chapter 22 - diatomic molecules - grandinetti€¦ · p. j. grandinetti chapter 22: diatomic...

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


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...


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:


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?


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.



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.



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.


Two sheet hyperboloid



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.



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 ...


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.


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.


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.


Exact solutions for 1𝜎g and 1𝜎u of H+2 as a function of R

(A)

(D)

(E)

(F)

(B) (C)


Shape of H+2 wave functions

When R = 0 solution becomes identical to He+ wave function.


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 .


Linear Combination of Atomic Orbitals (LCAO)


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


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



Linear Combination of Atomic Orbitals (LCAO)Taking the derivative of both sides

𝜕𝜕cA

(c2

A + c2B + 2cAcBSAB

)E = 𝜕

𝜕cA

(c2


)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


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



(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


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.


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.


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.


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).


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


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)


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.


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.


chapter 22 - diatomic molecules - grandinetti€¦ · p. j. grandinetti chapter 22: diatomic...

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