chemical essay creativity portion
Jichenyan 
Chapter9BCHEM101PreLesson.pdf
11/16/20
1
Chapter 9B: Advanced Bonding Theory
CHEM 101 Fall 2020
Dr. Lauren Genova
1
2
Chapter 9
By the end of this chapter, you will be able to:
• Differentiate between a sigma (σ) bond and pi (𝜋) bond
• Determine the hybrid orbitals (hybridization) associated with various molecular geometries
Chapter 9B: Learning Objectives
2
Fig. 10-1, p.439
• Valence bond theory states that covalent bonds are formed by overlapping and pairing electrons in atomic orbitals:
H: 1s
H: 1s
Valence Bond Theory
• We say that the orbitals on 2 different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space.
• If the overlap is constructive, electron density will increase in the region between the two nuclei, resulting in a bonding orbital (system is stabilized)
(max stability)
Chapter 9
3
• Two types of bonds are permitted in valence bond theory: – Sigma (σ) bonds – formed by end-to-end overlap of two
atomic orbitals (which may be s, p, or hybrid)
– Pi (𝛑) bonds – formed by side-to-side overlap of two atomic p orbitals
1.) Sigma and Pi Bonding
4
Chapter 9
4
Sigma Bonding • Sigma (σ) bond: Covalent bond in which the highest
electron density lies between the two atoms along the bond axis. – All single bonds are sigma bonds – Free rotation is allowed along the axis of a sigma bond
bond axis
bond axis
bond axis 5
Chapter 9
5
• Pi (𝛑) bond: Covalent bond in which electron density is greatest around—not along—the bonding axis. – A double bond contains 1 sigma and 1 pi bond – A triple bond contains 1 sigma and 2 pi bonds – Free rotation is not allowed with a pi bond (thus adding
rigidity to molecules which contain them) • This rigidity imparted by pi bonds has a major influence on
the conformations of complex molecules in solution
Pi Bonding
Double bond Triple bond
𝛑 σσ
σ 𝛑 σσ σ 𝛑
6
Chapter 9
6
11/16/20
2
Free Rotation Differences between σ and 𝛑
Image courtesy of: https://chemistry.stackexchange.com/questions/68480/why-can-a-sigma-bond-rotate
(σ)
(σ)
(𝛑)
bond axis
bond axis
bond axis
7
Chapter 9
7
Practice: Identifying σ and 𝛑 bonds 2.) How many sigma (σ) and pi (𝛑) bonds are in this molecule?
8
Chapter 9
8
• The overlap of atomic orbitals from valence bond theory results in predicted bond angles that differ from the observed bond angles (from VSEPR theory)
• Example: H2O
• We require a more detailed model to explain this discrepancy
H: 1s
O:H: 1s
Expected bond angle from p orbitals (valence bond theory) = 90o
Problem with Valence Bond Theory
Actual bond angle (VSEPR) = 104.5o104.5
o
Chapter 9
9
Observed reality: Tetrahedral (109.5o)
Hybridization • Hybridization is the concept of mixing atomic orbitals into new
hybrid orbitals (with different energies/shapes from the original atomic orbitals) suitable for the pairing of e– – Atomic orbitals can hybridize to maximize
distances between e–
– This helps to account for the problem with valence bond theory: namely, that the straight overlap of atomic orbitals doesn’t account for all observed molecular shapes.
Valence bond theory prediction:
Bond angles = 90o apart
Chapter 9
10
A Closer Look at Hybridization
Hybridization – the mixing of atomic orbitals to generate new sets of orbitals that form covalent bonds with other atoms.
Chapter 9
11
Scientist Spotlight Dr. Herdeline “Digs” Ardoña Assistant Professor, Department of Chemical and Biomolecular Engineering, University of California, Irvine (Irvine, CA)
Topic: Hybridization (with a focus on pi-stacking)
Favorite quote: “Winners never quit on something that makes and/or will make them happy.”
Email: [email protected]
Chapter 9
12
11/16/20
3
Hybridization and Geometry • Goal for this unit:
to determine the hybridization of a molecule
• How? By identifying its electron-pair geometry!
Recall: electron-pair geometry includes bonding pairs (electrons shared between 2 atoms) AND nonbonding pairs (lone pairs)
Chapter 9
13
Number of connections
Type of hybridization
Molecular geometry
2 sp Linear
3 sp2 Trigonal planar
4 sp3 Tetrahedral
5 sp3d Trigonal bipyramidal
6 sp3d2 Octahedral
Dr. G’s method for determining hybridization: 1.) Count the number of connections (number of orbitals) around the central atom whose hybridization you have been asked to identify
• Order when counting number of connections: s, p, p, p, d, d • INCLUDE THE LONE PAIRS while you’re counting!
2.) Number of connections = hybridization!
3.) Hybridization and Geometry
• Goal for this unit: to determine the hybridization of a molecule
• How? By identifying its electron-pair geometry!
Chapter 9
14
These two unhybridized p orbitals ultimately lead to a triple bond (one σ and two 𝛑 bonds)
Linear (2 connections): sp Hybridization
• Formed by mixing one s and one p orbital
• Example: each C in H–C≡C–H
s p
s p
s, p, p, p, d, d
sp hybridized
15
Chapter 9
15
Trigonal Planar (3 connections): sp2 Hybridization This unhybridized p orbital ultimately leads to a double bond (one σ and one 𝛑 bond)
• Formed by mixing one s and two p orbitals 16
Chapter 9
16
Trigonal Planar (3 connections): sp2 Hybridization
• Formed by mixing one s and two p orbitals • Examples: the C in H2CO, the O in H2CO
s
p p
s p
p
s, p, p, p, d, d sp2 hybridized
This unhybridized p orbital ultimately leads to a double bond (one σ and one 𝛑 bond)
17
Chapter 9
17
Tetrahedral (4 connections): sp3 Hybridization
s
p
p
p
• Formed by mixing one s and three p orbitals • Example: CH4 – four sigma bonds
s, p, p, p, d, d
sp3 hybridized
18
Chapter 9
18
11/16/20
4
Other sp3 Hybrid Examples
• Formed by mixing one s and three p orbitals • Example: NH3 (3 sigma bonds, 1 lone pair)
Remember to count the lone pairs (non-bonding e–)!s
p
p
p
s, p, p, p, d, d
sp3 hybridized
19
Chapter 9
19
Other sp3 Hybrid Examples
• Formed by mixing one s and three p orbitals • Example: H2O (2 sigma bonds, 2 lone pairs)
Remember to count the lone pairs (non-bonding e–)!s
pp
p
s, p, p, p, d, d
sp3 hybridized
20
Chapter 9
20
Trigonal Bipyramidal (5 connections): sp3d Hybridization
• Formed by mixing one s, three p, and one d orbital
• Example: PF5 – six sigma bonds
s
p
p p
d s, p, p, p, d, d
sp3d hybridized
21
Chapter 9
21
• Formed by mixing one s, three p, and two d orbitals
• Example: SF6 – six sigma bonds
Octahedral (6 connections): sp3d2 Hybridization
s
p
p p
d
d s, p, p, p, d, d
sp3d2 hybridized
22
Chapter 9
22
4.) Describe the hybridization around each central atom in the amino acid glycine.
O
C O H
H
H NH H
sp3
sp3
sp3
sp2 ••
•• ••C
1 2
4
3
Hybridization Practice
23
Chapter 9
23
