Chapter 22Diatomic Molecules
P. J. Grandinetti
Chem. 4300
P. J. Grandinetti Chapter 22: Diatomic Molecules
The Hydrogen Molecular IonSimplest molecule to consider is H+
2 , with only 1 electron. Hamiltonian is
H+2= β β2
2mp
(β2
A + β2B)
βββββββββββββββββββββ1
β β2
2meβ2
e
βββββ2
βZAq2
e4ππ0rA
βββββ3
βZBq2
e4ππ0rB
βββββ4
+ZAZBq2
e4ππ0RABβββββ
5
1 is kinetic energy of nuclei2 is kinetic energy of eβ
3 is Coulomb attraction between eβ and nucleus A4 is Coulomb attraction between eβ and nucleus B5 is Coulomb repulsion between nuclei A and B
Written in terms of atomic units
H+2= β1
2memp
(β2
A + β2B)β 1
2β2
e βZArA
βZBrB
+ZAZBRAB
P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer (B-O) ApproximationSince nuclei are much heavier than eβ we separate motion into 2 timescales:
fast time scale of eβ motion and slow time scale of nuclear motion.Born-Oppenheimer approximation assumes nuclei are fixed in place and solve for eβ wave functionin potential of 2 fixed nuclei.We then change internuclear spacing and repeat process.Not allowing nuclei to move while solving for eβ wave function has 2 effects:
1 nuclear kinetic energy terms: 1 go away2 nuclearβnuclear repulsion potential energy term 5 becomes constant and can be simply
added to energy eigenvalue.With this approximation wave equation for eβ (in atomic units) becomes[
β12β2
e βZArA
βZBrB
]βββββββββββββββββββββββββββ
el
πel(r,RAB) = E(RAB)πel(r,RAB).
Solving this wave equation gives eβ wave function, πel(r,RAB), and its energy for given internucleardistance, RAB.P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer (B-O) ApproximationNext in B-O approximation we take total wave function as
π(r,RAB) β πel(r,RAB)πnuc(RAB)
Next we assume that πel(r,RAB) varies so slowly with RAB that
β12
memp
(β2
A + β2B)πel(r,RAB)πnuc(RAB) β πel(r,RAB)
[β1
2memp
(β2
A + β2B)πnuc(RAB)
]In other words we assume
(β2
A + β2B)πel(r,RAB) β 0
Putting B-O wave function approximation
H+2π(r,RAB) = Eπ(r,RAB)
into full SchrΓΆdinger equation
H+2= β1
2memp
(β2
A + β2B)β 1
2β2
e βZArA
βZBrB
+ZAZBRAB
we obtain...P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer (B-O) Approximation
πel(r,RAB)[β1
2memp
(β2
A + β2B)]πnuc(RAB) +
[β1
2β2
e βZArA
βZBrB
]βββββββββββββββββββββββββββ
el
πel(r,RAB)πnuc(RAB)
+ZAZBRAB
πel(r,RAB)πnuc(RAB) = Eπel(r,RAB)πnuc(RAB)
Making the replacement elπel(r,RAB) = E(RAB)πel(r,RAB) gives
πel(r,RAB)[β1
2memp
(β2
A + β2B)+ E(RAB) +
ZAZBRAB
]πnuc(RAB) = πel(r,RAB)Eπnuc(RAB)
Dividing both sides by πel(r,RAB) gives...
P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer ApproximationDividing both sides by πel(r,RAB) and obtain wave equation for nuclei:[
β12
memp
(β2
A + β2B)
βββββββββββββββββββββββnuclear kinetic energy
+ E(RAB) +ZAZBRAB
βββββββββββββββββnuclear effective potential
]πnuc(RAB) = Eπnuc(RAB)
General strategy is to
fix nuclei in position and calculate πel(r,RAB) and energy, E(RAB). Do this for all possible values ofRAB, and
use E(RAB) + ZAZBβRAB as effective nuclear potential energy (Ground state looks like Morsepotential) in nuclear wave equation to obtain πnuc(RAB) and energies:
P. J. Grandinetti Chapter 22: Diatomic Molecules
Solving one electron SchrΓΆdinger equation for the H2+ ion
With B-O approximation out of way letβs look at solutions for πel(r,RAB) of H+2 , given the
electronic Hamiltonian in atomic units[β1
2β2
e βZArA
βZBrB
]βββββββββββββββββββββββββββ
el
πel(r,RAB) = E(RAB)πel(r,RAB).
