chapter 22 - diatomic molecules - grandinetti€¦ · p. j. grandinetti chapter 22: diatomic...
TRANSCRIPT
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= − ℏ2
2mp
(∇2
A + ∇2B)
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟1
− ℏ2
2me∇2
e
⏟⏞⏟⏞⏟2
−ZAq2
e4𝜋𝜖0rA
⏟⏞⏟⏞⏟3
−ZBq2
e4𝜋𝜖0rB
⏟⏞⏟⏞⏟4
+ZAZBq2
e4𝜋𝜖0RAB⏟⏞⏟⏞⏟
5
1 is kinetic energy of nuclei2 is kinetic energy of e−
3 is Coulomb attraction between e− and nucleus A4 is Coulomb attraction between e− and nucleus B5 is Coulomb repulsion between nuclei A and B
Written in terms of atomic units
H+2= −1
2memp
(∇2
A + ∇2B)− 1
2∇2
e −ZArA
−ZBrB
+ZAZBRAB
P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer (B-O) ApproximationSince nuclei are much heavier than e− we separate motion into 2 timescales:
fast time scale of e− motion and slow time scale of nuclear motion.Born-Oppenheimer approximation assumes nuclei are fixed in place and solve for e− wave functionin potential of 2 fixed nuclei.We then change internuclear spacing and repeat process.Not allowing nuclei to move while solving for e− wave function has 2 effects:
1 nuclear kinetic energy terms: 1 go away2 nuclear–nuclear repulsion potential energy term 5 becomes constant and can be simply
added to energy eigenvalue.With this approximation wave equation for e− (in atomic units) becomes[
−12∇2
e −ZArA
−ZBrB
]⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
el
𝜓el(r,RAB) = E(RAB)𝜓el(r,RAB).
Solving this wave equation gives e− wave function, 𝜓el(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
𝜓(r,RAB) ≈ 𝜓el(r,RAB)𝜓nuc(RAB)
Next we assume that 𝜓el(r,RAB) varies so slowly with RAB that
−12
memp
(∇2
A + ∇2B)𝜓el(r,RAB)𝜓nuc(RAB) ≈ 𝜓el(r,RAB)
[−1
2memp
(∇2
A + ∇2B)𝜓nuc(RAB)
]In other words we assume
(∇2
A + ∇2B)𝜓el(r,RAB) ≈ 0
Putting B-O wave function approximation
H+2𝜓(r,RAB) = E𝜓(r,RAB)
into full Schrödinger equation
H+2= −1
2memp
(∇2
A + ∇2B)− 1
2∇2
e −ZArA
−ZBrB
+ZAZBRAB
we obtain...P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer (B-O) Approximation
𝜓el(r,RAB)[−1
2memp
(∇2
A + ∇2B)]𝜓nuc(RAB) +
[−1
2∇2
e −ZArA
−ZBrB
]⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
el
𝜓el(r,RAB)𝜓nuc(RAB)
+ZAZBRAB
𝜓el(r,RAB)𝜓nuc(RAB) = E𝜓el(r,RAB)𝜓nuc(RAB)
Making the replacement el𝜓el(r,RAB) = E(RAB)𝜓el(r,RAB) gives
𝜓el(r,RAB)[−1
2memp
(∇2
A + ∇2B)+ E(RAB) +
ZAZBRAB
]𝜓nuc(RAB) = 𝜓el(r,RAB)E𝜓nuc(RAB)
Dividing both sides by 𝜓el(r,RAB) gives...
P. J. Grandinetti Chapter 22: Diatomic Molecules
Born-Oppenheimer ApproximationDividing both sides by 𝜓el(r,RAB) and obtain wave equation for nuclei:[
−12
memp
(∇2
A + ∇2B)
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟nuclear kinetic energy
+ E(RAB) +ZAZBRAB
⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟nuclear effective potential
]𝜓nuc(RAB) = E𝜓nuc(RAB)
General strategy is to
fix nuclei in position and calculate 𝜓el(r,RAB) and energy, E(RAB). Do this for all possible values ofRAB, and
use E(RAB) + ZAZB∕RAB as effective nuclear potential energy (Ground state looks like Morsepotential) in nuclear wave equation to obtain 𝜓nuc(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 let’s look at solutions for 𝜓el(r,RAB) of H+2 , given the
electronic Hamiltonian in atomic units[−1
2∇2
e −ZArA
−ZBrB
]⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
el
𝜓el(r,RAB) = E(RAB)𝜓el(r,RAB).
Problem is no longer spherically symmetric. So, what coordinate system should we use?
