homework class

4.1 Determine the sets of specific atomic orbitals that

can be combined to form hybrid orbitals with the

following geometries: (a) trigonal planar, (b) square

planar,(c) trigonal bipyramidal,(d) octahedral

Group table are attached.

4.1 Carter

What are the point groups?

How do you determinepoint group Symmetry?

ABB

B

trigonal planar

(a) D3h

Hybrid orbital construction, Carter, page 100

“To construct the reducible representation for any set of hybrid

orbitals, count the number of vectors in the basis set that remain

nonshifted by a respresentative operation of each class in the point

group of the system. The number of unshifted vectors is the

character for the class in each case”

4.1a-- Determine the sets of specific atomic orbitals that can be combined

to form hybrid orbitals

(a) trigonal planar -- D3h

What is the reducible representation for the hybrid orbital vectors?

ABB

B

trigonal planar

t

Count # unshifted vectors

3 0 1 3 0 1

t = A’1

A1':

(x axis coincident with C'2 axis)

Which orbitals?

+ E’

s, dz2 E': (px, py) , (dxy, dx2-y2)

4.1a


ABB

B

trigonal planar

A1': s, dz2 E': (px, py), (dxy, dx2-y2)

Use 3 atomic orbitals to make 3 hybrid orbitals can have

What are the hybrid orbitals?

s (px, py) sp2

s

(px, py)

sd2(dxy, dx2-y2)

(dxy, dx2-y2)

p2d dp2dz2

dz2 d3

A1' E' Notation

4.1b --Determine the sets of specific atomic orbitals that can be

combined to form hybrid orbitals for a square planar geometry-- XeF4

What is the point group?

D4h


4.1b What is the reducible representation for the hybrid orbital vectors

for a square planar geometry-- XeF4?

D4h

4 0 0 2 0 0 0 4 2 0

(x axis

coincident

with C'2 axis)

What are the irreducible representations??

tReducible representation

ni =1h

gc i rc

ni = number of times irreducible representation i occurs in the reducible

representation

h = order of the group = number of elements in the group

c = class of operations

gc = number of operations in the class

i = character of the irreducible representation for the operations of the

class

r = character for the reducible representation for the operations of the

class


USE

By inspection or……

4.1b Irreducible representations for square planar hybrids

/16t 4 0 0 2 0 0 0 4 2 0

14 0 0 4 0 0 0 4 4 0 16

04 0 0 -4 0 0 0 4 -4 0 0

14 0 0 4 0 0 0 4 4 0 16

04 0 0 -4 0 0 0 4 -4 0 0

08 0 0 0 0 0 0 -8 0 0 0

04 0 0 4 0 0 0 -4 -4 0 0

04 0 0 -4 0 0 0 -4 4 0 0

04 0 0 4 0 0 0 -4 -4 0 0

04 0 0 -4 0 0 0 -4 4 0 0

18 0 0 0 0 0 0 8 0 0 16

t = A1g Which orbitals?+ B1g + Eu

4.1b -- Which orbitals can be used?

t = A1g + B1g + Eu

A1g:

What are the

hybrid orbitals?

A1g B1g Eu Notation

s dx2-y2 (px, py) dsp2

dz2dx2-y2 (px, py) d2p2

s, dz2 B1g: Eu:dx2-y2(px, py)

4.1c --Determine the sets of specific atomic orbitals that can be

combined to form hybrid orbitals for trigonal bipyramidal


t


5 2 1 3 0 3

t = 2A 1 + A 2 + E

A1':


Which orbitals?

-- D3h

Determine Irreducible reps

A2": E':, dz2 pz ,(dxy, dx2-y2)(px, py) s

4.1c

(a) trigonal bipyramidal -- D3h

Use 5 atomic orbitals to make 5 hybrid orbitals

What are the hybrid orbitals?

2A1': s, dz2 A2": pz E': (px, py), (dxy, dx2-y2)

2A1' A2" E' Notation

s, dz2 pz (px, py)

(dxy, dx2-y2)

dsp3

d3sppzs, dz2

4.1d Determine the sets of specific atomic orbitals to form hybrid

orbitals for octahedral geometry

6 0 0 2 2 0 0 0 4 2t

t = + T1uA1g + Eg

Determine reducible

representation

– A1g 5 -1 -1 1 1 -1 -1 -1 3 1

– Eg 3 0 -1 1 -1 -3 -1 0 1 1

Irreducible rep’s?