Problem is no longer spherically symmetric. So, what coordinate system should we use?
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : πel(r,RAB) to πel(π, π, π,RAB)
We can derive exact solution for πel(r,RAB) using spheroidal coordinates,where π = (rA + rB)βR, π = (rA β rB)βR, and R is internuclear distance.Lines of constant π are ellipses which share foci rA and rB.Lines of constant π are hyperbolas with rA and rB as foci.Ellipses and hyperbolas form orthogonal system of curves.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : πel(r,RAB) to πel(π, π, π,RAB)
Variable π varies over range 1 β€ π β€ β, and plays role analogous to r in usual polar coordinatesystem.Variable π varies over range β1 β€ π β€ 1.As π changes point (π, π) moves around origin, so π plays role similar to quantity cos π in polarcoordinates.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : πel(r,RAB) to πel(π, π, π,RAB)
Three dimensional prolate ellipsoidal coordinates are obtained by rotating figure around z axis.Ellipses generate set of confocal ellipsoidsHyperbolas generate family of hyperboloids with 2 sheets.Surface of constant π are half-planes though x axis.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Two sheet hyperboloid
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : πel(r,RAB) to πel(π, π, π,RAB)
Prolate ellipsoidal coordinates in 3D space are obtained by rotating figure around z axis.Ellipses generate set of confocal ellipsoidsHyperbolas generate family of hyperboloids with 2 sheets.Surface of constant π are half-planes though x axis.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : πel(r,RAB) to πel(π, π, π,RAB)
Spheroidal Coordinates allows us to separate wave function into product
π(π, π, π) = L(π)M(π)Ξ¦(π)
Substituting π(π, π, π) into electronic wave equation gives 3 ODEs.Weβll do no derivations, just look at results ...
P. J. Grandinetti Chapter 22: Diatomic Molecules
Solutions to Ξ¦(π)Solutions to Ξ¦(π) which are eigenfunctions of Lz,
Ξ¦(π) = 1β2π
eimπ
Each value of |m| leads to different energy. States associated with Β±m are degenerate.We refer to states by their m value:
m = 0 π state,m = Β±1 π state,m = Β±2 πΏ state,
β«βͺβ¬βͺβthese follow same lettersequence as π usingGreek letters instead.
States are also labeled by their inversion symmetry.
when πu(r) = βπu(βr), odd symmetry,when πg(r) = πg(βr), even symmetry,
Use subscript u for odd wave functions (ungerade)Use subscript g for even wave functions (gerade).Wave functions labeled as πg, πu, πg, πu, πΏg, πΏu, and so on.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Lowest energy levels of H+2 as function of internuclear R
with internuclear repulsive energy.
Minimum in 1πg energy is Re β 2a0, corresponding toequilibrium length of 1πg ground state of H+
2 .
As R β β energy of 1πg state approaches β0.5Eh.As expected, this is energy of electron in 1s state ofH-atom infinitely separated from isolated proton.Difference between this energy and energy at equilibriumbond length is binding energy,E1πg
(Re) β E1πg(β) = 0.1Eh.
Both equilibrium distance and binding energy from thisexact solution are in excellent agreement withexperimentally determined values of 2.00a0 and 0.102Eh,respectively.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Lowest energy levels of H+2 as function of internuclear R
without internuclear repulsive energy.
As R β 0, i.e., both protons at origin form He nucleus, wefind energy of β2Eh. This is ground state energy of singleelectron bound to He nucleus.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Exact solutions for 1πg and 1πu of H+2 as a function of R
(A)
(D)
(E)
(F)
(B) (C)
P. J. Grandinetti Chapter 22: Diatomic Molecules
Shape of H+2 wave functions
When R = 0 solution becomes identical to He+ wave function.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Shape of H+2 wave functions
When R = 8a0 observe 2 sharp peaks at Β±4a0 where nucleiare located.
