1. Chapter 6. Orbitals and Chemical Bonding I: The Valence Bond Model and Molecular Geometry
Hydrogen Molecule
Valence Bond Model
Atomic Hybridization
Electron-Pair Repulsion:
Molecular Shapes
Electric Dipole Moment
σ and π Bond Types
Shortcomings of the Valence Bond Model
How to make Lewis electron-dot structures somehow consistent with quantum mechanics?

2. 6.1 Hydrogen Molecule: The Simplest Electron-Pair Bond
• H2: simplest covalent bond
• When the neutral atoms are
very far apart, two electron-
nuclear attractions are
vanishingly small.
• Then, the system can be approximated to a system of two independent H-atoms.
• The wave function for the
ground state becomes the
product of the 1s orbitals
on the two atoms:
(6.1)
Nevertheless, QM solution for the exact ψ is impossible.

3. • Substituting Eq.(6.1) into the Schrødinger equation, the output:
Fig.6.2 Bond potential energy curve & orbital overlaps
Eq(6.1)
Heitler-London Treatment of H2
• Allowing nuclei a and b to share
electrons 1 and 2 (or, the electrons
exchange between a and b), we
should consider both 1sa(r1)1sb(r2)
and 1sa(r2)1sb(r1). Therefore,
Eq(6.2)
(Spectroscopy)
(6.2)
a normalization const.
Importance of orbital overlap Electrons → can only be exchanged between atoms if their orbitals have an overlap
← overlap integral ψa= valence orbital of atom a
S ≠ 0 for exchange

4. Exchange energy:
Non-classical – it cannot be described as an electrostatic interaction among charges.
Formation of a covalent chemical bond is a fundamentally quantum-mechanical phenomenan.
These orbital products allow for swapping or exchange of the electrons
What about Pauli’s exclusion principle?:
Electron spins had to be paired just as in an atomic orbital.
Lewis’ ideas fit nicely with this => The electron pair bond is justified from the orbital analysis.
This is not an orbital (wave function) for one electron case, but
a two-electron wave function !

5. 6.2 Polyatomic Molecules: Pauling’s Valence Bond Model
• Pauling: all electron sharing within a Lewis molecular structure
~ can be accounted for by valence-bond wave functions
like the Hitler-London treatment
• covalent bond electron pair of
F (1s22s22p5)
(6.4)
Fluorine uses valence 2p orbital for the overlap bond.
All other F electrons remain paired
in their filled AOs

6. Fig. 6.3 Valence bond model for H2O
• A VB for H2O employs two mutually perpendi-
cular 2px and 2py of O in making overlaps with
the 1s orbital of H.
• Two identical VB wave functions obtained from Eq.(6.4) replacing pF by 2pO are the two covalent bonds pointing at a right angle.
• The lone pairs on O then occupy the remaining 2pZ and the 2s.
• It may be thought that, due to the polar covalent bonding, each H bears a partial positive charge +δ, and the two H’s therefore repel each other by Coulomb’s law. Correct? No! Why?

7. Encountered Problems of VB Model
• Methane, CH4: Tetrahedral (109.5°)
• C (1s22s22p2)  How?
• 4 bonds from VB model
 breakup of 2s2
but, 3 at 90° angles (2p)
1 with no direction (2s)
• Directional 2p orbitals
 more effective at overlapping the H 1s than 2s of C
 expects stronger and shorter bonds than 2s
How adjust the reality ? What is the Next ??

8. Pauling (1931) propsed ‘Hybrid Orbital ’
A new set of orbitals
• 4 orbitals  identical except for spatial orientation,
• hybrid orbitals  formed by linear combinations of the
original 2s and 2p orbitals
• one of these hybrid orbitals: sp3 orbital
h2 = ½(2s -2px-2py+2pz)
h3 = ½(2s -2px+2py-2pz)
h4 = ½(2s +2px-2py-2pz)
• sp3 hybridization can be extended further
← VB model: sp3 hybridization for 2 C

9. BeH2 is a linear molecule (sp hybrid)
* 2 equivalent Be-H bonds from Be(1s22s2) 
2 linear hybrid orbitals 
contributed from 2s and 2px of Be
Boundary surfaces for 2s and 2px
sp hybrids
Overlaps of the hybrids

10. The hybrids:
(6.6)
Overlapping each sp hybrids with H 1s then produces the requisite bonding characteristics.

11. BF3 (sp2 hybrid)
• B (1s22s22p1)
• hybridization (sp2)
 Planar structure (x-y plane)
 usable orbitals are 2s, 2px, and 2py
* A trigonal-planar structure, with a central boron and the fluorines located at the vertices of an equilateral triangle, yielding bond angles of 120°. The usable orbitals are 2s, 2px, and 2py . Since these orbitals are not adapted to 120°, the VB model requires hybridization, sp2. (Fig. 6.6)
The boron lacks an octet but is relatively electronegative, making BF3 a strong Lewis acid and an important catalyst for many organic reactions.

