺F ionic bonds. (The ionic repulsion between outer F atoms was ignored. See the Supplemental MaterialW for further discussion on this approximation.)

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E ionic =

(2)(332.0)(1) (−0.557

× 0. 5)

1.605

= −115 kcal/mol

The energy costs and gains used to calculate the net ∆E for the reaction PF3 + F2 → PF5 are summarized in the thermochemical cycle in Scheme II. The ESP charges in the scheme are rounded for presentation purposes. Actual numbers with more significant figures were used in the energy calculations. The result for PF3 + F2 → PF5 is that ∆E = ᎑128 kcal兾mol. As expected, the formation of PF5 from PF3 and F2 is energetically favorable. For reference, the experimental enthalpy change for this reaction calculated from ∆fH ⬚ values is ᎑150 kcal兾mol (12). When these steps are repeated for PH5 (using the data in Tables 2 and 3), the result for PH3 + H2 → PH5 is ∆E = 625 − 554 = +71 kcal兾mol, suggesting that the formation of


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PH5 is disfavored. The costs and gains for this process are summarized in the thermochemical cycle in Scheme III. The P⫺H bond energy in PH3 is 79 kcal兾mol; its ionic and covalent contributions are 4 and 75 kcal兾mol, respectively. Note that the fractional IP and EA values are larger than the full IP and EA since a net of 1.230 units of charge needs to be moved from P to the outer H atoms. (The experimental IP of PH3 and EA of H are 228 and 17 kcal兾mol, respectively.) Also in step 2e, because P and H are now both positively charged, the ionic contribution is worsened. This is more than compensated for in step 4b. The two major reasons why there is such a large difference in ∆E contrasting the fluoride and the hydride are (i) in step 1, the F⫺F bond (32 kcal兾mol) is significantly weaker than the H⫺H bond (108 kcal兾mol) and (ii) the overall cost of moving electron density away from the central P (to attain a +1 charge on P) in the hydride is not sufficiently compensated by the attractive ionic contribution from the P⫺H bond. In the cycle, this cost corresponds to the sum of energy changes for steps 2d, 2e, 3, and 4b. (The contribution of steps 4a, difference in P⫺F and P⫺H bond energies, and 3, difference in electron affinities of F and H, are minor.) To evaluate how well this works, we applied the same method to calculate the hydrides and fluorides of sulfur and chlorine. Using the raw data from Tables 2 and 3, we obtain:

SF2 + F2 SH2 + H2 ClF + F2 ClH + H2

SF4 SH4 ClF3 ClH3

∆E = −112 kcal/mol ∆E = +117 kcal/mol ∆E = −28 kcal/mol ∆E = +139 kcal/mol

The scheme successfully predicts that SF4 and ClF3 are stable, but their hypervalent hydride counterparts are not. The experimental enthalpy changes calculated from ∆fH ⬚ values are ᎑113 kcal兾mol for SF4 and ᎑26 kcal兾mol for ClF3 (12). SH4 and ClH3 do not exist as stable molecules in nature. This is in surprisingly good agreement with our calculated results. Discussion Upon completion of the exercise students should be able to conclude that an alternative description of PF5 with four P⫺F covalent bond pairs and net one ionic bond is superior to the textbook Lewis structure with five P⫺F covalent bonds, at least in terms of predicting the stability of PF5 and the instability of PH5. We think it may be instructive to discuss the approximations used in our scheme. Without getting into computational issues such as the choice of density functional and basis set, the following points can be raised in a postexercise discussion. Further discussion on some of these approximations is found in the Supplemental Material.W • We assumed that the charge on P is +1 to maintain the octet rule. This is the lower limit of “ionicity” in the P⫺F bond (3). Since a goal of the exercise is to predict the stability of PF5 without actually computing PF5, we choose this lower limit as our reference. The actual calculated charge on PF5 may be higher than +1, in which case, there would be “less than eight electrons” around the central P.

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• The axial and equatorial P⫺F bonds are assumed to be equivalent in distance, charge, and mix of ionic– covalent character in PF5. We also assumed that the P⫺F bonds in PF5 would be the same distance as in PF3, and that the covalent contribution is the same in both molecules. (The actual computed distances in PF 5 are 1.5706 Å for the equatorial P⫺F bonds and 1.6042 Å for the axial P⫺F bonds.) • The covalent contribution was assumed to be the total bond energy minus the ionic contribution (calculated using a Coulomb law approximation). F⫺F ionic contributions were also ignored. • An increase in Pauli repulsion in the hypervalent molecule was not included in the energy cost calculation. • Resonance stabilization was not included in the energy gain calculation. • While explaining the exclusion of zero point energy in our values might be beyond the level of general chemistry, the point that our calculated values are at absolute zero rather than 298 K can be discussed. • Fractional IP and EA values were used as an approximation to model charge transfer from P to F. • The calculation of the ionic contribution and the fractional IP and EA values are dependent on the value of the calculated charges. While we have chosen the widely used ESP charges, there are a variety of other methods for calculating atomic charges.