When R β β two peaks correspond to 1s orbital centeredon each nucleus.
In case of H+2 only one of these 1s orbitals is occupied.
Difference between 1πg and 1πu is in how two 1s orbitalsare combined.
Normalization factors aside, in R β β limit we find (in atomic units)
1πg = eβrA + eβrB , and 1πu = βeβrA + eβrB .
Results suggest approximate approach to describe bonding wave functions as a linear combinationof atomic orbitals (LCAO) on each nucleus.LCAO approach more useful than exact solutionβwhich only works for H+
2 .
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)Use variational theorem with LCAO as trial H+
2 wave function
πguess(r,RAB) = cAπ1sA+ cBπ1sB
π1sAand π1sB
are atomic orbitals associated with eβ in 1s orbital on nuclei A and B, respectively.There are 2 adjustable parameters, cA and cB, in πguess.
β¨β© = β« πβguessπguessdπ β₯ E0
E0 is true ground state energy. Canβt assume trial wave function is normalized so need to minimizeenergy for
E =β«V π
βguessπguessdπ
β«V πβguessπguessdπ
β₯ E0
Even though atomic orbitals are normalized, LCAO wave function is not. Substituting πguess(r,RAB) weobtain
E =c2
A β«Vπβ
1sAπ1sA
dπ + c2B β«V
πβ1sB
π1sBdπ + 2cAcB β«V
πβ1sA
π1sBdπ
c2A + c2
B + 2cAcB β«Vπβ
1sAπ1sB
dπβ₯ E0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)To simplify equations define
HAB β‘ β«Vπβ
1sAπ1sB
dπ, and SAB β‘ β«Vπβ
1sAπ1sB
dπ
SAB is called overlap integral. These definitions allow us to write
E =c2
AHAA + c2BHBB + 2cAcBHAB
c2A + c2
B + 2cAcBSABβ₯ E0
Next, find values of cA and cB where E is at minimum by taking derivative of E wrt cA and cB andsetting equal to zero,
πEπcA
= 0, and πEπcB
= 0
To make this easier letβs move the denominator to the left(c2
A + c2B + 2cAcBSAB
)E = c2
AHAA + c2BHBB + 2cAcBHAB
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)Taking the derivative of both sides
ππcA
(c2
A + c2B + 2cAcBSAB
)E = π
πcA
(c2
AHAA + c2BHBB + 2cAcBHAB
)gives
(2cA + 2cBSAB)E +(c2
A + c2B + 2cAcBSAB
) πEπcA
= 2cAHAA + 2cBHAB
Doing same with πβπcB gives
(2cB + 2cASAB)E +(c2
A + c2B + 2cAcBSAB
) πEπcB
= 2cBHBB + 2cAHAB
Setting πEβπcA = πEβπcB = 0 leads to two simultaneous equations
cA(HAA β E) + cB(HAB β ESAB) = 0
cA(HAB β ESAB) + cB(HBB β E) = 0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)Writing these in matrix form givesββββ
HAA β E HAB β ESAB
HAB β ESAB HBB β E
ββββ ββββ
cA
cB
ββββ = 0
Matrix diagonalization problem can be solved with determinant,|||||||HAA β E HAB β ESAB
HAB β ESAB HBB β E
||||||| = 0
In homonuclear example make it little easier since HAA = HBB = πΌ.Also set HAB = π½ and S = SAB|||||||
πΌ β E π½ β ES
π½ β ES πΌ β E
||||||| = 0, which gives (πΌ β E)2 β (π½ β ES)2 = 0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)
(πΌ β E)2 β (π½ β ES)2 = 0
which leads toπΌ β E = Β±(π½ β ES) = Β±π½ β ES
and we find 2 solutions for E:E+ =
πΌ + π½1 + S
and Eβ =πΌ β π½1 β S
Putting solution for E+ back into simultaneous Eqs one can show that cA = cB.Put solution for Eβ into 2 simultaneous equations and obtain cA = βcB.Thus, 2 solutions for wave function are
ππg= c
(π1sA
+ π1sB
), and ππu
= c(π1sA
β π1sB
)Normalizing these two wave functions gives
ππg= 1β
2 + 2S
(π1sA
+ π1sB
)and ππu
= 1β2 β 2S
(π1sA
β π1sB
)P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)
Bring two 1s orbitals together in phase for ππgand out of phase for ππu
(A) (B)
Above is comparison of Exact (solid lines) and LCAO (dashed lines) wave functions ππgand
ππufor H+
2 with R = 2 for (A) bonding and (B) anti-bonding states.
