Quantum Chemistry and Spectroscopy Text book Engel: Quantum Chemistry and Spectroscopy (3 ed., QCS) or Engel and Reid: Physical Chemistry (3 ed.)

Chapter 6. Many electron molecules 2 (Engel: QCS chapter 13) The two atomic molecules are very small part of chemistry and we need to extend the QM treatment to larger molecules. Naturally the wf is more complicated and simple models are less reliable. But the simple model still describe many of the essential features of the molecules. One simple model is the Lewis’ VSEPR model. This model states · · · ·

Ligands (here atoms) and electron pairs will repel each other Electron pairs takes more space than ligands (atom) Ligand (atom) increases with the increase of the electronegativity Multiple bonds will take more space that single bond

Hybrid orbitals Let us look C-H and O-H bonds. As we saw in the Huckel calculation of the HF the F:2p and H:1s orbitals will form a sbond. In general any two atomic orbital that are close in energy (and space) will form a bonding orbital. Now we have a small problem in the methane molecule, CH4. The carbon have electronic structure of 2s2 2p2 clearly the p-orbitals can form bonds with H:1s orbitals but there is only 2 p electrons and they are in 90o angle and what the 2s orbital are doing. Bonds with 90 o angle and two different type of bond is not in agreement with the structure of methane. Explanation: the 2s and 2p orbitals are rather close in energy and they will hybridize. They will form 4 new equivalent orbitals which are called sp3-orbitals. These orbitals will form a tetrahedron.

h1(sp3)= (-f2s + f2px + f2py + f2pz)/2 h2(sp3)= (-f2s - f2px - f2py + f2pz)/2 h3(sp3)= (-f2s + f2px - f2py - f2pz)/2 h4(sp3)= (-f2s - f2px + f2py - f2pz)/2 These hybrid orbitals will now bind with H:1s orbitals 2p

-↑- -↑- -sp3

=> 2s

-↑- -↑- -↑- -↑-

-↓↑-

Carbon can also form other hybrids. In the sp2 hybrid the 2s and two 2p’s are hybridized h1(sp2)= -f2s/√3 + f2px√2/√3 h2(sp2)= -f2s/√3 - f2pz/√2 - f2px/√6 h3(sp2)= -f2s/√3 + f2pz/√2 - f2px/√6

Now the orbitals will form 120o angle 2p

-↑- -↑- -=>

2s

-↓↑-

sp2

-↑- -↑- -↑-

+

2p -↑-

The last hybrid is sp where 2s and one 2p will hybridize h1(sp)= (-f2s + f2pz)/√2 h2(sp)= (-f2s - f2pz)/√2

The methane molcules electronic structure is now easy. The hydrogen will drive the carbon to form sp3 hybrid and then the hybrid orbitals will form s-bond with H:1s, s(h,1s) --sp3 -↑- -↑- -↑- -↑- + 4–↑- 1s

---

---

---

s*

=> -↑↓- -↑↓- -↑↓- -↑↓- s

The ethene (ethylene) is an interesting case. It have a double bond, H2C=CH2, and we can use the hybrid orbital concept to explain its structure. Now the carbon is sp2 hybridized

Two of the hybrid orbital are used to form the C-H s-bonds. One hybrid orbital will for a s-bond between the carbons (C-C). The remaining 2px orbitals will form p-bond between the carbons. The p-bond is the second CC bond and explains the CC double bond (C=C).

---

---

s*

sp2 -↑- -↑- -↑- + 2px-↑- + 41s –↑- => -↑↓-

p(2px,2px)

-↑↓-

s(sp2,sp2)

-↑↓- -↑↓-

s(sp2,1s)

You very likely know that the double bond is rigid and it cannot be rotated. This is easy to explain with the orbitals. The nonhybridized px orbital is defined with the CHH plane. Now if we rotate the CH2 group the px orbitals become orthogonal (px and py) and they don’t form a bond. The rotated ethene will have a single bond which is energetically unfavorable.

rotation

Water molecule Water molecule is a rather delicate case. Let us first see what happen is we assume the sp-hybridization. Like methane water has also 8 valence electrons. Analogous to methane we can assume that the orbitals of oxygen will form sp3-hybridization. In water two of the hybrid orbital will form bond to hydrogens and

two are occupied with the electron pairs. According the VSEPR the electron pairs will take more space than H so the H-O-H angle will be smaller than 109.4o. The experimental angle is 104.5o.

