Post on 20-Jun-2020
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= â â2
2mp
(â2
A + â2B)
âââââââââââââââââââââ1
â â2
2meâ2
e
âââââ2
âZAq2
e4ðð0rA
âââââ3
âZBq2
e4ðð0rB
âââââ4
+ZAZBq2
e4ðð0RABâââââ
5
1 is kinetic energy of nuclei2 is kinetic energy of eâ
3 is Coulomb attraction between eâ and nucleus A4 is Coulomb attraction between eâ and nucleus B5 is Coulomb repulsion between nuclei A and B
Written in terms of atomic units
H+2= â1
2memp
(â2
A + â2B)â 1
2â2
e âZArA
âZBrB
+ZAZBRAB
P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer (B-O) ApproximationSince nuclei are much heavier than eâ we separate motion into 2 timescales:
fast time scale of eâ motion and slow time scale of nuclear motion.Born-Oppenheimer approximation assumes nuclei are fixed in place and solve for eâ wave functionin potential of 2 fixed nuclei.We then change internuclear spacing and repeat process.Not allowing nuclei to move while solving for eâ wave function has 2 effects:
1 nuclear kinetic energy terms: 1 go away2 nuclearânuclear repulsion potential energy term 5 becomes constant and can be simply
added to energy eigenvalue.With this approximation wave equation for eâ (in atomic units) becomes[
â12â2
e âZArA
âZBrB
]âââââââââââââââââââââââââââ
el
ðel(r,RAB) = E(RAB)ðel(r,RAB).
Solving this wave equation gives eâ wave function, ðel(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
ð(r,RAB) â ðel(r,RAB)ðnuc(RAB)
Next we assume that ðel(r,RAB) varies so slowly with RAB that
â12
memp
(â2
A + â2B)ðel(r,RAB)ðnuc(RAB) â ðel(r,RAB)
[â1
2memp
(â2
A + â2B)ðnuc(RAB)
]In other words we assume
(â2
A + â2B)ðel(r,RAB) â 0
Putting B-O wave function approximation
H+2ð(r,RAB) = Eð(r,RAB)
into full Schrödinger equation
H+2= â1
2memp
(â2
A + â2B)â 1
2â2
e âZArA
âZBrB
+ZAZBRAB
we obtain...P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer (B-O) Approximation
ðel(r,RAB)[â1
2memp
(â2
A + â2B)]ðnuc(RAB) +
[â1
2â2
e âZArA
âZBrB
]âââââââââââââââââââââââââââ
el
ðel(r,RAB)ðnuc(RAB)
+ZAZBRAB
ðel(r,RAB)ðnuc(RAB) = Eðel(r,RAB)ðnuc(RAB)
Making the replacement elðel(r,RAB) = E(RAB)ðel(r,RAB) gives
ðel(r,RAB)[â1
2memp
(â2
A + â2B)+ E(RAB) +
ZAZBRAB
]ðnuc(RAB) = ðel(r,RAB)Eðnuc(RAB)
Dividing both sides by ðel(r,RAB) gives...
P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer ApproximationDividing both sides by ðel(r,RAB) and obtain wave equation for nuclei:[
â12
memp
(â2
A + â2B)
ââââââââââââââââââââââânuclear kinetic energy
+ E(RAB) +ZAZBRAB
ââââââââââââââââânuclear effective potential
]ðnuc(RAB) = Eðnuc(RAB)
General strategy is to
fix nuclei in position and calculate ðel(r,RAB) and energy, E(RAB). Do this for all possible values ofRAB, and
use E(RAB) + ZAZBâRAB as effective nuclear potential energy (Ground state looks like Morsepotential) in nuclear wave equation to obtain ðnuc(RAB) and energies:
P. J. Grandinetti Chapter 22: Diatomic Molecules
Solving one electron Schrödinger equation for the H2+ ion
With B-O approximation out of way letâs look at solutions for ðel(r,RAB) of H+2 , given the
electronic Hamiltonian in atomic units[â1
2â2
e âZArA
âZBrB
]âââââââââââââââââââââââââââ
el
ðel(r,RAB) = E(RAB)ðel(r,RAB).
Problem is no longer spherically symmetric. So, what coordinate system should we use?
