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Chapter9BCHEM101PreLesson.pdf

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Chapter 9B: Advanced Bonding Theory

CHEM 101 Fall 2020

Dr. Lauren Genova

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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

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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)

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• 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

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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

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• 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

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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

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Practice: Identifying σ and 𝛑 bonds 2.) How many sigma (σ) and pi (𝛑) bonds are in this molecule?

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• 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

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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

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A Closer Look at Hybridization

Hybridization – the mixing of atomic orbitals to generate new sets of orbitals that form covalent bonds with other atoms.

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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: hardona@uci.edu

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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)

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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!

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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

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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

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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)

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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

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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

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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

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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

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• 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

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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

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Hybridization Practice

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