The University of Illinois Session 3 – Molecules The University of Illinois Bonding Module David E. Woon, Beth Lindquist, Lina Chen, Tyler Takeshita, Lu Xu, Jeff Leiding, & Thom H. Dunning, Jr.

Session 3 Overview – Preliminaries Outline for Session 3 – Molecules 1. Preliminary: Traditional Approach vs. New Approach Some Questions 2. Module Material on Molecules (a) Oxygen compounds (b) Carbon compounds

discard information about orbitals Concept Map for the Traditional Approach discard information about orbitals Atomic Configurations from Aufbau principle and Hund’s rules # of valence electrons octet rule Lewis Diagrams Molecular Structure Apply VSEPR often rationalized with hybridization O: 2s2 2px2 2py 2pz 2p 2s O or

Lewis Structures: Pros and Cons Positives: Lewis structures are based on electron pair bonding. They provide a framework for determining bond orders and resonance structures and are a useful symbolic shorthand. Negatives: They’re sometimes misused to deduce structure. Identities of the atomic electronic orbitals are lost. Students must memorize/apply a complex set of rules. Lewis structures are driven by the octet rule, for which there are many exceptions. The octet rule also ignores radicals. They provide false intuition for molecules such as O2.

VSEPR: Pros and Cons Positives: VSEPR provides a simple, 3D approach to characterize common structural motifs. Molecular structures are partially determined by pair overlap repulsion. Negatives: VSEPR is a post hoc model that works reasonably well for some cases but generates tortuous and often incorrect explanations for other cases. Electrons don’t behave like balloons! Orbitals adjust to one another in a hierarchical manner that depends upon their energies and symmetries. VSEPR is much less useful for understanding the structures of radicals and ions: it gets species like CH3 wrong.

Pauling Hybridization: Pros and Cons Positives: Orbitals of different angular symmetries are able to mix. Pauling hybridization provides a framework for understanding molecular structure. Negatives: It’s difficult for students to understand: this form of hybridization is arbitrary. In some cases, the assumed hybridization is not observed: hypervalency in P, S, and Cl is not due to sp3d or sp3d2 hybridization, which has been known for over 30 years.

Concept Map for the GVB Orbital Approach Infer Aufbau principle and Hund’s rules Atomic Configurations from electron-by-electron approach convert Orbital Diagrams for Atoms suggest structure Identify Bond Types by combining different atoms H + H+ 1 e- bond H + H 2 e- covalent H + F 2 e- polar covalent Li + F 2 e- ionic H+ + He 2 e- dative H + Be 3 e- recoupled pair A “What happens if…” approach Build Molecules Atom-by-Atom from Diagrams Lewis diagrams by inspection Octet rule as observation refine structure radical vs. stable reactivity by observation

Session 3 – Question from the Instrument 1. In which manner does a nitrogen atom form a bond with a hydrogen atom? (a) A covalent bond forms when the electron on the H atom interacts with a 2s electron on the N atom. (b) A covalent bond forms when the electron on the H atom interacts with a 2p electron on the N atom. (c) The N atom removes the electron from the H atom to form an ionic bond. (d) None of the above: NH is not stable.

Session 3 – Question from the Instrument 1. In which manner does a nitrogen atom form a bond with a hydrogen atom? (a) A covalent bond forms when the electron on the H atom interacts with a 2s electron on the N atom. (b) A covalent bond forms when the electron on the H atom interacts with a 2p electron on the N atom. (c) The N atom removes the electron from the H atom to form an ionic bond. (d) None of the above: NH is not stable.

