py3p05 lecture 13: diatomic orbitals ohydrogen molecule ion (h 2 + ) ooverlap and exchange integrals...
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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
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PY3P05
Molecular orbitalsMolecular orbitals
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PY3P05
Molecular orbitalsMolecular orbitals
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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.
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PY3P05
Molecular orbitalsMolecular orbitals
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