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : ðel(r,RAB) to ðel(ð, ð, ð,RAB)
We can derive exact solution for ðel(r,RAB) using spheroidal coordinates,where ð = (rA + rB)âR, ð = (rA â rB)âR, and R is internuclear distance.Lines of constant ð are ellipses which share foci rA and rB.Lines of constant ð 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 : ðel(r,RAB) to ðel(ð, ð, ð,RAB)
Variable ð varies over range 1 †ð †â, and plays role analogous to r in usual polar coordinatesystem.Variable ð varies over range â1 †ð †1.As ð changes point (ð, ð) moves around origin, so ð plays role similar to quantity cos ð in polarcoordinates.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : ðel(r,RAB) to ðel(ð, ð, ð,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 ð 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 : ðel(r,RAB) to ðel(ð, ð, ð,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 ð are half-planes though x axis.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : ðel(r,RAB) to ðel(ð, ð, ð,RAB)
Spheroidal Coordinates allows us to separate wave function into product
ð(ð, ð, ð) = L(ð)M(ð)Ί(ð)
Substituting ð(ð, ð, ð) into electronic wave equation gives 3 ODEs.Weâll do no derivations, just look at results ...
P. J. Grandinetti Chapter 22: Diatomic Molecules
Solutions to Ί(ð)Solutions to Ί(ð) which are eigenfunctions of Lz,
Ί(ð) = 1â2ð
eimð
Each value of |m| leads to different energy. States associated with ±m are degenerate.We refer to states by their m value:
m = 0 ð state,m = ±1 ð state,m = ±2 ð¿ state,
â«âªâ¬âªâthese follow same lettersequence as ð usingGreek letters instead.
States are also labeled by their inversion symmetry.
when ðu(r) = âðu(âr), odd symmetry,when ðg(r) = ðg(âr), even symmetry,
Use subscript u for odd wave functions (ungerade)Use subscript g for even wave functions (gerade).Wave functions labeled as ðg, ðu, ðg, ðu, ð¿g, ð¿u, 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 1ðg energy is Re â 2a0, corresponding toequilibrium length of 1ðg ground state of H+
2 .
As R â â energy of 1ðg state approaches â0.5Eh.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,E1ðg
(Re) â E1ðg(â) = 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 â 0, i.e., both protons at origin form He nucleus, wefind energy of â2Eh. This is ground state energy of singleelectron bound to He nucleus.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Exact solutions for 1ðg and 1ðu 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 ±4a0 where nucleiare located.
When R â â two peaks correspond to 1s orbital centeredon each nucleus.
In case of H+2 only one of these 1s orbitals is occupied.
Difference between 1ðg and 1ðu is in how two 1s orbitalsare combined.
Normalization factors aside, in R â â limit we find (in atomic units)
1ðg = eârA + eârB , and 1ðu = âeârA + eârB .
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 solutionâwhich 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
ðguess(r,RAB) = cAð1sA+ cBð1sB
ð1sAand ð1sB
are atomic orbitals associated with eâ in 1s orbital on nuclei A and B, respectively.There are 2 adjustable parameters, cA and cB, in ðguess.
âšâ© = â« ðâguessðguessdð ⥠E0
E0 is true ground state energy. Canât assume trial wave function is normalized so need to minimizeenergy for
E =â«V ð
âguessðguessdð
â«V ðâguessðguessdð
⥠E0
Even though atomic orbitals are normalized, LCAO wave function is not. Substituting ðguess(r,RAB) weobtain
E =c2
A â«Vðâ
1sAð1sA
dð + c2B â«V
ðâ1sB
ð1sBdð + 2cAcB â«V
ðâ1sA
ð1sBdð
c2A + c2
B + 2cAcB â«Vðâ
1sAð1sB
dð⥠E0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)To simplify equations define
HAB â¡ â«Vðâ
1sAð1sB
dð, and SAB â¡ â«Vðâ
1sAð1sB
dð
SAB is called overlap integral. These definitions allow us to write
E =c2
AHAA + c2BHBB + 2cAcBHAB
c2A + c2
B + 2cAcBSAB⥠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,
