Chemistry 2

Lecture 2

Particle in a box approximation

Be able to predict the geometry of a hydrocarbon from its structure

and account for each valence electron. Predict the hybridization of

atomic orbitals on carbon atoms.

Assumed knowledge

Learning outcomes

• Be able to explain why confining a particle to a box leads to

quantization of its energy levels

• Be able to explain why the lowest energy of the particle in a box is

not zero

• Be able to apply the particle in a box approximation as a model for

the electronic structure of a conjugated molecule (given equation for

En).

ππππ orbitals of CO

π*

More carbon character

C 2p

O 2p

π*

πThe unequal energies of C and O

2p orbitals means that the mixing is O 2pπ2p orbitals means that the mixing is

incomplete. This results in the p

bonding orbital being polarized to

the oxygen, and the antibonding

orbital polarized to the carbon. This

is important for inorganic

chemistry…More oxygen character

σσσσ orbitals of CO

“2sσ” “2sσ*” “2pσ”

σ-bonding orbitalσ-bonding orbital

Polarized to oxygenlone pairs. Highest occupied σ orbital is localized on

carbon lone pair… this is important for inorganic chemistry

:C≡O:

sp hybridization

• When one mixes the 2s and the 2p to make an sp hybrid orbital, one must obtain, afterwards, two orbitals. These are the “plus” and the “minus” combinations.“minus” combinations.

acetylene (ethyne)

•There are 10 valence electrons

•2 go into C-C bonding σ orbital (sp + sp)σ

•2 go into one of the C-H bonding σorbitals (sp + 1s)

•2 go into the other C-H bond

•We are left with pristine (unhybridized) 2px and 2py orbitals.

•2 electrons go into 2px π bonding orbital

σ

σ

•2 electrons go into 2px π bonding orbital

•2 electrons go into 2py π bonding orbital

•CC bond is triple – σ and 2 × π

•“minus” combinations are anti-bonding and are not occupied.

π

π

H C C H

sp2 hybridization

• In order to make two σ bonds, sp hybrids are constructed. In some molecules, such as ethylene (ethene), the carbon atom must make three sigma bonds. To achieve this, the 2s orbital is mixed with two three sigma bonds. To achieve this, the 2s orbital is mixed with two 2p orbitals to make three sp2 hybrid orbitals.

ethylene (ethene)

•There are 12 valence electrons

•2 go into C-C bonding σ orbital (sp2 + sp2)

σsp2)

•2 go into each the C-H bonding σorbitals (sp2 + 1s) (8 electrons)

•We are left with pristine (unhybridized) 2pz orbitals.

•2 electrons go into 2p π bonding orbital

σ

•CC bond is double – σ and π

•“minus” combinations are anti-bonding and are not occupied.

π

sp3 hybridization

• In order to make four bonds, sp3 hybrids are constructed. All saturated carbon atoms are sp3 hybridized . To achieve this, the 2s orbital is mixed with two 3p orbitals to make four sp3 hybrid orbitals.orbital is mixed with two 3p orbitals to make four sp3 hybrid orbitals.

•There are four sp3 hybrids since they are constructed from four atomic orbitals, •There are four sp3 hybrids since they are constructed from four atomic orbitals, (2s + 3 × 2p).

•They point at the corners of a tetrahedron and are 109°apart.

sp3 hybridization

• The bonding orbitals in methane are sums of sp3 hybrids on the carbon, and 1s orbitals of the hydrogen.on the carbon, and 1s orbitals of the hydrogen.

1s

sp3C

H

H

HH

sp3 hybridization

• All tetrahedral atoms are considered to be sp3 hybridized.

O

H

: N

H

:O :

:H

N :

HH

water ammonia

σ-framework

• The electronic structure of most molecules may be simplified by considering the σ framework to be constructed from σ-bondingorbitals, with anti-bonding combinations unoccupied.orbitals, with anti-bonding combinations unoccupied.

• Where carbons are sp2, or sp hybridized, there are unhybridized p orbitals which may be used to construct π-bonds.

