General Physical Chemistry II
 
 Lecture 9
 
 
 
 Aleksey Kocherzhenko
 
 
 
 
 October 2, 2014"

Last time…"

Valence bond theory" Valence bond theory:" Ø  A bond is formed when an electron in an atomic orbital on one atom pairs its spin with that of an electron in an atomic orbital on another atom"

σ- and π-bonds" z!

σ-bonds symmetric about
 the internuclear axis (usually labeled z )"

What if atoms share more than one electron pair?"

x!

[He] 2s2 2px1 2py1 2pz1"

y! π-bonds formed by two side-by-side p-orbitals"

z!

z!

N2 molecule: ! σ-bond formed by pz– pz orbitals N2 molecule " π-bonds formed by px– px and py– py orbitals"

z!

Polyatomic molecules" What does valence bond theory tell us about water?"

1s1"

O" H"

104°"

H"

[He] 2s2 2px2 2py1 2pz1"

Valence bond theory predicts two σ-bonds, between 2py and 2pz orbitals of O and the
 1s orbital of H" 2py ⊥  2pz à predicted bond angle = 90°, experimental value = 104°" So far, valence bond theory does a poor job predicting bond angles
 (and, in some cases, the number of possible bonds)"

How do we accurately predict 
 the possible number of bonds for an atom 
 and the magnitude of bond angles?"

Promotion" How many bonds can carbon form! Before it can form no more?!

Bob Dylan"

H"

[He] 2s2 2px1 2py1 2pz0"

H"

promotion"

Methane"

H"

C" H"

Valence electron can be promoted from full to empty atomic orbital (AO)"

2 bonds?!

[He] 2s1 2px1 2py1 2pz1 – ? "

Energy required for promotion is recovered
 by the atom’s ability to form 4 bonds instead of 2" s-orbitals and p-orbitals
 are very different, but:"

2 unpaired electrons"

4 unpaired electrons!"

Ø  All bonds in CH4 are of equal length! Ø  All angles between bonds are the same!

Hybridization" s-orbitals and p-orbitals
 are very different, but:"

Methane"

H"

Ø  All bonds in CH4 are of equal length! Ø  All angles between bonds are the same! Orbitals are standing waves!

H"

(and thus, can interfere with each other)"

H"

C" H"

Ø  Hybrid orbitals are interference patterns formed by orbitals of different symmetry" Ø  Electrons in a promoted atom occupy hybrid orbitals" Ø  Mathematically, hybrid orbitals are 
 linear combinations of s- / p- / d-orbitals"

You are familiar with different types of hybrid orbitals from GenChem:
 sp3, sp2, sp, …" How are these orbitals defined mathematically in terms of s, px, py, pz?"

sp3 hybridization" sp3 hybrid orbitals are built from one s-orbital and three p-orbitals" 1

=

1 2

2s

+

2px

+

2py

+

2pz

2

=

1 2

2s

+

2px

2py

2px

2py

2pz

Hybrid orbitals are identical
 but differently oriented:"

3

=

1 2

2s

2px

+

2py

2pz

4

=

1 2

2s

As many hybrid orbitals
 as non-hybrid orbitals
 that they are built from:" s, px, py, pz à four sp3 orbitals"

+

2pz

Hybridization results from interference" Radial part
 of wavefunction" Ø  The s- and p-orbital are added interfere constructively (add)
 where they have the same sign
 and interfere destructively (subtract) where they have opposite signs"

Ø  Angular dependence of sp3 orbital
 is determined by the p-orbitals
 (s-orbital is spherically symmetric) "

Bonding for sp3 orbitals" Methane"

H"

H"

H"

C" H"

Each sp3 orbital on C atom forms a σ-bond with 1s orbital of H atom"

109.47

All σ-bonds are identical,
 apart from their orientation in space " Tetrahedral structure"

Orthogonality"

Definition:"

ZZZ

⇤ n

m dV

=0

for"

n 6= m

V

Ø  Eigenfunctions that correspond to different eigenvalues
 of a quantum mechanical operator are orthogonal " Ø  Applies to the orbitals that belong to the same atom"

sp2 hybridization" sp2 hybrid orbitals are built from one s-orbital and two p-orbitals" Three sp2 orbitals
 " "(lie in xy-plane)"

py" s"

px"

sp2 hybrid orbitals participate in σ-bonds " H 1s"