Session 3 – Question from the Instrument 2. We asked 49 Gen Chem students the following question. What do think was their most frequent response? In which manner does a nitrogen atom form a bond with a hydrogen atom? (a) A covalent bond forms when the electron on the H atom interacts with a 2s electron on the N atom. (b) A covalent bond forms when the electron on the H atom interacts with a 2p electron on the N atom. (c) The N atom removes the electron from the H atom to form an ionic bond. (d) None of the above: NH is not stable. 1. In which manner does a nitrogen atom form a bond with a hydrogen atom? (a) A covalent bond forms when the electron on the H atom interacts with a 2s electron on the N atom. (b) A covalent bond forms when the electron on the H atom interacts with a 2p electron on the N atom. (c) The N atom removes the electron from the H atom to form an ionic bond. (d) None of the above: NH is not stable.

Session 3 – Question from the Instrument 2. We asked 49 Gen Chem students the following question. What do think was their most frequent response? In which manner does a nitrogen atom form a bond with a hydrogen atom? % (a) A covalent bond forms when the electron on the H atom interacts with a 2s electron on the N atom. 12.1 (b) A covalent bond forms when the electron on the H atom interacts with a 2p electron on the N atom. 32.7 (c) The N atom removes the electron from the H atom to form an ionic bond. 4.1 (d) None of the above: NH is not stable. 51.0

Session 3 – Question from the Pilot Set 3. Which of the following is closest to the bond angle of H2S? (a) 109 (b) 180 (c) 90 (d) 120

Session 3 – Question from the Pilot Set 3. Which of the following is closest to the bond angle of H2S? (a) 109 (b) 180 (c) 90 (d) 120

Session 3 – Question from the Pilot Set 4. We asked 50 Gen Chem students the following question. What do think was their most frequent response? Which of the following is closest to the bond angle of H2S? (a) 109 (b) 180 (c) 90 (d) 120

Session 3 – Question from the Pilot Set 4. We asked 50 Gen Chem students the following question. What do think was their most frequent response? Which of the following is closest to the bond angle of H2S? % (a) 109 38 (b) 180 28 (c) 90 6 (d) 120 14

Session 3 – Question from the Pilot Set 5. What is the shape of the PF3 molecule? (a) T-shaped (b) Linear (c) Trigonal pyramid (d) Trigonal planar

Session 3 – Question from the Pilot Set 5. What is the shape of the PF3 molecule? (a) T-shaped (b) Linear (c) Trigonal pyramid (d) Trigonal planar

Session 3 – Question from the Pilot Set 6. We asked 19 Gen Chem students the following question. What do think was their most frequent response? What is the shape of the PF3 molecule? (a) T-shaped (b) Linear (c) Trigonal pyramid (d) Trigonal planar

Session 3 – Question from the Pilot Set 6. We asked 19 Gen Chem students the following question. What do think was their most frequent response? What is the shape of the PF3 molecule? % (a) T-shaped 21.1 (b) Linear 0.0 (c) Trigonal pyramid 63.2 (d) Trigonal planar 15.8

One Last Preliminary Question 7. What is the shape of the methyl (CH3) radical? (a) T-shaped (b) Tetrahedral (c) Trigonal pyramid (d) Trigonal planar

One Last Preliminary Question 7. What is the shape of the methyl (CH3) radical? (a) T-shaped (b) Tetrahedral (c) Trigonal pyramid (d) Trigonal planar

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Building Molecules Last week we explored the ways in which different combi-nations of orbitals (s+s, p+p, s+p) and occupations led to bonds between two atoms chosen from the elements H through F. This week we will turn that around and explore what we can build if we start with an atom and keep adding other atoms to it.

Review Question 8. Which type of bonding involves one electron contributed by each of the entities involved in the bond? (a) A coupled pair bond. (b) A recoupled pair bond. (c) A dative bond.

Review Question 8. Which type of bonding involves one electron contributed by each of the entities involved in the bond? (a) A coupled pair bond. (b) A recoupled pair bond. (c) A dative bond. What is the condition necessary to form a coupled pair bond involving two electrons?

