PY3P05

Lecture 13: Diatomic orbitalsLecture 13: Diatomic orbitals

o Hydrogen molecule ion (H2+)

o Overlap and exchange integrals

o Bonding/Anti-bonding orbitals

o Molecular orbitals

PY3P05

Schrödinger equation for hydrogen molecule ionSchrödinger equation for hydrogen molecule ion

o Simplest example of a chemical bond is the hydrogen molecule ion (H2

+).

o Consists of two protons and a single electron.

o If nuclei are distant, electron is localised on one nucleus. Wavefunctions are then those of atomic hydrogen.

rab

rbra

++

-

ab

€

−h2

2m∇ 2 −

e2

4πε0ra

⎡

⎣ ⎢

⎤

⎦ ⎥φa (ra ) = Ea

0(ra )φa (ra )

a is the hydrogen atom wavefunction of electron belonging to nucleus a. Must therefore satisfy

and correspondingly for other wavefunction, b. The energies are therefore

Ea0 = Eb

0 = E0

(1)

Ha

PY3P05

Schrödinger equation for hydrogen molecule ionSchrödinger equation for hydrogen molecule ion

o If atoms are brought into close proximity, electron localised on b will now experience an attractive Coulomb force of nucleus a.

o Must therefore modify Schrödinger equation to include Coulomb potentials of both nuclei:

where

o To find the coefficients ca and cb, substitute into Eqn. 2:

€

−h2

2m∇ 2 −

e2

4πε0ra

−e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥ψ = Eψ

€

ψ =caφa + cbφb

(2)

€

−h2

2m∇ 2 −

e2

4πε0ra

−e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥caφa +

−h2

2m∇ 2 −

e2

4πε0rb

−e2

4πε0ra

⎡

⎣ ⎢

⎤

⎦ ⎥cbφb = E(caφa + cbφb )

Ha

Hb

PY3P05

Solving the Schrödinger equationSolving the Schrödinger equation

o Can simplify last equation using Eqn. 1 and the corresponding equation for Ha and Hb.

o By writing Ea0 a in place of Ha a gives

o Rearranging,

o As Ea0 = Eb

0 by symmetry, we can set Ea0 - E = Eb

0 - E = - E,

€

Ha −e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥caφa + Hb −

e2

4πε0ra

⎡

⎣ ⎢

⎤

⎦ ⎥cbφb = E(caφa + cbφb )

€

Ea0 −

e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥caφa + Eb

0 −e2

4πε0ra

⎡

⎣ ⎢

⎤

⎦ ⎥cbφb = E(caφa + cbφb )

€

Ea0 − E −

e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥caφa + Eb

0 − E −e2

4πε0ra

⎡

⎣ ⎢

⎤

⎦ ⎥cbφb = 0

€

−ΔE −e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥caφa + −ΔE −

e2

4πε0ra

⎡

⎣ ⎢

⎤

⎦ ⎥cbφb = 0 (3)

PY3P05

Overlap and exchange integralsOverlap and exchange integrals

a and b depend on position, while ca and cb do not. Now we know that for orthogonal wavefunctions

but a and b are not orthoganal, so

o If we now multiply Eqn. 3 by a and integrate the results, we obtain

o As -e|a|2 is the charge density of the electron => Eqn. 4 is the Coulomb interaction energy between electron charge density and nuclear charge e of nucleus b.

o The -ea b in Eqn. 5 means that electron is partly in state a and partly in state b => an exchange between states occurs. Eqn. 5 therefore called an exchange integral.

€

φa*∫ φadV = φb

*∫ φbdV =1 Normalisation integral

€

φa∫ φbdV = S(rab ) Overlap integral

€

− φa∫ (ra )e2

4πε0rb

⎛

⎝ ⎜

⎞

⎠ ⎟φa (ra )dV = C(rab )

− φa∫ (ra )e2

4πε0ra

⎛

⎝ ⎜

⎞

⎠ ⎟φb (rb )dV = D(rab )

(4)

(5)

PY3P05

Overlap and exchange integralsOverlap and exchange integrals

o Eqn. 4 can be visualised via figure at right.

o Represents the Coulomb interaction energy of an electron density cloud in the Coulomb field on the nucleus.

o Eqn. 5 can be visualised via figure at right.

o Non-vanishing contributions are only possible when the wavefunctions overlap.

o See Chapter 24 of Haken & Wolf for further details.

PY3P05

Orbital energiesOrbital energies

o If we mutiply Eqn 3 by a and integrate we get

o Collecting terms gives,

o Similarly,

o Eqns. 6 and 7 can be solved for c1 and c2 via the matrix equation:

o Non-trivial solutions exist when determinant vanishes:

€

φa −ΔE −e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥caφa + −ΔE −

e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥cbφb

⎛

⎝ ⎜

⎞

⎠ ⎟∫ dV

€

φb −ΔE −e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥caφa + −ΔE −

e2

4πε0rb

⎡

⎣ ⎢

⎤

⎦ ⎥cbφb

⎛

⎝ ⎜

⎞

⎠ ⎟∫ dV

€

(−ΔE + C)ca + (−ΔE ⋅S + D)cb = 0

€

=>(−ΔE ⋅S + D)ca + (−ΔE + C)cb = 0

(6)

(7)

€

−ΔE + C −ΔE ⋅S + D

−ΔE ⋅S + D −ΔE + C

⎛

⎝ ⎜

⎞

⎠ ⎟ca

cb

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

€

−ΔE + C −ΔE ⋅S + D

−ΔE ⋅S + D −ΔE + C= 0

PY3P05

Orbital energiesOrbital energies

o Therefore (-ΔE + C)2 - (-ΔE·S + D)2 = 0

=> C - ΔE = ± (D - ΔE ·S)

o This gives two values for ΔE: and

o As C and D are negative, ΔEb < ΔEa

ΔEb correspond to bonding molecular

orbital energies. ΔEa to anti-bonding MOs.

o H2+ atomic and molecular orbital energies

are shown at right.

€

ΔEa =C − D

1− S

€

ΔEb =C + D

1+ S

Energy

(eV)

-13.6

PY3P05

Bonding and anti-bonding orbitalsBonding and anti-bonding orbitals

o Substituting for ΔEa into Eqn. 6 => cb = -ca = -c.The total wavefunction is thus

o Similarly for ΔEb => ca = cb = c, which gives

o For symmetric case (top right), occupation probability for ψ is positive. Not the case for a asymmetric wavefunctions (bottom right).

o Energy splits depending on whether dealing with a bonding or an anti-bonding wavefunction.

€

ψ−=c(φa − φb )

€

ψ+ =c(φa + φb )

Anti-bonding orbital

Bonding orbital

PY3P05

Bonding and anti-bonding orbitalsBonding and anti-bonding orbitals

o The energy E of the hydrogen molecular ion can finally be written

o ‘+’ correspond to anti-bonding orbital energies, ‘-’ to bonding.

o Energy curves below are plotted to show their dependence on rab

€

E = E 0 + ΔE = E 0 +C ± D

1± S

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

PY3P05

Molecular orbitalsMolecular orbitals

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

PY3P05

Molecular orbitalsMolecular orbitals

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

PY3P05

Molecular orbitalsMolecular orbitals

o Energies of bonding and anti-bonding molecular orbitals for first row diatomic molecules.

o Two electrons in H2 occupy bonding molecular orbital, with anti-parallel spins. If irradiated by UV light, molecule may absorb energy and promote one electron into its anti-bonding orbital.

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

PY3P05

Molecular orbitalsMolecular orbitals

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.