H H

π-system

C C

H

H

H

C

H

H

H

σ-framework

σσσσ interactions: ethylene

C-C σ bond appears to be positive overlap of sp2 hybridsC-C σ bond appears to be positive overlap of sp2 hybrids

σσσσ interactions

no nodes

(…ignoring inside)

Same energy,

interact and mix

one node

one node

two nodes

two nodes

σσσσ orbitals do not interact with ππππ orbitals

0ˆ =∫ τψψ dH 0ˆ12 =∫ τψψ dH

σψ πψ

cancels

There is no net interaction between these orbitals.

The positive-positive term is cancelled by the positive-negative term

ππππ orbital of ethylene

C-C π bond appears to be positive overlap of unhybridized p orbitals

σσσσ orbitals of cis-butadiene

The σ molecular orbitals are mixtures of the sp2 hybrids

π π π π orbitals of cis-butadiene

three nodes

The four p orbitals all

have the same energy

two nodes

one nodehave the same energy

… interact and mix

no node

(ignore plane)

π π π π orbitals of hexatriene

The six p orbitals all

have the same energy

… interact and mix… interact and mix

Maybe we can model the

π-electrons as confined to

a one dimensional “box”…?

The Schrödinger equation

• The total energy is extracted by the Hamiltonian operator.Hamiltonian operator.

• These are the “observable” energy levels of a quantum particle

Energy eigenfunction

Energy eigenvalueHamiltonian operator

The Schrödinger equation

• The Hamiltonian has parts corresponding to Kinetic Energy and Potential Energy. In to Kinetic Energy and Potential Energy. In one dimension, x:

Kinetic EnergyHamiltonian operator

Potential Energy

“The particle in a box”

• The box is a 1d well, with sides of infinite potential, where the electron cannot be…potential, where the electron cannot be…

E

x

0

0 L

“The particle in a box”

• The box is a 1d well, with sides of infinite potential, where the electron cannot be…potential, where the electron cannot be…

E

x

0

0 L

“The particle in a box”

• The particle cannot exist outside the box… Ψ = 0 {x<0;x>L (boundary conditions)Ψ = 0 {x<0;x>L (boundary conditions)

E

x

0

0 L

? ?

“The particle in a box”

• Let’s try some test solutionsΨ = sin(πx/L) {x>0;x<LΨ = sin(πx/L) {x>0;x<L

E

x

0

0 L

“The particle in a box”

Zero potential inside box

=εΨ !!!εΨ !!!εΨ !!!εΨ !!!

“The particle in a box”

• Lowest energy possible is ε = ℏ2 π2/2mL2

• This is called zero point energy (ZPE)

V

• This is called zero point energy (ZPE)

x

0

0 L

“The particle in a box”

• Other solutions?Ψ = sin(2πx/L) {x>0;x<LΨ = sin(2πx/L) {x>0;x<L

V

x

0

0 L

“The particle in a box”

• Other solutions?Ψ = sin(3πx/L) {x>0;x<LΨ = sin(3πx/L) {x>0;x<L

V

x

0

0 L

“The particle in a box”

• Other solutions?Ψ = sin(4πx/L) {x>0;x<LΨ = sin(4πx/L) {x>0;x<L

V

x

0

0 L

“The particle in a box”

• Other solutions?Ψ = sin(nπx/L) {x>0;x<LΨ = sin(nπx/L) {x>0;x<L

V εn = ℏ2n2π2/2mL2

x

0

0 L

?

“The particle in a box”

Ψ = sin(nπx/L) {x>0;x<L;n>0ℏεn = ℏ2n2π2/2mL2

Philosophical question: why is n = 0 not an appropriate solution?

Hint: what’s the probability of observing the particle?

“The particle in a box”Ψ = sin(nπx/L) {x>0;x<L;n>0 εn = ℏ2n2π2/2mL2

εn

0

Consequences of confinement

• For a particle in a box, there exist solutions to the Schrödinger equation (eigenfunctions) of the the Schrödinger equation (eigenfunctions) of the form Ψ = sin(nπx/L) with energies (eigenvalues)

εn = ℏ2n2π2/2mL2

• These are quantized energy levels!

• Increasing L – decreases spacing… • Increasing L – decreases spacing… • Increasing m – decreases spacing…

(correspondence principle)

Boundary conditions

We impose the boundary conditions on imaginary atoms one bond length beyond the edge of the π-system.

