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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 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 ).

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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Reaction of O, S and N with H Atoms

• The complete electron configurations for S could be written as

• Sulfur 1s22s22p63px23py

13pz1

• Again, we can “pair up” all electrons if S and two H atoms combine to form H2S. The valence bond picture suggests that all bond angles in H2O, NH3 and H2S should be 90o. This is close to the value seen in H2S (92o) but significantly underestimates bond angles in H2O (105o) and NH3 (107o).

Bonding in H2S represented by atomic orbital overlapFIGURE 11-3

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

The Methane “Problem”

• The ground state configurations for C can be written as 1s22s22px

12py1 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 CH2. The CH2 molecule does form but is unstable (a transient species). However, carbon “happily” reacts with H to form the methane, CH4.

Methane and Hybridization

• By experiment, as previously discussed, methane has a regular tetrahedral geometry – four equal bond distances and all bond angles of 109.5o. 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?

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: 1s22s22px12py

1

• Carbon Excited State: 1s22s12px12py

12pz1

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 sp3 hybrid orbitals (each containing a single unpaired electron). In step three the “hybridized C atom” reacts with four H atoms to form a CH4 molecule. The process is represented on the next few slides.

Hybridization of Atomic Orbitals

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

The sp3 hybridization schemeFIGURE 11-6

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 8 of 57

Bonding and structure of CH4

FIGURE 11-7

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 9 of 57

Organic Compounds and Structures: An Overview

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26 Slide 10 of 75

FIGURE 26-1

•Representations of the methane molecule

Organic Chemistry

• The next two slides illustrate what starts to happen when two or more sp3 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.

The ethane molecule C2H6

FIGURE 26-2

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26 Slide 12 of 75

The propane molecule, C3H8

FIGURE 26-3

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26 Slide 13 of 75

Alkanes from petroleum

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

Hybridization in NH3 and H2O

• The sp3 hybridization picture can also be used to discuss the bonding in NH3 and H2O. 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 sp3 hybrid orbitals. The following slides represent the process for N (NH3).

Bonding in H2O and NH3

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 16 of 57

Notice that hybrid orbitals can accommodate lone-pair electrons as well as bonding electrons.

sp3 hybrid orbitals and bonding in NH3

FIGURE 11-8

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 17 of 57

Class Examples

• 1. Jumping the gun “just a bit” let’s draw structural formulas for ethane (H3C-CH3), methyl amine (H3C-NH2), methanol (CH3-OH) and propane (H3C-CH2-CH3). What structural features do these molecules have in common? Does the “octet rule” still hold?

“Finding” sp3 Hybridized Atoms

• 2. Using the molecular structure of the morphine molecule shown on the next slide, find (a) an sp3 hybridized C atom, (b) an sp3 hybridized N atom and (c) an sp3 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.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26 Slide 20 of 75

Morphine , a very powerful and addictive painkiller, can be isolated from the opium poppy (Papaver somniferum).

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 1s22s22px

1 (say) to its first excited state 1s22s12px

12py1. The hybrid

orbitals formed here (from the combination of a single s orbital and two p orbitals are called sp2 hybrid orbitals. The sp2 hybridization scheme is invoked for the BF3 molecule.

sp2 Hybrid Orbitals

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

The sp2 hybridization scheme

FIGURE 11-9

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 23 of 57

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.

sp Hybridization

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

The sp hybridization schemeFIGURE 11-10

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 26 of 57

Class Example

• 3. The sp3 hybridization scheme can be invoked to explain the bonding in the silane (SiH4) molecule. What atomic orbitals on silicon would be used to construct hybrid orbitals?

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.

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?

Bonding in H2S represented by atomic orbital overlap

FIGURE 11-3

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 30 of 57

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).

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, H2C=CH2. Ethylene is a planar symmetric molecule (four identical C-H bonds and all bond angles near 120o).

• Aside: Hydrocarbons containing C=C double bonds are called unsaturated. They can react with H2 to from saturated hydrocarbons.

Carbon-Carbon Double Bonds – cont’d:

• In saturated hydrocarbons (such as propane H3C-CH2-CH3 the bonding behaviour of all C atoms is well explained using the sp3 hybridization scheme. In many unsaturated hydrocarbons the bonding of C atoms joined by double bonds is rationalized using an sp2 hybridization scheme.

Carbon-Carbon Double Bonds – cont’d:

• For the sp3 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 sp2 hybridization scheme we will “scramble” the single 2s orbital and two 2p orbitals to form three hybrid sp2 orbitals. A single 2p orbital remaining is used to from a pi bond.

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 11 Slide 35 of 57

Bonding in C2H4

H

H

H

H

C C

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 sp2 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 sp2 orbitals.

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.

Sigma (s) and pi (π) bonding in C2H4

FIGURE 11-14

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11 Slide 38 of 57

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?)

Esters

CH3CO2CH2(CH2)6CH3

The distinctive aroma and flavor of oranges are due in part to the ester octyl acetate,

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26 Slide 40 of 75