Simple LCAO approximation is not bad, and is good starting point for refining LCAO method.
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Overlap Integral STo finish derivation need to evaluate overlap integral S and energies. Starting with S we find
S = β«Vπβ
1sAπ1sB
dπ = eβRAB
(1 + RAB +
R2AB3
)
0 1 2 3 40.0
0.2
0.4
0.6
0.8
1.0
As expected, overlap integral goes to zero in limit that R β β.With decreasing R overlap integral increases and reaches value of 1 at R = 0.
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Coulomb IntegralπΌ integral is called Coulomb Integral
πΌ = β«Vπβ
1sAπ1sA
dπ
To evaluate πΌ start with electronic Hamiltonian in atomic units = β1
2β2
e β1rA
β 1rB
+ 1RAB
which can be written = A β 1rB
+ 1RAB
or = B β 1rA
+ 1RAB
A or B are Hamiltonians for eβ in H-atom alone. Thus,
πΌ = β«Vπβ
1sA
[A β 1
rB+ 1
RAB
]π1sA
dπ = β«Vπβ
1sAAπ1sA
dπ β β«Vπβ
1sA
1rBπ1sA
dπ + 1RAB
which gives πΌ = E1s +2E1sRAB
[1 β eβ2RAB(1 + RAB)
]+ 1
RABCoulomb Integral contains energy of eβ in 1s orbital of H-atom, attractive energy of nucleus Bfor eβ, and repulsive force of nuclei B with A.
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Coulomb Integral
10 2 3 4-1
0
1
2
3
4
πΌ decreases monotonically (i.e., no minimum) from β at RAB = 0 to β1β2 at RAB = β. In other words,πΌ, which is leading term in
E+ =πΌ + π½1 + S
and Eβ =πΌ β π½1 β S
does not give any stability to H+2 over 2 infinitely separated nuclei (recall H atom has energy of βEhβ2).
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Exchange Integral
Finally, examine π½ integral, also called the resonance or Exchange Integral
π½ = β«Vπβ
1sAπ1sB
dπ
which becomes
π½ = β«Vπβ
1sA
[B β 1
rA+ 1
RAB
]π1sB
dπ = β«Vπβ
1sABπ1sB
dπββ«Vπβ
1sA
1rAπ1sB
dπ+β«Vπβ
1sA
1RAB
π1sBdπ
to obtainπ½ = E1sS + 2E1seβRAB(1 + RAB) +
SRAB
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Exchange Integral
10 2 3 4-1
0
1
2
3
4 π½ integral goes through a minimum inenergy.It is stabilization energy from allowing eβto move (exchange) between 2 nuclei.Since both πΌ and π½ are negative, E+ willbe lowest energy,
E1πg= E+ =
πΌ + π½1 + S
, (bonding)
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Energy
-1.0
-0.5
0.0
0.5
1.0
10 2 3 4
LCAO model predicts that energy of groundstate has minimum at bond length ofRe = 2.50a0 and has binding energy ofE+(Re) β E(β) = 0.0648Eh.
Predicted bond length is longer thanexperimentally observed Re = 2.00a0
Predicted binding energy is lower thanexperimentally observed value of 0.102Eh.
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Energy
Anti-bonding orbital energy is
E1πu= Eβ =
πΌ β π½1 β S
, (anti-bonding)
This orbital gives no stability since π½ raises total energy in this case.Putting lone electron into π1πu
would destabilize H+2 molecule and cause it to break apart.
P. J. Grandinetti Chapter 22: Diatomic Molecules