Review – Bond Categories We found three categories of bonds that could occur: 1  1 coupled pair bonds: covalent, polar covalent, ionic condition: different spins 1  2 recoupled pair bond condition: reorganization is favorable (not repulsive) 0  2 dative bond conditions: lone pair on one entity, unoccupied valence orbital on the other entity (e.g., H+)

Consider An Oxygen Atom Starting with the familiar ground state O atom, how many molecules can we build? or

Consider An Oxygen Atom 1) Name as many compounds as you can that contain O and H, with at least one O atom. 2) Which of the three types of chemical bonds we’ve talked about might be present? a) What is the maximum number of coupled pair bonds that could form? b) What is the maximum number of recoupled pair bonds that could form? c) What is the maximum number of dative bonds that could form?

Nominal Valence & Nominal Bond Angles It is straightforward to infer how many coupled pair bonds an atom can form: just count the unpaired electrons in its ground state configuration. We call this the nominal valence of the element. The nominal bond angles of an atom are the angles between any singly occupied orbitals.

Anticipation Question 9. Given the configuration diagram shown below, what do you think is the nominal valence of O? 1 (b) 2 (c) 3 (d) Not enough information.

Anticipation Question 9. Given the configuration diagram shown below, what do you think is the nominal valence of O? 1 (b) 2 (c) 3 (d) Not enough information.

Anticipation Question 10. Given the configuration diagram for O shown below, what do you think the nominal bond angle of H2O will be? 180 (b) 120 (c) 90 (d) Not enough information.

Anticipation Question 10. Given the configuration diagram for O shown below, what do you think the nominal bond angle of H2O will be? 180 (b) 120 (c) 90 (d) Not enough information. Sulfur has the same configuration as O and thus also has a nominal valence of two. Like water, H2S has a nominal bond angle of 90.

Beyond the Nominal The nominal geometry of a molecule is only a starting point. We can determine the actual bond lengths with experiments or accurate calculations. We will soon compare H2O and H2S. Some elements can exceed their nominal valence by undergoing recoupled pair bonding or dative bonding.

Adding H to O: H1

Anticipation Question 11. If a H atom interacts with an O atom in the two orientations shown below, three combinations of orbitals are possible. Which of the scenarios do you think yields bound forms of OH? A. H 1s()  O 2p() B. H 1s()  O 2p() C. H 1s()  O 2p() A and C (b) A only (c) A, B, and C (c) C only

Anticipation Question 11. If a H atom interacts with an O atom in the two orientations shown below, three combinations of orbitals are possible. Which of the scenarios do you think yields bound forms of OH? A. H 1s()  O 2p() B. H 1s()  O 2p() C. H 1s()  O 2p() A and C (b) A only (c) A, B, and C (c) C only

O + H Potential Energy Curves As we saw in HF, the only combination on the previous slide that isn’t repulsive is the covalent bond formed by coupling the H 1s and O 2pz orbitals as . 1s1  2px1 () 1s1  2px1 () 1s1  2px2

OH Orbitals and Potential Energy Curve Can you tell from the animation if the bond is polarized? If so, toward which atom?

Comparing Orbitals for OH and SH The bonding electrons in OH are polarized more toward O then is the case for HS, consistent with electronegativities: O: 3.44 S: 2.58 H: 2.20 The bond length of SH is longer than that of OH, 1.34 Å vs 0.97 Å.

Comparing Orbitals for OH and SH Also, the greater polarization in OH with respect to SH yields a larger partial charge on H on OH: OH: qH = 0.259 SH: qH = 0.137

Comparing Orbitals for OH and SH The 2s2 orbital of OH is pushed farther way from the nucleus than the 3p2 orbital of SH.

Comparing Orbitals for OH and SH The singly occupied p orbitals of OH and SH are both directed perpendicular to the internuclear axis.

Anticipation Question 12. Which of the following statements is true? The bond angle of H2O is larger than the bond angle of H2S because the two bond pairs are polarized more toward the O nucleus, forcing the angle to open more. (b) The bond angle of H2O is larger than the bond angle of H2S because there is more repulsion between the H atoms due to their greater partial charges. (c) Both a and b.

