Lecture 11 Covalent Bonding Pt 3: Hybridization (Ch. 9.5-9.13)

1. Lecture 11Covalent Bonding Pt 3: Hybridization (Ch. 9.5-9.13) Dr. Harris Suggested HW: Ch 9: 23, 25, 29, 39, 43, 72, *75 (For 25 and 43, you are illustrating the hybridization of the atomic orbitals into hybrid orbitals and the overlapping of these hybrid orbitals as described in the examples provided. *(For 75, show all carbon atoms. Refer to pg 215)

2. Introduction • We now know that atoms can bond covalently through the sharing of electrons • VSEPR theory helps us predict molecular shapes. But, it does not explain what bonds are, how they form, or why they exist. • In ch 9, chemical bonding will be explained in terms of orbitals

3. Covalent Bonding Is Due to Orbital Overlap • In a covalent bond, electron density is concentrated between the nuclei. • Thus, we can imagine the valence orbitals of the atoms overlapping • The region of orbital overlap represents the covalent bond

4. Overlapping Valence Orbitals • Recall s and p orbitals (ch 5) • S orbitals are spherical. L = 0, mL = 0 • Max of 2 electrons S • P orbitals consist of two lobes of electron density on either side of the nucleus. • L= 1, mL = -1, 0, 1 (3 suborbitals) • Max of 6 electrons px py pz

5. Forming Sigma (σ) Bonds Energy σ + + + + H H H H Covalent bond stabilization (energy drop) 1s1 1s1 • Two overlapping atomic orbitals form a molecular bonding orbital. Plus sign indicates phase of electron wave, NOT CHARGE • Asigma(σ) bonding orbital forms when s-orbitals overlap. σ

6. Hybridization • Imagine the molecule BeH2. We know that Be has a valence configuration of [He] 2s2 (Be-H bond is polar covalent). • However, when we fill our orbitals in order as according to Hund’s rule, we notice that there are no unpaired electrons. Hence, we can not make any bonds. Stay mindful of the fact that a covalent bond involves the sharing of unpaired electrons X Be 2p0 2 H ENERGY 2s2 1s1 1s1

7. Hybridization • So how does BeH2 form? How can Beryllium make 2 bonds? • To bond with 2 hydrogen atoms, Beryllium mixes (hybridizes) two of its atomic orbitals. This creates two sphybrid orbitals. Energy sp hybridization 2p sp hybrid orbital 2s2 • The name “sp” originates from the fact that the orbitals form from the combination of 1 s-orbital and 1 p-orbital. Each sp orbital is 50% s character and 50% p character

8. Hybridization pz s z + = z • The addition of an s-orbital to a pz orbital is shown above. The s orbital adds constructively to the (+)lobe of the pz orbital and adds destructivelyto the lobe that is in the opposite phase (-). The symbols indicate phase, not charge. • Remember, we are adding two atomic orbitals, so we will obtain two hybrid orbitals 2 sp hybrid orbitals

9. sp-Hybridized Bonding • Small negative lobes not shown. Recall that we have two empty, unused p-orbitals along the x and y axes(blue). • Now, 2 Hydrogen s-orbitals can overlap with the Be sp-hybrid orbitals to form BeH2. We would expect BeH2 to be linear, as is predicted by VSEPR

10. sp2 Hybridization • Whenever we mix a certain number of s and p atomic orbitals, we get the same number of molecular orbitals. This is called the principle of conservation of orbitals. • The BH3 molecule gives us an example of sp2 hybrid orbitals. • Once again, we have a situation where we don’t have enough bonding sites to accommodate all of the hydrogens. Energy 3 H 2p1 2s2 B 1s1 1s1 1s1

11. Hybridization • So, to make 3 bonding sites, 3 hybrid molecular orbitals are formed by mixing the 2s-orbital with two 2p-suborbitals. • This forms an sp2 orbital. Each of these three hybrid orbitals are one-thirds-character, and two-thirds p-character. unused 2p suborbital 2p1 ENERGY B sp2 hybrid orbitals 2s2

12. sp2 orbitals The result of adding the s and p orbitals together is a trigonal planar arrangement of electron domains This figure illustrates the 3 hybrid orbitals combined with the unused 2p orbital, which is perpendicular to the hybrid orbitals.