H 1s"

C"

H 1s"

C"

H 1s"

The remaining non-hybridized orbital participates in π-bond " 2pz"

2pz"

C"

C"

Constructing sp2 hybrid orbitals" Three sp2 orbitals
 " "(lie in xy-plane)"

py" s"

y"

px"

c

a a

= c1

2s

+ c2

2px

+ c3

2py

x"

b

= c4

2s

+ c5

2px

+ c6

2py

cn ?

c

= c7

2s

+ c8

2px

+ c9

2py

n = 1, 9

b

s-orbital is spherically symmetric à contributes equally to all hybrid orbitals:"

c1 = c4 = c7 = a is oriented along the x-axis à "c3

Symmetry requires that" c5

p1 3

=0

= c8 , c6 =

c9

Find the rest of the coefficients using normalization and orthogonality conditions"

Let’s do it!" a

=

b

=

c

=

Orbital:"

p1 2s + c2 2p x 3 p1 2s + c5 2p x 3 p1 2s + c5 2p x 3

Radial part:"

y"

c

a

x"

+ c6

2py

c6

2py

c2 , c5 , c6 ? Angular part:"

b

Coefficients for sp2 hybrid orbitals: Results" y"

This is the result that you should get:"

a

b

c

= = =

p1 3 p1 3 p1 3

⇣

✓

✓

2s

2s

2s

+

p

2

q

q

1 2 1 2

2px

a

⌘

x"

q

2px

2px

c

+

q

3 2 3 2

2py

2py

◆

◆

b

sp hybridization" sp hybrid orbitals are built from one s-orbital and one p-orbital" Two sp orbitals" s" pz" sp hybrid orbitals participate in σ-bonds " H 1s"

The non-hybridized p-orbitals participate in π-bonds"

H 1s" C"

1

2

= =

q q

2py"

2py"

C" 2px" 1 2

(

2s

1 2

(

2s

+

2pz ) 2pz )

C" 2p x"

C"

Hybridization type and molecular shape"

Hybridization of N atomic orbitals always results in formation of N hybrid orbitals"

Hybridization ratio" Pure sp2"

Pure p"

Hybrid orbitals can have an intermediate proportion of different type of pure orbital character"

Ionic bonds" Select so as to minimize total energy"

Covalent-ionic resonance" HCl

=

H Cl

Covalent wavefunction: e–’s equally distributed between H and Cl" 2

Small

+

H+ Cl

Ionic wavefunction: both e–’s are on Cl, no e–’s on H"

– relative proportion of ionic contribution"

: primarily covalent bond"

Large

: primarily ionic bond"

Resonance" Resonance:" a wavefunction is expressed as a superposition of wavefunctions corresponding to different electron distributions with the nuclei at the same location, !

= c1 |cn |

2

1

+ c2

– relative contribution of

2

n to wavefunction!

Accounting for resonance structures results in a more accurate wavefunction that better describes the electronic distribution (“Resonance stabilization”)" Famous example:
 Kekule structures of benzene!

Adding more structures gives more accurate description of e– distribution"

Summary" Ø  Promotion of a valence electron from a full orbital to an empty orbital allows atoms to form more bonds (this recovers the energy required for promotion)" Ø  Electrons in atoms where promotion occurs occupy hybrid orbitals that result from interference of orbitals with different symmetry (s / p/ d)" Ø  Mathematically, hybrid orbitals are linear combinations of s- / p- / d-orbitals" Ø  Hybridization of N atomic orbitals results in formation of N hybrid orbitals" Ø  Common hybridizations: sp3, sp2, sp (intermediate ratios also possible)" Ø  By definition, two wavefunctions are orthogonal if"

ZZZ

⇤ n

m dV

=0

for"

n 6= m

V

Eigenfunctions corresponding to different eigenvalues of operator are orthogonal Ø  Resonance: wavefunction expressed as a superposition of wavefunctions with different electron distributions for nuclei at the same location! Ø  Accounting for resonance structures à more accurate wavefunctions!