Hybridization

Dr. HarrisLecture 11 (Ch 9.5-9.13)9/25/12HW: Ch 9: 21, 23, 39, 43, *75 *(show all carbon atoms, refer to pg 215)

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

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

Overlapping Valence Orbitals

• Recall s and p orbitals (ch 5)

• S orbitals are spherical. L = 0, mL = 0 • Max of 2 electrons

pz

• 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

S

Forming Sigma (σ) Bonds

σ

Covalent bond

1s1 1s1

H+

H+

H+

H+

σ

• Two overlapping atomic orbitals form a molecular bonding orbital.

• A sigma (σ) bonding orbital forms when s-orbitals overlap.

stabilization (energy drop)

Energy

Hybridization

• Imagine the molecule BeH2. We know that Be has a valence configuration of [He] 2s2 (Be-H bond is not ionic).

• 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 electron sharing

2s2

2p0

Be

1s1 1s1

2 H

X

ENER

GY

sp hybrid orbital

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 sp hybrid orbitals.

2s2

2psp hybridization

• Each sp orbital is 50% s character and 50% p character

Energy

• 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 destructively to 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

s pz

+ =

2 sp hybrid orbitals

z

z

Hybridization

sp-Hybridized Bonding

• Small negative lobes not shown. Recall that we have two unused p-orbitals along the x and y axes.

• 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

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.

2s2

2p1

B 1s1 1s1 1s1

3 H

Energy

Hybridization

• So, to make 3 bonding sites, 3 hybrid molecular orbitals are formed by mixing the 2s-orbital with two 2p-suborbitals.

• Each of these three hybrid orbitals are one-third s-character, and two-thirds p-character.

2s2

2p1

Bsp2 hybrid orbitals

unused 2p suborbital

ENER

GY

This figure illustrates the 3 hybrid orbitals combined with the unused 2p orbital, which is perpendicular to the hybrid orbitals.

sp2 orbitals

The result of adding the s and p orbitals together is a trigonal planar arrangement of electron domains

H+

empty 2p orbital

H+

H+

B

H

σ bond

HH

sp2 Geometry and Bonding

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)

2s2

2p2

C

Four sp3 hybrid orbitals

ENER

GY

Formation of sp3 Hybrid Orbitals

4 σ-bonds

The four hybrid orbitals arrange themselves tetrahedrally. Do you notice a trend yet?

What about molecules with lone electron pairs?

• Ex. What is the hybridization of H2O?• The valence electron configuration of O is [He]2s2 2p4

2p4

2s2

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??

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.

ENER

GY

Water has sp3 hybridization

2p4

2s2

1s1 1s1

2 H

ENER

GY

covalent bonding

O

σ bondsH2O

lone pair bonding electrons

HO

H

•• ••

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

sp3d and sp3d2 hybridization

• Atoms like S, Se, I, Xe … etc. can exceed an octet because of sp3d and sp3d2 hybridization (combination of an ns, np, and nd orbitals where n>3).

• This results in either trigonal bipyramidal or octahedral geometry

sp3d sp3d2

Exceeding an Octet. Example: SF6

3p4

3s2

Energy

3d0

sp3d2 hybrid orbitals

sp3

6 F

S

SF6

Fluorine lone pair

S

Exceeding an Octet. Example: SF6

F

3 lone pair

unpaired electronoverlap

x 6

Look Familiar ???

Examples:

• What is the hybridization of the central atom?• NH3

• NH4+

• PCl5

• CH3Cl

• SeF6

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

C CH

H

H

H

The hybridization of each carbon is sp2 because each is surrounded by three electron domains. The geometry around each C is trigonal planar.

sp2 sp2

2s2

2p2

sp2 hybrid orbitals

unhybridized p-electron

C CH

H

H

H

Forming Double Bonds

Carbon atoms

• We can see that for each carbon atom, the three sp2 electrons will be used to make 3 bonds. But how is the double bond formed?

All double bonds consist of 1 σ-bond and 1 π-bond

Double Bonds formed by simultaneous σ and π interaction

The remaining 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.

H+

H+

H+

H+

• •

Triple Bonds formed by 1 σ-bond and 2 π-bonds.Ex. HCN

H C N ••

sp sp

• Can you draw the orbital diagram for this molecule?

Examples

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

• CH3CH2CHCHCH3

• CH3CCCHCH2