O:sp3 -↑- -↑- -↑↓- -↓↑- + 2–↑- H:1s

---

---

s*

---

---

s*

=> -↑↓- -↑↓-

s(Oe)

-↑↓- -↑↓-

s(OH)

The measured molecular orbital of water molecule do not agree well with this model. The other possibility is to assume that oxygen does not hybridize. The 2s-2p energy separation is larger than in carbon so the hybrid orbital is more expensive to build. Again the naïve interpretation would lead to linear molecule. --O:p4 -↑- -↑- -↑↓- + 2–↑- H:1s s2

-↓↑-

=>

---

-↑↓- -↑↓-

s* O:px,py

-↑↓-

s(O:2pz,H:1s)

-↑↓-

s(O:2s,H:1s)

Now both of the models are wrong and the correct electronic structure is between of them. The later picture is more correct but there is little hybridization.

The oxygen linear “water” molecule have pure 2s and 2p orbitals.

Exercise: explain the structure and electronic structure of an ammonia molecule.

Huckel model 2

We discussed a bit of Huckel model in previous chapter but let us return to it. It is a very simple but general model to calculate general LCAO wf’s. For four orbital system (8 electrons) the trial wf is F = c1f1+c2f2+c3f3+c4f4 If we apply that to methane we would need 8 orbitals since C has 4 valence orbitals and the hydrogens 1 each. The generalization is simple so let us look first the four orbital system. The Huckel matrix can be built from the Schrödinger equation Hϕ=Eϕ. ⟨Φ| |Φ⟩ = ⟨Φ|Φ⟩, Φ =

+

+.. ⟹

=

(6.1)

Where the Hij and Sij are the energy expectation and overlap matrix. The typical LCAO orbitals are non-orthonormal so S ij≠0. The solution of the equation (6.1) can be found by solving E from the determinant

H 11 - E H - ES 21 det 21 H 31 - ES31 H 41 - ES 41

H 21 - ES 21 H 22 - E H 32 - ES 32 H 42 - ES 42

H 31 - ES 31 H 32 - ES 32 H 33 - E H 43 - ES 43

H 41 - ES 41 H 42 - ES 42 =0 H 43 - ES 43 H 44 - E

This have 4 solutions and once the E’s are known the corresponding ci,n can be solved. The most complex part is the solution of the matrix elements Hij. In the Huckel model strong approximation for Hij and Sij are made. Huckel model All Sij‘s =0 Diagonal Hii’s are the ionization potentials For Hij only nearest neighbor interactions are included

Extended Huckel model: Nearest neighbor Sij‘s ≠ 0

The Huckel model for methane is. The symbols are Hss – 2s energy of C, Hpx – 2px energy of C, HHs – 1s energy of H, h – interaction of C:2p and H:1s, the angle θn is the angle between C-H vector and coord n, h’ - interaction of C:2s and H:1s. The p orbital will not interact similarly with all the hydrogens. − 0 0 0 ℎ′ ℎ′ ℎ′ ℎ′

ℎ ℎ ℎ ℎ

0 − 0 0

ℎ ℎ ℎ ℎ

0 0 − 0

ℎ ℎ ℎ ℎ

0 0 0 −

ℎ ℎ ℎ

ℎ′ − 0 0 0



ℎ ℎ ℎ

ℎ′ 0 − 0 0

ℎ ℎ ℎ

ℎ′ 0 0 − 0

ℎ ℎ ℎ

ℎ′ 0 0 0 −

=0

This matrix will not give correct answer since the Hss and Hpp energies are different. We need to change to the hybrid orbitals. Now the Hnn means one of the hybrid orbitals. Also the interaction parameter is easier. Now the h1 hybrid orbital will interact with only one hydrogen. We assume the tetrahedral geometry. In general geometry the angles need to be include.

− 0 0 0 ℎ 0 0 0

0 − 0 0 0 ℎ 0 0

0 0 − 0 0 0 ℎ 0

0 0 0 − 0 0 0 ℎ

ℎ 0 0 0 − 0 0 0

0 ℎ 0 0 0 − 0 0

0 0 ℎ 0 0 0 − 0

0 0 0 ℎ 0 0 0 −

=0

The methane example shows that the simplest Huckel model is a bit problematic. But when the Huckel parameters are reasonable it will give qualitatively correct results and often even quantitatively reasonable results. For example simple Huckel (or Tight Binding) model will give good results for carbon materials like fullerenes or carbon nanotubes.