In light of the points above, there is no strong reason to expect anything other than ballpark agreement between the calculated ∆E values and experimental ∆H ⬚ values. This is the case for PF5 where our calculated ∆E of ᎑128 kcal兾mol is ∼15% lower than the experimental ∆H ⬚ value of ᎑152 kcal兾mol. Since our scheme was originally worked out for PF5, the very good (1–2 kcal兾mol) agreement between this scheme and experimental ∆H ⬚ values for SF4 and ClF3 was unexpected. Although one could substitute S or Cl into the exercise instead of P to get better agreement, we think that P is instructive because the discrepancy leads to discussion of the approximations used in our scheme. Interestingly, the largest source of error in the computation of bond energies is the P⫺F bond in PF3. Experimental ∆fH ⬚ values (12) suggest that the bond energy should be 121 kcal兾mol rather than our calculated value of 110 kcal兾mol. In the scheme we have presented, the charges of the equatorial and axial F atoms in the final structure of PF5 are different. However given the approximations in our scheme (in particular, the partitioning of ionic and covalent contributions, the Coulomb approximation for the ionic part and ignoring F⫺F ionic terms), forming a final structure with P(+1) and five equivalent F(᎑0.2) is numerically equivalent for the calculation of ∆E. An alternative version of the exercise handout that leads to charge-equivalent F but does not formally use a thermochemical cycle is included in the Supplemental Material.W An important goal of this exercise is for students to get comfortable thinking about PF5 as a set of resonance structures (rather than the one textbook structure that exceeds the octet rule) because the resonance structures more accurately describe P as being a center of positive charge. Although there


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In the Classroom

are substantial approximations in the exercise, the goal is not to be quantitatively accurate (even though the numbers work out quite well compared to experiment). Rather we aim to give students a sense of how qualitative concepts of electronegativity differences and chemical bonding can be related to energy and stability.

This exercise introduces students to using thermochemistry to predict a priori the relative stability of a hypervalent molecule. Students explore issues surrounding the octet rule, partial ionic character in polar covalent bonds, and resonance structures. As part of the exercise, they integrate and apply a variety of sources including their own computations, textbook, and the Internet to answer the question at hand. Acknowledgments We thank T. J. Dwyer, D. Tahmassebi, and H. Joshi for their critique of the manuscript and the W. M. Keck Foundation for funding of our computational resources. Note 1. Different levels of theory can be used provided that some electron correlation is taken into account. Different basis sets can also be used as long as the basis of P contains enough flexibility (polarizability is important) and the basis sets are not too small. This is less important for the outer atoms; see ref 2 for details.


Supplemental Material

Student handouts, an alternative version of the exercise, and further discussion of the approximations are available in this issue of JCE Online. Literature Cited

Conclusion

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1. Lewis, G. N. J. Am. Chem. Soc. 1916, 38, 762–785. 2. Reed, A. E.; Schleyer, P. von R. J. Am. Chem. Soc. 1990, 112, 1434–1445. 3. Cioslowski, J.; Mixon, S. T. Inorg. Chem. 1993, 32, 3209– 3216. 4. Noury, S.; Silvi, B.; Gillespie, R. J. Inorg. Chem. 2002, 41, 2164–2172. 5. Gillespie, R. J.; Silvi, B. Coord. Chem. Rev. 2002, 233–234, 53–62. 6. Curnow, O. J. J. Chem. Educ. 1998, 75, 910–915. 7. Montgomery, C. D. J. Chem. Educ. 2001, 78, 844–846. 8. Shusterman, G. P.; Shusterman, A. J. J. Chem. Educ. 1997, 74, 771–776. 9. Gaussian98; Gaussian, Inc.: Pittsburgh, PA, 1998. 10. Spartan’02; Wavefunction, Inc.: Irvine, CA, 2002. 11. Besler, B. H.; Merz, K. M., Jr.; Kollman, P. A. J. Comp. Chem. 1990, 11, 431–439. 12. NIST Chemistry WebBook. http://webbook.nist.gov/chemistry/ (accessed Dec 2006). 13. Olmsted, J, III,.; Williams, G. M. Chemistry, 3rd ed.; John Wiley & Sons, Inc.: Hoboken, NJ, 2002; p 394 (Table 9-2).


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Predicting the Stability of Hypervalent Molecules

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