13. sp2 Geometry and Bonding empty 2p orbital σ bond + + + H H H H B H H

14. sp3 Hybridization • Involves the mixing of an s-orbital and 3 p-orbitals. The resulting hybrid orbitals are one-fourth s-character and three-fourths p-character. • Ex. CH4. To accommodate 4 hydrogen atoms, 4 hybrid orbitals are created (C: [He] 2s22p2) C 2p2 ENERGY Four sp3 hybrid orbitals 2s2

15. Formation of sp3 Hybrid Orbitals The four hybrid orbitals arrange themselves tetrahedrally. Do you notice a trend yet? 4 σ-bonds

16. What about molecules with lone electron pairs? • Ex. What is the hybridization of H2O? • The valence electron configuration of O is [He]2s2 2p4 As you see, there are two unpaired O electrons. Does this mean that these two p-suborbitals can overlap with the two H 1s orbitals without hybridizing?? 2p4 ENERGY 2s2 No. The reason is that we now have two sets of lone pairs of electrons that are substantially different in energy. The orbitals will hybridize to form degenerate(equal energy) sets of electrons.

17. Water has sp3 hybridization •• •• O H H 2p4 covalent bonding bonding electrons lone pair 2s2 2 H ENERGY O 1s1 1s1 σ bonds H2O

18. Water has sp3 hybridization • The angle between the sp3 hybrid orbitals in water is 104.5o, NOT 109.5o as expected in a normal tetrahedron • LONE PAIR REPEL THE ELECTRONS IN THE O-H BONDS Strength of Repulsion Lone pair – Lone pair > Lone pair – Bonding pair > Bonding pair- Bonding pair

19. Examples • Determine the hybridization of the central atom in each of the following molecules? • NH3 • CO2 • N2 • NO2-

20. sp3d and sp3d2 hybridization • Atoms like S, Se, I, Xe… etc. can exceed an octet because of sp3d and sp3d2 hybridization (combination of ns, np, and nd orbitals where n>3). • This results in either trigonalbipyramidal or octahedral skeletal geometry sp3d sp3d2

21. Exceeding an Octet. Example: SF6 Energy 3d0 sp3d2 hybrid orbitals 3p4 Fluorine lone pair 3s2 S sp3 6 F SF6

22. Exceeding an Octet. Example: SF6 unpaired electron overlap F x 6 S 3 lone pair

23. Look Familiar ???

24. Examples: • What is the hybridization of the central atom? • IF5 • PCl5 • SeOF4

25. Double and Triple Bonding • How can orbital overlap be used to explain double and triple bonds? What kind of interactions are these? • Lets look at ethene, C2H4 The hybridization of each carbon is sp2because each is surrounded by three electron domains. The geometry around each C is trigonal planar. H H sp2 sp2 C C H H

26. Forming Double Bonds Carbon atoms unhybridized p-electron 2p2 sp2 hybrid orbitals H H 2s2 C C H H • We can see that for each carbon atom, three sp2 orbitals are required, each containing an unpaired electron, to make three sigma bonds. But how is the double bond formed?

27. Double Bonds formed by simultaneous σ and π interaction • • + + + + H H H H The remaining p-electrons form a π bond. This bond forms due to attraction between the parallel p-orbitals. The like-phase regions are drawn toward one another and overlap. All double bonds consist of 1 σ-bond and 1 π-bond

28. Triple Bonds formed by 1 σ-bond and 2 π-bonds.Ex. HCN sp sp • • H C N • Can you draw the orbital diagram for this molecule?

29. Examples • How many σ and π bonds are in each of the following molecules? Give the hybridization of each carbon. • CH3CH2CHCHCH3 • CH3CCCHCH2