Anticipation Question 12. Which of the following statements is true? The bond angle of H2O is larger than the bond angle of H2S because the two bond pairs are polarized more toward the O nucleus, forcing the angle to open more. (b) The bond angle of H2O is larger than the bond angle of H2S because there is more repulsion between the H atoms due to their greater partial charges. (c) Both a and b. 104.2 92.3 The bond angle of H2S is very close to the nominal angle of 90.

Adding H to OH: H2 OH(2p) H(1s) first bond pair

Adding H to OH: H2 formation of second bond pair first bond pair is largely unperturbed

Adding H to H2O: H3? Our exploration with adding H to O showed that H cannot recouple the 2p2 pair of O. If we try to add a third H to H2O, we only get a weakly bound structure.

Bonding Diagrams – 2D + O H OH H2O H + OH

Lewis Structures from Bonding Diagrams + O H O H OH (by inspection) O H + OH H H2O

Bonding Diagrams – 3D O

Bonding Diagrams – 3D O H

Bonding Diagrams – 3D OH O H

Bonding Diagrams – 3D H H2O O H

Bonus: NHn Family NH NH2 NH3 N H linear reactive diradical bent reactive radical pyramidal stable closed-shell N H NH2 NH3

Anticipation Question 13. Based on its 2s2 2px 2py configuration, what is the nominal valence of Carbon? (a) 1 (b) 2 (c) 3 (d) 4

Anticipation Question 13. Based on its 2s2 2px 2py configuration, what is the nominal valence of Carbon? (a) 1 (b) 2 (c) 3 (d) 4 Carbon has two unpaired electrons and can thus only form two coupled pair bonds.

three unpaired electrons Carbon Compounds: CH There are two ways H can approach C. It can form a covalent bond in this configuration: one unpaired electron + However, we can rotate C so that the H approaches the 2s2 pair: then it behaves like Be, and the 2s2 pair is recoupled: three unpaired electrons +

Orbitals for CH with Recoupled Pair Bond singly occupied 2px 2py pair Re + 1.4 Å singly occupied 2px 2py

Carbon Compounds: CH Potential Energy Curves The recoupled pair bond state of CH is almost as stable as the covalently bonded state. CH 1st bond: 82.4 (kcal/mol) 1st bond: 66.0

The state of CH2 with recoupled pair bonding is more stable! Carbon Compounds: CH + H  CH2 We can add H to each of these forms of CH. In the first case, we get this bonding diagram and optimized structure: 102 1st bond: 82.4 2nd bond: 99.5 Net: 177.9 (kcal/mol) If we add a second H to the form of CH with three unpaired electrons, we get this structure: 134 1st bond: 66.0 2nd bond: 121.3 Net: 187.3 (kcal/mol) The state of CH2 with recoupled pair bonding is more stable!

Carbon Compounds: CH + H  CH2 We will adopt the following 2D bonding diagram for this form of CH2. Note that it has one unpaired electron in the plane of the screen and one perpendicular to the screen. This is the in-plane orbital. (The out-of-plane orbital is a 2p orbital.)

Carbon Compounds: CH2 + H  CH3 If we add a third H, it forms a covalent bond to the unpaired electron in the in-plane orbital to yield a trigonal planar structure. 120 This is the remaining singly occupied orbital.

Carbon Compounds: CH3 + H  CH4 When the 4th H adds to CH3, its bond pair pushes the other three bond pairs into a tetrahedral arrangement as the bond forms.

3D Bonding Diagrams for CHn Compounds Start with a C atom.

3D Bonding Diagrams for CHn Compounds Add H.

3D Bonding Diagrams for CHn Compounds Uncouple the C pair.

3D Bonding Diagrams for CHn Compounds Add H to form CH2. Recouple to C to form CH. Add H to form CH3. Add H to form CH4.