ðEðcA
= 0, and ðEðcB
= 0
To make this easier letâs 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
ððcA
(c2
A + c2B + 2cAcBSAB
)E = ð
ðcA
(c2
AHAA + c2BHBB + 2cAcBHAB
)gives
(2cA + 2cBSAB)E +(c2
A + c2B + 2cAcBSAB
) ðEðcA
= 2cAHAA + 2cBHAB
Doing same with ðâðcB gives
(2cB + 2cASAB)E +(c2
A + c2B + 2cAcBSAB
) ðEðcB
= 2cBHBB + 2cAHAB
Setting ðEâðcA = ðEâðcB = 0 leads to two simultaneous equations
cA(HAA â E) + cB(HAB â ESAB) = 0
cA(HAB â ESAB) + cB(HBB â E) = 0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)Writing these in matrix form givesââââ
HAA â E HAB â ESAB
HAB â ESAB HBB â E
ââââ ââââ
cA
cB
ââââ = 0
Matrix diagonalization problem can be solved with determinant,|||||||HAA â E HAB â ESAB
HAB â ESAB HBB â E
||||||| = 0
In homonuclear example make it little easier since HAA = HBB = ðŒ.Also set HAB = ðœ and S = SAB|||||||
ðŒ â E ðœ â ES
ðœ â ES ðŒ â E
||||||| = 0, which gives (ðŒ â E)2 â (ðœ â ES)2 = 0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)
(ðŒ â E)2 â (ðœ â ES)2 = 0
which leads toðŒ â E = ±(ðœ â ES) = ±ðœ â ES
and we find 2 solutions for E:E+ =
ðŒ + ðœ1 + S
and Eâ =ðŒ â ðœ1 â S
Putting solution for E+ back into simultaneous Eqs one can show that cA = cB.Put solution for Eâ into 2 simultaneous equations and obtain cA = âcB.Thus, 2 solutions for wave function are
ððg= c
(ð1sA
+ ð1sB
), and ððu
= c(ð1sA
â ð1sB
)Normalizing these two wave functions gives
ððg= 1â
2 + 2S
(ð1sA
+ ð1sB
)and ððu
= 1â2 â 2S
(ð1sA
â ð1sB
)P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)
Bring two 1s orbitals together in phase for ððgand out of phase for ððu
(A) (B)
Above is comparison of Exact (solid lines) and LCAO (dashed lines) wave functions ððgand
ððufor 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 = â«Vðâ
1sAð1sB
dð = eâRAB
(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 â â.With decreasing R overlap integral increases and reaches value of 1 at R = 0.
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Coulomb IntegralðŒ integral is called Coulomb Integral
ðŒ = â«Vðâ
1sAð1sA
dð
To evaluate ðŒ start with electronic Hamiltonian in atomic units = â1
2â2
e â1rA
â 1rB
+ 1RAB
which can be written = A â 1rB
+ 1RAB
or = B â 1rA
+ 1RAB
A or B are Hamiltonians for eâ in H-atom alone. Thus,
ðŒ = â«Vðâ
1sA
[A â 1
rB+ 1
RAB
]ð1sA
dð = â«Vðâ
1sAAð1sA
dð â â«Vðâ
1sA
1rBð1sA
dð + 1RAB
which gives ðŒ = E1s +2E1sRAB
[1 â eâ2RAB(1 + RAB)
]+ 1
RABCoulomb Integral contains energy of eâ in 1s orbital of H-atom, attractive energy of nucleus Bfor eâ, 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
ðŒ decreases monotonically (i.e., no minimum) from â at RAB = 0 to â1â2 at RAB = â. In other words,ðŒ, which is leading term in
E+ =ðŒ + ðœ1 + S
and Eâ =ðŒ â ðœ1 â S
does not give any stability to H+2 over 2 infinitely separated nuclei (recall H atom has energy of âEhâ2).
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Exchange Integral
Finally, examine ðœ integral, also called the resonance or Exchange Integral
ðœ = â«Vðâ
1sAð1sB
dð
which becomes
ðœ = â«Vðâ
1sA
[B â 1
rA+ 1
RAB
]ð1sB
dð = â«Vðâ
1sABð1sB
dðââ«Vðâ
1sA
1rAð1sB
dð+â«Vðâ
1sA
1RAB
ð1sBdð
to obtainðœ = E1sS + 2E1seâRAB(1 + RAB) +
SRAB
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Exchange Integral
10 2 3 4-1
0
1
2
3
4 ðœ integral goes through a minimum inenergy.It is stabilization energy from allowing eâto move (exchange) between 2 nuclei.Since both ðŒ and ðœ are negative, E+ willbe lowest energy,
E1ðg= E+ =
ðŒ + ðœ1 + 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) â E(â) = 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
E1ðu= Eâ =
ðŒ â ðœ1 â S
, (anti-bonding)
This orbital gives no stability since ðœ raises total energy in this case.Putting lone electron into ð1ðu
would destabilize H+2 molecule and cause it to break apart.
P. J. Grandinetti Chapter 22: Diatomic Molecules