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : 𝜓el(r,RAB) to 𝜓el(𝜆, 𝜇, 𝜙,RAB)
We can derive exact solution for 𝜓el(r,RAB) using spheroidal coordinates,where 𝜆 = (rA + rB)∕R, 𝜇 = (rA − rB)∕R, and R is internuclear distance.Lines of constant 𝜆 are ellipses which share foci rA and rB.Lines of constant 𝜇 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 : 𝜓el(r,RAB) to 𝜓el(𝜆, 𝜇, 𝜙,RAB)
Variable 𝜆 varies over range 1 ≤ 𝜆 ≤ ∞, and plays role analogous to r in usual polar coordinatesystem.Variable 𝜇 varies over range −1 ≤ 𝜇 ≤ 1.As 𝜇 changes point (𝜆, 𝜇) moves around origin, so 𝜇 plays role similar to quantity cos 𝜃 in polarcoordinates.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : 𝜓el(r,RAB) to 𝜓el(𝜆, 𝜇, 𝜙,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 𝜙 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 : 𝜓el(r,RAB) to 𝜓el(𝜆, 𝜇, 𝜙,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 𝜙 are half-planes though x axis.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Spheroidal Coordinates : 𝜓el(r,RAB) to 𝜓el(𝜆, 𝜇, 𝜙,RAB)
Spheroidal Coordinates allows us to separate wave function into product
𝜓(𝜆, 𝜇, 𝜙) = L(𝜆)M(𝜇)Φ(𝜙)
Substituting 𝜓(𝜆, 𝜇, 𝜙) into electronic wave equation gives 3 ODEs.We’ll do no derivations, just look at results ...
P. J. Grandinetti Chapter 22: Diatomic Molecules
Solutions to Φ(𝜙)Solutions to Φ(𝜙) which are eigenfunctions of Lz,
Φ(𝜙) = 1√2𝜋
eim𝜙
Each value of |m| leads to different energy. States associated with ±m are degenerate.We refer to states by their m value:
m = 0 𝜎 state,m = ±1 𝜋 state,m = ±2 𝛿 state,
⎫⎪⎬⎪⎭these follow same lettersequence as 𝓁 usingGreek letters instead.
States are also labeled by their inversion symmetry.
when 𝜓u(r) = −𝜓u(−r), odd symmetry,when 𝜓g(r) = 𝜓g(−r), even symmetry,
Use subscript u for odd wave functions (ungerade)Use subscript g for even wave functions (gerade).Wave functions labeled as 𝜎g, 𝜎u, 𝜋g, 𝜋u, 𝛿g, 𝛿u, 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 1𝜎g energy is Re ≈ 2a0, corresponding toequilibrium length of 1𝜎g ground state of H+
2 .
As R → ∞ energy of 1𝜎g state approaches −0.5Eh.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,E1𝜎g
(Re) − E1𝜎g(∞) = 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 → 0, i.e., both protons at origin form He nucleus, wefind energy of −2Eh. This is ground state energy of singleelectron bound to He nucleus.
P. J. Grandinetti Chapter 22: Diatomic Molecules
Exact solutions for 1𝜎g and 1𝜎u 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 → ∞ two peaks correspond to 1s orbital centeredon each nucleus.
In case of H+2 only one of these 1s orbitals is occupied.
Difference between 1𝜎g and 1𝜎u is in how two 1s orbitalsare combined.
Normalization factors aside, in R → ∞ limit we find (in atomic units)
1𝜎g = e−rA + e−rB , and 1𝜎u = −e−rA + e−rB .
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 solution—which 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
𝜓guess(r,RAB) = cA𝜙1sA+ cB𝜙1sB
𝜙1sAand 𝜙1sB
are atomic orbitals associated with e− in 1s orbital on nuclei A and B, respectively.There are 2 adjustable parameters, cA and cB, in 𝜓guess.
⟨⟩ = ∫ 𝜓∗guess𝜓guessd𝜏 ≥ E0
E0 is true ground state energy. Can’t assume trial wave function is normalized so need to minimizeenergy for
E =∫V 𝜓
∗guess𝜓guessd𝜏
∫V 𝜓∗guess𝜓guessd𝜏
≥ E0
Even though atomic orbitals are normalized, LCAO wave function is not. Substituting 𝜓guess(r,RAB) weobtain
E =c2
A ∫V𝜙∗
1sA𝜙1sA
d𝜏 + c2B ∫V
𝜙∗1sB
𝜙1sBd𝜏 + 2cAcB ∫V
𝜙∗1sA
𝜙1sBd𝜏
c2A + c2
B + 2cAcB ∫V𝜙∗
1sA𝜙1sB
d𝜏≥ E0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)To simplify equations define
HAB ≡ ∫V𝜙∗
1sA𝜙1sB
d𝜏, and SAB ≡ ∫V𝜙∗
1sA𝜙1sB
d𝜏
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,
𝜕E𝜕cA
= 0, and 𝜕E𝜕cB
= 0