A word about nitrogen

When a nitrogen is adjacent to a π-system, it may possess sp2 hybridized resonance forms which allow it to participate in that π-system.

e.g. amides

If the resonance forms are not favourable, they may only contribute in a minor way and the nitrogen will still have a pyramidal structure to some extent. In amides, the oxygen stabilizes the resonance structure and the nitrogen is planar.

Application: Polymethine dyes

Adapted from Brooker, JACS 62, 1116 (1940).

Application: Polymethine dyes

Question: how many π-electrons in the a-Brooker dye?

A = ε.c.l (Beer’s Law)

Answer: firstly, we should recognize that the π -system extends over both nitrogens. There are two resonance forms. By symmetry these will involve double bonds with either nitrogen.

Looking carefully at the structure, there are 2a+3 carbon atoms which each contribute one electron each. There are two nitrogens, one which contributes two electrons, and one which contributes one electron, since it bears a positive charge. Therefore, there are 2a+6 electrons.

Polymethine dyes

Question: what is the length over which the π-electrons are delocalized, if the average bond length is 1.40 Å?

Question: if the energy levels of the electrons are given by εn = ℏ2n2π2/2mL2,

Answer: Looking carefully at the structure there are 2a+4 bonds in the π -system. If we add a bond at each end of the system, we get 2a+6 bonds in length, which equates to L = (2a+6)×1.40 Å.

Question: if the energy levels of the electrons are given by εn = ℏ2n2π2/2mL2, what is the energy of the HOMO in eV if a=1?

Answer: since there are 2a+ 6 π-electrons, there must be 8 electrons, and therefore the HOMO must have n=4. We know that L = 8 × 1.40 Å = 11.2 Å. From these numbers, we get εn = 4.80×10-20 n2 in Joules. The energy of the HOMO is thus ε4 = 7.68×10-19J = 4.79 eV.

Polymethine dyesQuestion: what is the energy of the LUMO, and thus the HOMO-LUMO transition? How does this compare to experiment (see figure)?

Answer: ε = 4.80×10-20 n2 in Joules. The Answer: εn = 4.80×10-20 n2 in Joules. The energy of the LUMO is thusε5 = 1.20×10-18J = 7.49 eV. The energy of the HOMO-LUMO transition is thus 2.69 eV. This corresponds to photons of wavelength

λ = hc/(2.69 × 1.602×10-19) ~ 460nm. This is not so far from the experimental value (about 550nm), which is about as good as we can

Indeed, the R groups in the picture are, in fact, t hemselves conjugated. Inclusion of these as part of the “elec tron track” will improve the model considerably by further delocalizing the electrons and thus bringing the energies down closer to the e xperimental values.

550nm), which is about as good as we can expect.

Effect of chromophore size

400 450 500 550 600 650 700 750 800

Wavelength (nm)

Chromophore

Something to think about

• Particle on a ring

Must fit even wavelengths into whole cycle

Summary

• The particle in a box problem can be solved exactly and is a

good first approximation for the electrons in a delocalized π-good first approximation for the electrons in a delocalized π-

system.

• Confining a particle in a box leads to quantization of its

energy levels due to the condition that its wavefunction is

zero at the edges of the box

• The lowest energy (ZPE) of a particle in a box is not zero

• Be able to apply the particle in a box approximation as a • Be able to apply the particle in a box approximation as a

model for the electronic structure of a conjugated molecule

(given equation for En).

Next lecture

• Particle-on-a-ring model

Week 10 tutorials

• Particle in a box approximation• Particle in a box approximation– you solve the Schrödinger equation.

Practice Questions1. The energy levels of the particle in a box are given by εn = ℏ2n2p2/2mL2.

(a) Why does the lowest energy correspond to n = 1 rather than n = 0?(a) Why does the lowest energy correspond to n = 1 rather than n = 0?

(b) What is the separation between two adjacent levels?(Hint: ∆ε = εn+1 - εn)

(c) The π chain in a hexatriene derivative has L = 973 pm and has 6 πelectrons. What is energy of the HOMO – LUMO gap?

(d) What does the particle in a box model predicts happens to the HOMO – LUMO gap of polyenes as the chain length increases?