Conjugated hydrocarbons The conjugated hydrocarbons is also a good example where the Huckel model works. One example is 1,3-butadiene. It is interesting to know how the single-double bond structure comes from the Huckel model.

Now the carbons will form sp2 hybridization. The hybrid orbitals (and H:1s orbitals) will form the C-H and C-C bonds. We focus on the pz orbitals not included to the hybridization. There is one orbital and electron per C, so 4 orbitals and electrons. The full LCAO determinant is

H11 - E H 21 - ES21 Det H 31 - ES31 H 41 - ES41

H 21 - ES21 H11 - E H 32 - ES32 H 42 - ES42

H 31 - ES31 H 32 - ES32 H11 - E H 43 - ES43

H 41 - ES41 H 42 - ES42 =0 H 43 - ES43 H11 - E

With the Huckel approximations this simplifies to

H11 - E h Det 0 0

h H11 - E h 0

0 h H11 - E h

0 0 =0 h H11 - E

This is easy solve. (see Engels book, page 296) The energies are E = H11±1.62h, E = H11±0.62h and the coefficients can be solved. The Huckel matrix will lead to four orbitals but due to spin only two of the lowest ones are occupied. The lowest on is a bonding orbital between all the bonds but the next one is antibonding in middle of the molecule. The second orbital is bonding in the ends of the molecule. Together these orbital can form 2 bonds. One in both ends of the molecule and a weak bond in the middle. The quantum system have binding energy than the two separate bonds. In this model the p-bond will have energy of 2h (2 electrons). Now if we add the energies we get

2*(2H11+2.24h) (2 for 2 electrons) now the extra energy is 0.48h.

The same model can be used for benzene. Again we are interested only of the pz electrons.

Now the Huckel determinant is

Det

H11 - E h

h H11 - E

0 h

0 0

0 0

h 0

0

h

H11 - E

h

0

0

0

0

h

H11 - E

h

0

0

0

0

h

H11 - E

h

h

0

0

0

h

H11 - E

=0

This is easy to solve. The energies are E = H11±2h, E = H11±h. Again the lowest state is full bonding. The second state is degenerate. The left side wf will form a bond between atoms 2-3 and 5-6. The right side wf have some bonding between atoms 1-2, 3-4, 4-5, 6-1 and antibonding characterization between atoms 2-3 and 5-6. We can define the bond strength between atoms 1 and 2 to be proportional to the product c1*c2.

With this analysis the bonding and antibonding interactions will cancel each other and also the bonds due to the second energy level are equal. The simple Huckel model predict that in benzene all the C-C bonds are equal. The separate bond energy is 6h and the p-bond energy in benzene is 8h, so the delocalization energy is 2h.

We can also use the Huckel method for other (CH)m-type cyclic hydrocarbons. The orbital energies are show below. The benzene (m=6) have strong bonding orbital (well below the a level). The cyclo-pentane (C5H5) have also strong bonding orbital but this orbital have only 3 electrons. Thus the negative C5H5- is more stable than the neutral one and C5H5 is also rather electro negative. Similarly C5H5+ is less stable than neutral C5H5. Last the C5H5 binds well to metal like Fe (Ferrocene). C4H4 is diradical and thus very reactive. C3H3+ is more stable than neutral and negative C3H3. All these predictions have been confirmed with experiments so, even being very simple, the Huckel model can predict quite well some molecular properties. The Huckel model can be used to compare the energy of cyclic and linear molecules of the same composition. The Huckel energy of similar (poly)cyclic and linear molecules can be computed. The delocalization energy can be used a rough measure of the stability of the molecule. In the picture below the number is the coefficient of the h. So the large the number the more stable the molecule is (with respect to a linear molecule). Note that benzene and napthalene are particularly stable whereas cyclobutadiene is not stable.

Summary: The Huckel model is a useful computational method which in many cases can even predict chemical phenomena. The obtained energy levels are very often qualitatively correct. Also the wave functions are easy to interpret.