To make this easier let’s 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
𝜕𝜕cA
(c2
A + c2B + 2cAcBSAB
)E = 𝜕
𝜕cA
(c2
AHAA + c2BHBB + 2cAcBHAB
)gives
(2cA + 2cBSAB)E +(c2
A + c2B + 2cAcBSAB
) 𝜕E𝜕cA
= 2cAHAA + 2cBHAB
Doing same with 𝜕∕𝜕cB gives
(2cB + 2cASAB)E +(c2
A + c2B + 2cAcBSAB
) 𝜕E𝜕cB
= 2cBHBB + 2cAHAB
Setting 𝜕E∕𝜕cA = 𝜕E∕𝜕cB = 0 leads to two simultaneous equations
cA(HAA − E) + cB(HAB − ESAB) = 0
cA(HAB − ESAB) + cB(HBB − E) = 0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)Writing these in matrix form gives⎛⎜⎜⎝
HAA − E HAB − ESAB
HAB − ESAB HBB − E
⎞⎟⎟⎠⎛⎜⎜⎝
cA
cB
⎞⎟⎟⎠ = 0
Matrix diagonalization problem can be solved with determinant,|||||||HAA − E HAB − ESAB
HAB − ESAB HBB − E
||||||| = 0
In homonuclear example make it little easier since HAA = HBB = 𝛼.Also set HAB = 𝛽 and S = SAB|||||||
𝛼 − E 𝛽 − ES
𝛽 − ES 𝛼 − E
||||||| = 0, which gives (𝛼 − E)2 − (𝛽 − ES)2 = 0
P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)
(𝛼 − E)2 − (𝛽 − ES)2 = 0
which leads to𝛼 − E = ±(𝛽 − ES) = ±𝛽 ∓ ES
and we find 2 solutions for E:E+ =
𝛼 + 𝛽1 + S
and E− =𝛼 − 𝛽1 − S
Putting solution for E+ back into simultaneous Eqs one can show that cA = cB.Put solution for E− into 2 simultaneous equations and obtain cA = −cB.Thus, 2 solutions for wave function are
𝜓𝜎g= c
(𝜙1sA
+ 𝜙1sB
), and 𝜓𝜎u
= c(𝜙1sA
− 𝜙1sB
)Normalizing these two wave functions gives
𝜓𝜎g= 1√
2 + 2S
(𝜙1sA
+ 𝜙1sB
)and 𝜓𝜎u
= 1√2 − 2S
(𝜙1sA
− 𝜙1sB
)P. J. Grandinetti Chapter 22: Diatomic Molecules
Linear Combination of Atomic Orbitals (LCAO)
Bring two 1s orbitals together in phase for 𝜓𝜎gand out of phase for 𝜓𝜎u
(A) (B)
Above is comparison of Exact (solid lines) and LCAO (dashed lines) wave functions 𝜓𝜎gand
𝜓𝜎ufor 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 = ∫V𝜙∗
1sA𝜙1sB
d𝜏 = e−RAB
(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 → ∞.With decreasing R overlap integral increases and reaches value of 1 at R = 0.
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Coulomb Integral𝛼 integral is called Coulomb Integral
𝛼 = ∫V𝜙∗
1sA𝜙1sA
d𝜏
To evaluate 𝛼 start with electronic Hamiltonian in atomic units = −1
2∇2
e −1rA
− 1rB
+ 1RAB
which can be written = A − 1rB
+ 1RAB
or = B − 1rA
+ 1RAB
A or B are Hamiltonians for e− in H-atom alone. Thus,
𝛼 = ∫V𝜙∗
1sA
[A − 1
rB+ 1
RAB
]𝜙1sA
d𝜏 = ∫V𝜙∗
1sAA𝜙1sA
d𝜏 − ∫V𝜙∗
1sA
1rB𝜙1sA
d𝜏 + 1RAB
which gives 𝛼 = E1s +2E1sRAB
[1 − e−2RAB(1 + RAB)
]+ 1
RABCoulomb Integral contains energy of e− in 1s orbital of H-atom, attractive energy of nucleus Bfor e−, 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
𝛼 decreases monotonically (i.e., no minimum) from ∞ at RAB = 0 to −1∕2 at RAB = ∞. In other words,𝛼, which is leading term in
E+ =𝛼 + 𝛽1 + S
and E− =𝛼 − 𝛽1 − S
does not give any stability to H+2 over 2 infinitely separated nuclei (recall H atom has energy of −Eh∕2).
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Exchange Integral
Finally, examine 𝛽 integral, also called the resonance or Exchange Integral
𝛽 = ∫V𝜙∗
1sA𝜙1sB
d𝜏
which becomes
𝛽 = ∫V𝜙∗
1sA
[B − 1
rA+ 1
RAB
]𝜙1sB
d𝜏 = ∫V𝜙∗
1sAB𝜙1sB
d𝜏−∫V𝜙∗
1sA
1rA𝜙1sB
d𝜏+∫V𝜙∗
1sA
1RAB
𝜙1sBd𝜏
to obtain𝛽 = E1sS + 2E1se−RAB(1 + RAB) +
SRAB
P. J. Grandinetti Chapter 22: Diatomic Molecules
LCAO : Exchange Integral
10 2 3 4-1
0
1
2
3
4 𝛽 integral goes through a minimum inenergy.It is stabilization energy from allowing e−to move (exchange) between 2 nuclei.Since both 𝛼 and 𝛽 are negative, E+ willbe lowest energy,
E1𝜎g= E+ =
𝛼 + 𝛽1 + 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) − E(∞) = 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
E1𝜎u= E− =
𝛼 − 𝛽1 − S
, (anti-bonding)
This orbital gives no stability since 𝛽 raises total energy in this case.Putting lone electron into 𝜓1𝜎u
would destabilize H+2 molecule and cause it to break apart.
P. J. Grandinetti Chapter 22: Diatomic Molecules