12. Fig. 6.6

13. • N (1s22s22p3)  expects 3 H-N-H bonds at 90o
Trigonal pyramidal, with one lone pair electrons on N
NH3
• N (1s22s22p3)  expects 3 H-N-H bonds at 90o
• In reality, it is 107.5o  needs alternative view: hybridization
• Orbitals of N  sp3 hybrid
• One lone pair of N  occupying an sp3 orbital
• by making sp3 hybrid orbitals  roughly tetrahedral

14. • O (1s22s22p4)  expects 2 H-O-H bonds at 90o
V-shaped, with two lone pairs on O
H2O
• O (1s22s22p4)  expects 2 H-O-H bonds at 90o
• In reality, it is 104.5o  needs alternative view: hybridization
• Orbitals of O  sp3 hybrid
• Two lone pairs of O  occupying an sp3 orbital
• by making sp3 hybrid orbitals  roughly tetrahedral

15. 6.3 Double and Triple Bonds: σ and π Bond Types
The valence bond model for multiple bonding:
Ethylene (C2H4) – planar
• angles of C-C-H and H-C-H: ~ 120° trigonal planar
 suggests sp2 hybrid of each C
• C: 1s22s22p2
• if sp2  each C has one extra 2p orbital (not used for hybrids)
• leftover 2p  perpendicular to the bonding plane
• each leftover 2p orbital of each C  overlap and form π-bond

16. Valence bond σ-π picture
Cylindrically symmetric about the bond axis (σ-bond)
Overlapped orbitals form π-bond
two nonequivalent types of bonds (σ and π bonds)
C2H4 is relatively more reactive than C2H6
C2H4 is readily attacked by electrophiles:
C2H4 + Br2 → C2H4Br2
C2H6 + Br2 → no reaction

17. Acetylene (C2H2) • C2H2  more reactive than C2H4
• C2H2  Lewis structure H-C≡C-H (linear)
• C  sp hybridization
• C-C s bond  sp-sp overlap
• Two unused p orbitals on each C
 perpendicular to molecular axis
• each parallel pair of p’s
 overlap to form π bond
• C≡C  one σ and two π bonds

18. EXAMPLE 6.1
Give a valence bond description of and sketch the bonding in the
NO2-, given an ONO bond angle of 118°, and two equal bond lengths.
Solution:
Lewis structure
•Since both resonance forms are
identical in bonding characteristics,
only one of them is considered.
• Bond angle and double bond suggest sp2 hybridization of N.
• The bonding consists of a pair of N-O σ-bonds involving sp2 hybrids on N,
and either unhybridized 2p or sp3 hybrids on O.
• The unused 2p on N pointing perpendicular to the molecular plane and
corresponding 2pπ orbital on O’s
form a π bonding.

19. 6.4 Electron-Pair Repulsion: Predicting Molecular Shapes
• Two VB electron pairs repel each other
due to Coulomb repulsion
• Repulsion makes the orbitals swing away
to give lowest potential energy
 valence-shell electron-pair repulsion (VSEPR)
σ-Bonded Molecules
potential energy
three-pair case
four-pair case
Department of Chemistry, KAIST

20. Bond angles and Geometries
• The bond angles, HXH, of
CH4, NH3, OH2 
109°, 107°, 105°
• Lone pair electrons 
more repulsive than bonding
pair, because ???
equatorial position?
axial position?
the arrangement of electron pairs  atomic geometry

21. Expanded Octet Molecules
S atom accommodates 5 electron pairs
Trigonal-bipyramidal structure
• lone pair electron  equatorial (experimentally known)
 Sawhorse geometry

22. Molecules with π Bonds • C2H4 ~ trigonal planar about each C atom
with σ and π valence bonds between the carbons
• C=C p-electron density is concentrated above and below the molecular plane
 π bond exerts smaller effect than σ-bond for molecular geometry
• In the VSEPR scheme, π bonds are minor perturbations
 ignore them in classifying the geometry types
Molecules with π Bonds
CO2 is linear
SO2 is bent (O-S=O angle, 119°)

23. Heteroatom Effect (p 193-194)
We will skip this part !!!

24. Overall Assessment of VSEPR (p 195)
We will skip this part !!!

25. 6.5 Asymmetric Electron Sharing: Electric Dipole Moment
• Unbalanced electron sharing in pair bond between atoms with different electronegativity (χ)  constitute an electric dipole  the molecules possess a dipole moment (m)
6.5 Asymmetric Electron Sharing: Electric Dipole Moment
μ = q r distance between charges
magnitude of the charge separation

26. HF a heteronuclear diatomic molecule with ΔΧ ≠ 0
μcal = δere = (δ)(4.8 x 10-10esu)(1x10-8cm) = (δ)4.8 Debye (D)
μexp = 1.82 D
a heteronuclear diatomic molecule with ΔΧ ≠ 0
δe: charge separation (q), re: bond length
bond distance of HF
41% of the full electronic charge is transferred from H to F or
41% of HF is ionic

27. measuring dipole moments by capacitance
no E, polar molecules are randomly oriented
E, dipolar molecules are lined up
• total dipole moment  vector sum of individual bond dipole

30. Effect of Heteroatoms
VSEPR rules predict a bent molecule, making μ non-zero

31. Example 6.4 Predict whether or not the following molecules
will have nonzero dipole moments:
in VB-VSEPR model, start with Lewis structures !
1 of 3 isomers
trans-1,2-difluoroethylene
μ = 0
2-electron pair
linear, μ = 0
4-pair case
bent, μ≠0
3-pair case
m ≠ 0

32. Valence Bond Description of Partial Ionic Character (p 199)
We will skip this part !!!

33. 6.6 Shortcomings of the Valence Bond Model
Inadequate treatment of odd-electron molecules and
resonances
2) Magnetism of molecules O2 N2
• Paramagnetic: molecules with
unpaired electrons
• Diamagnetic: weakly repelled
by a magnetic field
both are expected to be diamagnetic !!

34. then, what would be the alternative ??
3) Reaction Mechanisms and Molecular Energy Levels
• not suitable to describe the reaction in σ-bonded molecules
• not giving a model for electron transferring to a molecule of
filled valance orbitals and all of its electrons already paired
• not giving a description of molecular energy levels that can be excited by UV.
then, what would be the alternative ??

35. Department of Chemistry, KAIST
Fig Molecular Orbitals
Department of Chemistry, KAIST