Three Bonding Motifs for Carbon Compounds Carbon has four bonds in most stable organic compounds, but there are there distinct motifs that are commonly found.

Three Bonding Motifs for Carbon Compounds Sometimes C is found in a linear configuration connected to two other atoms. Here is one example, acetylene: +

Three Bonding Motifs for Carbon Compounds Sometimes C is found in a linear configuration connected to two other atoms. Here is one example, acetylene: + Another example is HCN, shown here with a 3D diagram: H C N

Follow-up Question 14. How many s and p bonds are in acetylene? (a) 2 s bonds and 2 p bonds (b) 3 s bonds and 4 p bonds (c) 3 s bonds and 2 p bonds

Follow-up Question 14. How many s and p bonds are in acetylene? (a) 2 s bonds and 2 p bonds (b) 3 s bonds and 4 p bonds (c) 3 s bonds and 2 p bonds

Three Bonding Motifs for Carbon Compounds Another motif is trigonal planar, when C is connected to three atoms, as we saw with the methyl radical. Another example is ethylene, C2H4: +

Three Bonding Motifs for Carbon Compounds Another motif is trigonal planar, when C is connected to three atoms, as we saw with the methyl radical. Another example is ethylene, C2H4: +

Three Bonding Motifs for Carbon Compounds The final motif adopted by carbon when it is bonded to four other entities is the tetrahedral structure found in methane and compounds with a methyl group, such as methanol:

Some Examples Each response sheet has a number in the bottom right corner. I would like you to draw a bonding diagram for the molecule corresponding to your number from the list below: 1. Formaldehyde – H2CO 2. Dioxygen – O2 3. Ketene – H2CCO

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Session 3 – Question from the Instrument 15. Can a nitrogen atom and fluorine atom form a chemical bond? (a) No, NF does not satisfy the octet rule. (b) Yes, the singly occupied orbitals could form a bonding orbital. (c) No, the N atom and F atom are too electronegative and would repel each other. (d) Yes, F has a higher electronegativity than N, so they can form an ionic bond.

Session 3 – Question from the Instrument 15. Can a nitrogen atom and fluorine atom form a chemical bond? (a) No, NF does not satisfy the octet rule. (b) Yes, the singly occupied orbitals could form a bonding orbital. (c) No, the N atom and F atom are too electronegative and would repel each other. (d) Yes, F has a higher electronegativity than N, so they can form an ionic bond.

Session 3 – Question from the Instrument 16. We asked 53 Gen Chem students the following question. What do think was their most frequent response? Can a nitrogen atom and fluorine atom form a chemical bond? (a) No, NF does not satisfy the octet rule. (b) Yes, the singly occupied orbitals could form a bonding orbital. (c) No, the N atom and F atom are too electronegative and would repel each other. (d) Yes, F has a higher electronegativity than N, so they can form an ionic bond.

Session 3 – Question from the Instrument 16. We asked 53 Gen Chem students the following question. What do think was their most frequent response? Can a nitrogen atom and fluorine atom form a chemical bond? % (a) No, NF does not satisfy the octet rule. 20.8 (b) Yes, the singly occupied orbitals could form a bonding orbital. 41.5 (c) No, the N atom and F atom are too electronegative and would repel each other. 22.6 (d) Yes, F has a higher electronegativity than N, so they can form an ionic bond. 15.1

End of Session 3 Thanks for attending these three sessions! I appreciate all of your input about the module!

H2O and H2S at 90 If the bond angles of H2O and H2S are fixed at 90, there is more electronic congestion around O than S because the polarization is greater in H2O than in H2S.

H2O and H2S at 90 It’s even easier to see if the orbitals are superimposed on each other.

Reaction Network for O and H OH+ -e– H2O+ -e– H3O+ +H + +H+ +H + OH +H H2O +H O O– OH– +e–