Which orbitals do we use?

A1g:

Only possibility is

s Eg: (dz2, dx2-y2) T1u: (px, py, pz)

d2sp3

t = + T1uA1g+ Eg

4.3 Carter-- Borane

(a) Develop a general MO scheme for BH3. Assume that only the boron 2s and

2p orbitals interact with the hydrogen 1s orbitals (I.e., the boron 1s orbital is

nonbonding).

(b) The photoelectron spectrum of BH3 has not been observed. Nonetheless, if it

could be taken, what would you expect it to look like, based on your MO

scheme?

(c) Compare and contrast the general MO description of BH3 with a valence bond

(VB) model an its related loacalized MO model.

Borane, BH3, is an unstable compund produced by thermal

decomposition of H3B•PF3. Although it has not been isolated and

structurally characterized, it probably is trigonal planar.

MO’s for BH3

For simple molecule such as For simple molecule such as MXMXnn, match symmetries of the, match symmetries of theatomic atomic orbitals orbitals of M with symmetry adapted linear combinationsof M with symmetry adapted linear combinations((SALCSALC’’ss) of atomic) of atomic orbitals orbitals ofof thethe X X ligandsligands..

momo = a = a AOAO(M) ± b(M) ± b SALCSALC((nXnX))..

SALCSALC((nXnX)) = = iiccii ii where where ii are atomic are atomic orbitals orbitals of the X of the X ligandsligands

1. Use the directional properties of potentially bonding atomic orbitals on theouter atoms (depict as vectors) as a basis for a reducible representation ofthe SALC’s in the point group of the molecule.

2. Generate the reducible representation characters for these vectors.+1 if not shifted; -1 if shifted into the negative of itself (head becomes tail);and 0 if taken into another vector.

4. Determine symmetries of potentially bonding AO’s from central atom byinspecting the group table unit vector and direct product transformationslisted in the table.

3. Decompose the SALC reducible representation into its irreduciblecomponents. # of SALC’s must equal the number of starting basis AO’s

5. Central atom AO’s and pendant atom SALC’s of the same symmetrywill form both bonding and antibonding LCAO’s (molecular orbitals)

6. Central atom AO’s and pendant atom SALC’s form non-bonding MO’s if nosymmetry match

Molecular Orbital Generation BH3 Molecular Orbitals

What is the symmetry of BH3?

Expect D3h Symmetry -- no lone pair

3 – 3 = 0 for B

1–1 = 0 for H

3x1 + 3 = 6 valence electronsBH3

x

ox

oB

H

H

H

xo

Formal Charge

Draw Lewis dot structure…there are 3 boron valence electrons and

3 hydrogen valence electrons. The Lewis octet rule cannot be satisfied.

4.3-- Use the Hybrid orbital approach


What are the reducible representations for the H(1s) SALC’s?

ABB

B

trigonal planar

SALC


3 0 1 3 0 1

SALC =


Which Boron orbitals? A’1: E’ :

+ E’ A’1

s px, py A 2: pz

- A’1 2 -1 0 2 -1 0

General MO scheme

(a) Develop a general MO scheme for BH3. Assume that only the boron

2s and 2p orbitals interact with the hydrogen 1s orbitals (i.e., the

boron 1s orbital is nonbonding).

A’1: E’ :s px, py A 2: pz

Boron Hydrogen atoms

SALC = + E’ A’1

1. Bonding MOs always lie lower in energy than the anti-bonding MOs

formed from the same AOs.

2. Nonbonding MOs tend to have energies between those of bonding

and antibonding MOs formed from similar AOs.

3. interactions tend to have less effective overlap than sigma

interactions. -bonding MOs tend to have higher energies than -bonding MOs

formed from similar AOs. * MOs tend to be less anti-bonding and have lower energies than

* MOs formed from similar AOs

4. MO energies tend to rise as the number of nodes increases.

MOs with no nodes tend to lie lowest, and those with the greatest

number of nodes tend to lie highest in energy.

5. Among s-bonding MOs, those belonging to the totally symmetric

representation tend to lie lowest.

MO’s for BH3

pz

4.3 b The photoelectron spectrum of BH3 has not been observed.

Nonetheless, if it could be taken, what would you expect it to

look like, based on your MO scheme?

