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Reaction of O, S and N with H Atoms The complete electron configurations for S could be written as Sulfur 1s 2 2s 2 2p 6 3p x 2 3p y 1 3p z 1 Again, we.

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Presentation on theme: "Reaction of O, S and N with H Atoms The complete electron configurations for S could be written as Sulfur 1s 2 2s 2 2p 6 3p x 2 3p y 1 3p z 1 Again, we."— Presentation transcript:

1 Reaction of O, S and N with H Atoms The complete electron configurations for S could be written as Sulfur 1s 2 2s 2 2p 6 3p x 2 3p y 1 3p z 1 Again, we can “pair up” all electrons if S and two H atoms combine to form H 2 S. The valence bond picture suggests that all bond angles in H 2 O, NH 3 and H 2 S should be 90 o. This is close to the value seen in H 2 S (92 o ) but significantly underestimates bond angles in H 2 O (105 o ) and NH 3 (107 o ).

2 Bonding in H 2 S represented by atomic orbital overlap FIGURE 11-3 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 2 of 57

3 The Methane “Problem” The ground state configurations for C can be written as 1s 2 2s 2 2p x 1 2p y 1 Using the valence bond picture and the concept of paired electrons in molecular orbitals we might expect C to react with H atoms to form CH 2. The CH 2 molecule does form but is unstable (a transient species). However, carbon “happily” reacts with H to form the methane, CH 4.

4 Methane and Hybridization By experiment, as previously discussed, methane has a regular tetrahedral geometry – four equal bond distances and all bond angles of o. The regular geometry of methane and its ability to form four bonds can be explained using the concept of hybridization. How have we explained carbon’s tendency to form four bonds previously?

5 Methane and Hybridization – cont’d: In the hybridization picture we imagine methane being formed from C and H atoms in three steps. In the first step we take a ground state C atom and excite one electron (from the 2s orbital) to form the lowest lying ( or first) excited state. Carbon Ground State: 1s 2 2s 2 2p x 1 2p y 1 Carbon Excited State: 1s 2 2s 1 2p x 1 2p y 1 2p z 1

6 Methane and Hybridization – cont’d: In the second step we imagine “combining” the single occupied 2s orbital and the three occupied 3p orbitals in the excited to form four equivalent sp 3 hybrid orbitals (each containing a single unpaired electron). In step three the “hybridized C atom” reacts with four H atoms to form a CH 4 molecule. The process is represented on the next few slides.

7 Hybridization of Atomic Orbitals Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 7 of 57

8 The sp 3 hybridization scheme FIGURE 11-6 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 8 of 57

9 Bonding and structure of CH 4 FIGURE 11-7 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 9 of 57

10 Organic Compounds and Structures: An Overview Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 10 of 75 FIGURE 26-1 Representations of the methane molecule

11 Organic Chemistry The next two slides illustrate what starts to happen when two or more sp 3 hybridized carbons are linked. The chemistry of carbon is infinitely varied and organic compounds are part of all of living things, important energy sources, key pharmaceuticals and so on.

12 The ethane molecule C 2 H 6 FIGURE 26-2 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 12 of 75

13 The propane molecule, C 3 H 8 FIGURE 26-3 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 13 of 75

14 Alkanes from petroleum Copyright © 2011 Pearson Canada Inc. Slide 14 of 75General Chemistry: Chapter 26

15 Hybridization in NH 3 and H 2 O The sp 3 hybridization picture can also be used to discuss the bonding in NH 3 and H 2 O. The neutral N and O atoms have more valence electrons than does C. We thus end up putting either one lone pair of electrons (for N) or two lone pairs of electrons (for O) into sp 3 hybrid orbitals. The following slides represent the process for N (NH 3 ).

16 Bonding in H 2 O and NH 3 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 16 of 57 Notice that hybrid orbitals can accommodate lone- pair electrons as well as bonding electrons.

17 sp 3 hybrid orbitals and bonding in NH 3 FIGURE 11-8 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 17 of 57

18 Class Examples 1. Jumping the gun “just a bit” let’s draw structural formulas for ethane (H 3 C-CH 3 ), methyl amine (H 3 C-NH 2 ), methanol (CH 3 -OH) and propane (H 3 C-CH 2 -CH 3 ). What structural features do these molecules have in common? Does the “octet rule” still hold?

19 “Finding” sp 3 Hybridized Atoms 2. Using the molecular structure of the morphine molecule shown on the next slide, find (a) an sp 3 hybridized C atom, (b) an sp 3 hybridized N atom and (c) an sp 3 hybridized O atom. There may be more than one example of each. The example is a little unfair since many “non-terminal” H atoms are not shown in this structure.

20 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 20 of 75 Morphine, a very powerful and addictive painkiller, can be isolated from the opium poppy (Papaver somniferum).

