Hybridization and Delocalized Molecular

OrbitalsChemistry 2 AP

Why Hybridization?We cover molecular orbital theory next

How does observed bonding fit in with our orbital picture? Look at CH4:

Carbon make 4 bondsH-C-H bond angle is 109.5°–tetrahedralAll bonds are the same length (i.e. strength)

C HH

H

H

CHH H

H

C HH

H

H

CHH H

H

How is this Possible?Start with configuration: 1s22s22p2:

CH2 does not exist, so 2s electrons must not stay paired!One 2s electron promoted to 2p:Energy levels still not equal! What happens?

2p2p2p

2s

1s

Energy

HybridizationTo obtain 4 equivalent orbitals, mix one s andthree p orbitals to make four sp3 atomic orbitals:

2p2p2p

2s

Energy

HybridizationTo obtain 4 equivalent orbitals, mix one s andthree p orbitals to make four sp3 atomic orbitals:

Slightly higher energy than 2s and lower than 2pin energy4 orbitals in → 4 orbitals outAll 4 electrons are equivalent

2sp3

2p

2sEnergy

Geometry of sp3

Bonds spread out (VSEPR), symmetrically arranged: tetrahedral:We ignore small lobe opposite nucleusOverlap of H 1s and C 2sp3 makes a σ bond

All tetrahedral molecules (AB4, AB3E, AB2E2) are sp3 hybridized

Additional sp3 MoleculesNH3: lone pair occupies one 2sp3 orbital (H2O has 2 lone pairs in two 2sp3 orbitals)

CH3CH3 has two C sp3 orbitals overlapping

Trigonal Planar Electron Geometry3 electron regions, only need 3 equivalent orbitals: s + p + p → three sp2 hybridized orbitalsBH3 (2s22p1)

2p2p2p

2s

Energy

Trigonal Planar Electron Geometry3 electron regions, only need 3 equivalent orbitals: s + p + p → three sp2 hybridized orbitalsBH3 (2s22p1)

1 unoccupied p orbital remains unhybridized perpendicular to plane of moleculeAll trigonal planar molecules (AB3, AB2E) sp2 hybridized

2p2sp2

2p

2sEnergy

3 orbitals in → 3 orbitals out

Hybridization & Double BondsHow does hybridization explain multiple bonds?

Look at C2H4:

Lewis structure indicates each C is trigonal planarAccording to VBT, trigonal planar geometry and 120° bond angles indicates sp2 hybridization:

But what happens to the p electron?

2p2p2p

2s

Energy

2p

2sp2

2p

2sEnergy

C C

H

H

H

H120°

120°

π BondingSince each C is sp2 hybridized, each has 3 equivalent sp2 orbitals 120° apart.

Each C has 1 unpaired electron in a p orbital which interact above and below the plane of the molecule to form a π bond.

C C

H

H

H

H120°

120°

Double Bonds & Isomerizationcis (I) & trans (II) geometric isomers of 1,2-dichloroethene

π bond constrains rotation about the C-C bond, causing a high-energy barrier

Barrier is not present in 1,2-dichloroethane

Single bond relatively free to rotate

Cl

H

Cl

H

Cl

H

H

Cl

I II

C C

Cl

ClH

H

H

H

Double Bonds & Isomerizationcis (I) & trans (II) geometric isomers of 1,2-dichloroethene

π bond constrains rotation about the C-C bond, causing a high-energy barrier

Barrier is not present in 1,2-dichloroethane

Single bond relatively free to rotate

Cl

H

Cl

H

Cl

H

H

Cl

I II

C C

Cl

HH

Cl

H

H

Linear Electron Geometry2 electron regions, need 2 equivalent orbitals: s + p → two sp hybridized orbitals

BeH2 (2s2)

2p2p2p

2s

Energy

Linear Electron Geometry2 electron regions, need 2 equivalent orbitals: s + p → two sp hybridized orbitals

BeH2 (2s2)

2 unoccupied p orbitals remain unhybridized, perpendicular to sp orbitals and each otherAll AB2 atoms sp hybridized

2p2sp

2p

2sEnergy

2 orbitals in → 2 orbitals out

Triple BondsEthyne (acetylene) is linear: There are 2 electron regions on each C, each C geometry is linear, so hybridization must be sp.

Two C valence electrons occupy an sp orbital, two electrons occupy perpendicular p orbitals:

2p2p2p

2s

Energy

2p

2sp

2p

2sEnergy

π Bonding: Triple Bonds

Each set of p orbitals forms a π bond

π Bonding: Triple Bonds

Each set of p orbitals forms a π bond

π Bonding: Triple Bonds

Each set of p orbitals forms a π bond

π Bonding: Triple Bonds

Each set of p orbitals forms a π bondAny triple bond contains 1 σ and 2 π bondsAny atom with 2 π bonds must be sp hybridized

Can have 1 single and 1 triple bond: Or 2 double bonds: (e.g. CO2)Each π bond 90° from the other

Expanded OctetsFormerly invoked hybridization involving d orbitals: sp3d and sp3d2

New evidence indicates that this model is not valid for non-metalsWill still have d-orbitals involved in bonding for formation of complexes of transition metals (Chapter 16).

Current models out of scope of this course—will never be asked for hybridization of trigonal bipyramidal or octahedral geometry molecules.

Delocalized Molecular OrbitalsMolecules that exhibit resonance have an extended π-electron systemElectrons are delocalized over the molecule, e.g. benzene:

The orbital picture of benzene thus looks like:

Nitrate ion (NO3–)The true structure of bonding is the hybrid, or average of the 3 resonance structures, where the two π electrons are delocalized over the entire system:

The orbital picture looks like:This corresponds to the true structure in which all 3 N-O bonds are equivalent, with lengths indicating bond order between single and double.