Two bands, both with vibrational fine structure. The lower-energy band, from the doubly degenerate (e') level, should be

roughly twice as big as the higher-energy band, from thenondegenerate (a1') level.

4.3c Compare and contrast the general MO description of BH3 with a

valence bond (VB) model an its related localized MO model.

Similar overall electron distributions.

VB model ------ sp2 hybridized boron AOs forming 2c-2e

bonds with hydrogen 1s orbitals.

Localized MO description----- all three bonding pairs in

localized sigma-bonding MO’s.

VB suggestion of energetic equivalence of all three

pairs disagrees with the symmetry restrictions of the

MO’s.

Octahedral Coordination- Transition Metals

Octahedral MLOctahedral ML66 molecular molecular orbitalsorbitalswhere L is an arbitrary where L is an arbitrary donor donor ligandligand

Filled ligand

p orbitalsEmpty ligand

p orbitals

What about bonding?

Only difference is when we derive the reducible representation, have toinclude x and y vectors that are perpendicular to the bond vector.

Example: CO2 Carter, Page 119

Use Descent in Symmetry (subgroup) approach ---Carter page 73

Vector basis for a representation of oxygen SALCs of CO2

Projection Operator

How do we determine the SALC combinations in practice?

SALCSALC((nXnX)) = = iiccii ii where where ii are atomic are atomic orbitals orbitals of the X of the X ligandsligands

Use Projection operator: i jd = ij ij =1 i = j

ij = 0 i j

Molecular Orbital Orthonormal Properties

Orthogonality for wave functions that are solutions to the wave equationi.e. can be defined by symmetry

Slater overlap integral forwave functions on two atomsA and B

Sij = A Bd

=

Apply symmetry operations on all orbitals. If class, include all operationsin class. “Expanded” table may be useful. For our problem....

+1

b

vc

+1

c

vb

+1

c

S2

3

+1

b

C 2c

+1

c

C 2b

+1

c

C23(z)

+1

a

va

+1

b

S13

+1+1+1+1A’1

aabaRj

a

h(xy)C 2aC13(z)ED3h

Multiply, add, and normalize to get hydrogen A1 SALC

1(A1) =

1

3a + b + c( )

To get SALC’s for the H atoms---Apply Projection Operator

BHb

Hc

Ha

2( E ) =

1

62 a b c( )

Do same for E’ irreducible representation

Projection Operator for BH3

0

b

vc

0

c

vb

-1

c

S2

3

0

b

C 2c

0

c

C 2b

-1

c

C23(z)

0

a

va

-1

b

S13

20-1+2E’

aabaRj

a

h(xy)C 2aC13(z)ED3h

BHb

Hc

Ha

See page 143 Carter --- need to get other MO for this irreducible

representation.

This function must be orthogonal to

Get 4a

- 2b

- 2c Normalization gives:

1(A1) and 2( E )

Result:

3 E ( ) =

1

2b c( )

Midterm Monday, November 7

Carter, Chapter 3, pages 66-73 Note: Systematic Reduction ofReducible Representations!!Carter, Chapter 4 (Symmetry and Chemical Bonding)Carter Chapter 6 (Vibrational Spectroscopy)

Huheey Chapter 3, pp 71-74Huheey Chapter 5 (Bonding Models in Inorganic Chemistry: TheCovalent Bond)Huheey Chapter 6, pages 203-218 (The Structure of Molecules)

Class Notes

Homework

6.1 Carter Determine the number of frequencies, their symmetries , and the infraredand Raman activities of the normal modes for the following molecules. Indicate thenumber of polarized Raman bands and the number of frequencies that should becoincident between the two spectra. Representations of the normal modes for thesestructures can be found in Appendix C. (a) NH3 (b) FeCl6

3– (c) H2CO (d) PF5 (e)C2H6 (staggered configuration), (f) H2O2

Then, categorize symmetries of all of the 3n degrees of freedom by pointgroup irreducible representations……

Derive the reducible representation for all 3N degrees of freedom of the molecule.

Can simplify this greatly! (Carter, page 174)

Thus, to find R, the character for the overall operation, countthe number of atoms that remain nonshifted by the operation ,No , and multiply by the contribution per unshifted atom, i

R = Ni i

I = character of the 3 x 3 block matrix of which the operation iscomposed

The contribution per nonshifted atom for a particular operation isthe same regardless of the orientation of its associated symmetryelement.