21 Hybridization in B and Be Compounds A hybridization scheme can be invoked for B that involves exciting a B atom from its ground electronic state 1s 2 2s 2 2p x 1 (say) to its first excited state 1s 2 2s 1 2p x 1 2p y 1. The hybrid orbitals formed here (from the combination of a single s orbital and two p orbitals are called sp 2 hybrid orbitals. The sp 2 hybridization scheme is invoked for the BF 3 molecule.

22 sp 2 Hybrid Orbitals Copyright © 2011 Pearson Canada Inc. Slide 22 of 57General Chemistry: Chapter 11

23 The sp 2 hybridization scheme FIGURE 11-9 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 23 of 57

24 sp Hybridization sp hybridization is important for molecules such as H-C≡N and H-C≡C-H. Acetylene (ethyne) is the first member of an important series of organic compounds, the alkynes.

25 sp Hybridization Copyright © 2011 Pearson Canada Inc. Slide 25 of 57General Chemistry: Chapter 11

26 The sp hybridization scheme FIGURE Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 26 of 57

27 Class Example 3. The sp 3 hybridization scheme can be invoked to explain the bonding in the silane (SiH 4 ) molecule. What atomic orbitals on silicon would be used to construct hybrid orbitals?

28 Covalent Bonds – Orbital Overlap – Sigma and Pi Bonds The formation of both sigma bonds (σ bonds) and pi bonds (π bonds) is likely familiar. Sigma bonds are formed (for a pair of atoms) by the overlap of atomic orbitals “pointing” towards, in each case, the other bonded atom. We can use s and p orbitals (etc) to form sigma bonds.

29 Covalent Bonds – Orbital Overlap – Sigma and Pi Bonds – cont’d: In sigma bonds the two atomic orbitals used to “construct” the molecular orbital overlap in the most spatially “direct” manner possible. The overlap of a H 1s orbital and S 3p orbitals is shown on the next slide. As an aside, one can see that it difficult to “point” an s orbital in any direction. Why?

30 Bonding in H 2 S represented by atomic orbital overlap FIGURE 11-3 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 30 of 57

31 Pi Bonds Pi (π) bonds are formed when the bonding electron pair is placed in a molecular orbital formed (frequently) by p orbitals on adjacent atoms overlapping. The p orbitals on the bonded atoms are oriented perpendicular to the internuclear axis (which makes orbital overlap slightly less favourable).

32 Carbon-Carbon Double Bonds The carbon-carbon double bonds in common organic molecules are comprised of one σ bond and one π bond. The simplest molecule of this type is ethylene, H 2 C=CH 2. Ethylene is a planar symmetric molecule (four identical C- H bonds and all bond angles near 120 o ). Aside: Hydrocarbons containing C=C double bonds are called unsaturated. They can react with H 2 to from saturated hydrocarbons.

33 Carbon-Carbon Double Bonds – cont’d: In saturated hydrocarbons (such as propane H 3 C-CH 2 -CH 3 the bonding behaviour of all C atoms is well explained using the sp 3 hybridization scheme. In many unsaturated hydrocarbons the bonding of C atoms joined by double bonds is rationalized using an sp 2 hybridization scheme.

34 Carbon-Carbon Double Bonds – cont’d: For the sp 3 hybridization scheme (for carbon!) we imagined distributing four valence electrons among a 2s and the three 2p atomic orbitals and “scrambling” these orbitals together to form four hybrid orbitals. For the sp 2 hybridization scheme we will “scramble” the single 2s orbital and two 2p orbitals to form three hybrid sp 2 orbitals. A single 2p orbital remaining is used to from a pi bond.

35 Multiple Covalent Bonds Ethylene has a double bond in its Lewis structure. VSEPR says trigonal planar at carbon. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 35 of 57 Bonding in C 2 H 4 H H H H C C

36 Bonding in Ethylene – an Information Packed Slide! The next slide contains lots of information. Upper left corner – a representation (for C) showing the structure and disposition in space of the three hybrid sp 2 orbitals and the “left over” carbon p atomic orbital. Upper right corner – picture showing how the five sigma bonds in ethylene are formed using four H 1s orbitals and six carbon sp 2 orbitals.

37 Bonding in Ethylene – an Information Packed Slide – cont’d! Lower left corner. The sigma bonds are in place and the p orbitals on the two C atoms have “not yet” overlapped to from a pi bond. In fact the p orbitals have been drawn slightly “smaller than life” for clarity. Bottom center and right. Two representations of the pi bond in ethylene. Note the electron density well away from the C-C internuclear axis.

38 Sigma (  ) and pi (π) bonding in C 2 H 4 FIGURE Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 38 of 57

39 Class examples 4. The ethanoic acid (“vinegar”) moleculae and the methyl ethanoate molecule (an ester) shown on the next slide contain C=O double bonds. What is the hybridization of the C and O atoms in these double bonds? (Mention acetone, acetaldehyde, formaldehye?)

40 Esters CH 3 CO 2 CH 2 (CH 2 ) 6 CH 3 The distinctive aroma and flavor of oranges are due in part to the ester octyl acetate, Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 40 of 75

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