Moreover, the value of the contribution per nonshifted atom fora particular operation is the same in any point group in whichthe operation is found.

The character, I, that contributes to the reducible representationfor a given operation is universal for all point groups.

Atomic contributions, by symmetry operation, to the reduciblerepresentation for the 3N degrees of freedom for a molecule

ContributionOperation per atoma, I

E 3C2 -1C3 0C4 1C6 2

1i -3S3 -2S4 -1S6 0

aa CCnn = 1 +2cos(2= 1 +2cos(2 /n)/n)

SSnn = -1 + 2cos(2= -1 + 2cos(2 /n)/n)

Where do these come from?

6.1 a NH3 C3v

Determine reducible representations for 3N = 12 degrees of freedom for NH3

all atoms unchanged

1 + 1 + 1+ 1 = +4

1 atom unchanged

+1

2 atoms unshifted

+2

R = Ni i

4 1 2

3 0 1

12 0 2

Ni

A1 +1 +1 +1 z x2+y2, z2 A2 +1 +1 -1 Rz y(3x2-y2)E +2 -1 0 (x, y) (Rx, Ry) (x2-y2, xy)

(xz, yz)

CC3v3v E E 2C2C33 (z) (z) 33 vv

6.1 Carter Determine the number of frequencies, their symmetries , and the infrared

and Raman activities of the normal modes for the following molecules. Indicate the

number of polarized Raman bands and the number of frequencies that should be

coincident between the two spectra. (a) NH3

From the group table below, trans = A1 + E

rot = A2 + E

so that trans+ rot = 6 0 0

-( trans+ rot) 6 0 2

The remaining irreducible

rep’s can be obtained from

this reducible rep by using

ni =1h

gc i rc

to give 3n-6 = 2A1 + 2E

Ammonia -- C3v

tot

A1 +1 +1 +1 z x2+y2, z2 A2 +1 +1 -1 Rz y(3x2-y2)E +2 -1 0 (x, y) (Rx, Ry) (x2-y2, xy)

(xz, yz)

CC3v3v E E 2C2C33 (z) (z) 33 vv

4 1 2

Summary:

normal modes (irreducible rep of

3n-6) without x,y,z or quadratic

symmetry?

# totally symmetric normalmodes of 3n-6

mode active in both Raman and IR

(Carter, page 183)--exclusion rule?

pp 183-184

Carter

3n = 3A1 + A2 + 4E

trans =

rot =

3n-6 = 2A1 + 2E

4

4 (2A1+2E)

4 (2A1+2E)

2(2A1)

0

4 (2A1+2E)

A1 + E

A2 + EHow many independent

Raman and IR frequencies is

this?

6.1 Carter Determine the number of frequencies, their symmetries , and the

infrared and Raman activities of the normal modes for the following molecules.

Indicate the number of polarized Raman bands and the number of frequencies

that should be coincident between the two spectra. (a) NH3 (b) FeCl63– (c)

H2CO (d) PF5 (e) C2H6 (staggered configuration), (f) H2O2

7 1 1 3 3 1 1 1 5 3

3 0 -1 1 -1 -3 -1 0 1 1

21 0 -1 3 -3 -3 -1 0 5 3

R = Ni i

Atomic contributions, by symmetry operation, to the reduciblerepresentation for the 3N degrees of freedom for a molecule

ContributionOperation per atoma, I

E 3C2 -1C3 0C4 1C6 2

1i -3S3 -2S4 -1S6 0

aa CCnn = 1 +2cos(2= 1 +2cos(2 /n)/n)

SSnn = -1 + 2cos(2= -1 + 2cos(2 /n)/n)

Where do these come from?

ni =1h

gc i rc

ni = number of times irreducible representation i occurs in the reducible

representation

h = order of the group = number of elements in the group

c = class of operations

gc = number of operations in the class

i = character of the irreducible representation for the operations of the

class

r = character for the reducible representation for the operations of the

class


USE

By inspection or……

trans =

rot =

2(2T1u)

3(A1g+Eg+T2g)

1(A1g)

1(T2u)

0

T1u

T1g

3n = A1g + Eg + T1g + T2g + 3T1u + T2u

ni =1h

gc i rc

3n-6 = A1g + Eg + T2g + 2T1u + T2u

homework class

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