Spectroscopic Methods in Inorganic Chemistry

Dr.Chris UP July 2016

Part 1: IR Spectroscopy in Inorganic Chemistry

Interactions light - moleculesWe need to understand the different interactions dependent on the wavelenght or energy of light:

① ② ③ ④ ⑤

Wha

t are

the

spec

tros

copi

c m

etho

ds 1

- 4

?

Interaction light - matter

http://hyperphysics.phy-astr.gsu.edu/hbase/mod3.html

https://phet.colorado.edu/en/simulation/molecules-and-light

Check out the behaviour in all 4 lights:--> water vs CO2 vs OzoneWhat are the differences ?

(I) IR SPECTROSCOPY

Animationhttps://www.youtube.com/watch?v=3RqEIr8NtMI

Polarity and Dipole momentsDipole moment can be calculated as the product of the charge (abbreviated Q) times the distance (abbreviated r) between the charges.

Why do CO2 and CCl4 do not have a dipole moment ?

calculate dipole moment

H(+) Cl(-) ionic

μ = Q * r

1.60 10-19 C 127 pm

unit: 1 debye D = 3.335 × 10−30 C·m

Compare the dipole moment for completely ionic with the measured value of 1.03 D

Interaction light - dipoles

The change in electric dipole induced by change in electric field - but only at a resonance frequency

Absorption of energy

uv/vis

Raman vs. IR

• no sample preparation• non-destructive• can be used with a microscope

WHAT INFLUENCES THE WAVENUMBER OF A BOND ?

(1) Bond energy

https://www.youtube.com/watch?v=9HfJNnoRMPA

(2) Multiple bonds

(3) Atom weights

VIBRATION MODES FROM SYMMETRY

Normal modes of vibration

Triatomic molecules

Character tables – c2v

Vibration modes

https://www.youtube.com/watch?v=QnnAInt4mlM

=>2 A1 + B1

Both IR active-> 3 peaks

Examplesbendingstretch stretch

Which peaks belong to stretching or bending ?

Planar BF3 molecules

How many degrees of freedom ?(bond vibrations without rotation and translation)

1

2

3

4

5

Answers

1

2

3

4

5

Final result for BF3

E’ A2’’ E’

Examples

Identify stretching and bending modes !

Unknown anion

IR / Raman Simulationwww.molwave.com

ORGANO-METAL COMPOUNDSM(CO)X

CO bonding modes

https://books.google.co.th/books?id=oZeFG6QDNekC&pg=PA382&lpg=PA382&dq=M(CO)2L4&source=bl&ots=u9uyncbsDi&sig=Qr5CRFxT1cPpud5vnjs5PWgkkzc&hl=en&sa=X&ved=0ahUKEwjp84-n5eXKAhVSkY4KHQ5QAiUQ6AEIIDAC#v=onepage&q&f=false

Try to explain why ?

The C–O stretching wavenumbers are shifted to lower values when there are changes in the extent of backbonding in the compound. Removing positive charge from the metal causes the shift of electrons from the metal to the CO π orbitals causes the CO ∗wavenumber values to decrease. The highest excess of negative charge on the metal occurs in the [V(CO)6 ]− complex and so more backbonding occurs than in the other complexes. The next highest excess of electron density is in Cr(CO)6 , and then [Mn(CO)6 ]+.

Character of the M-CO bond

Slightly antibonding HOMO

Electron density on the metal

Higher C-O strength

Lower C-O strengthM=C=O character

Ligand donation effects

43

CO IR vibrationsWe compare cis- and trans-ML2(CO)2 complexes in IR:

What are the point groups ?

44

Tetrahedral Td Octahedral OhLinear: D∞h for A-B-A ( i )

C ∞h for A-B

http://en.wikibooks.org/wiki/Introduction_to_Mathematical_Physics/N_body_problem_in_quantum_mechanics/Molecules

45

Character Tables for cis and trans

46

Representations of 2 C-O groups

Which contains the irreducible representations :

Which contains the irreducible representations :

Conclusion: Number of IR peaks for cis and trans complex:

Metal-carbonyl compounds

M(II)hexamine complexes

N-H stretch

N-H bend

M-NH3 bend

M-NH3 rock

The spectra presented in Figure 5.6 show a trend in the wavenumber shifts for the three hexamine complexes; the N–H bands shift to lower wavenumbers from Co to Cr to Ni. This indicates that the N–H bond order (bond strength) decreases as the metal–N bond order increases in the stability order mentioned

FUNCTIONAL GROUPS

Find functional groupshttps://www.youtube.com/watch?v=ItW6Mj2CQKc

https://www.youtube.com/watch?v=9HfJNnoRMPA

Inorganic Compounds

Still of actual interest

Spectra Database:sdbs.db.aist.go.jp

http://www.chemicalbook.com/SpectrumEN_144-55-8_IR1.htm

Fe(CN)6 complexes

How can we explain the difference to Fe(III) ?

Raman

IR Spectrum of “Rennie”

Example: identify products in Antacidshttp://www.ptfarm.pl/pub/File/Acta_Poloniae/2000/2/083.p

df(1) Carbonate Compound

Ref. spectra

Mg CO3

Ca CO3

O-HCa-O

Mg-O

(2) Hydroxy Compounds

IR of “Maalox (an)”

IR of “Alusal”

Ref. spectra

Al(OH)3

Mg(OH)2

Al(OH)3

Mg(OH)2

SUPPLEMENTS

Overview

AM1 calculation“ArgusLab” Freeware

The dipole moment of HCl is 1.03 D, and the bond length is 127 pm. What is the percent ionic character of the HCl bond?

First we will assume that this molecule is 100% ionic. In this case, the charges are separated by the bond length, and we can calculate the dipole moment in this extreme case.

The actual dipole moment measured for this molecule is 1.03 D, so the molecule is not completely ionic.

http://wps.prenhall.com/wps/media/objects/4678/4791085/ch10_01.htm

Types of motions

Stretch:

symmetric asymmetric

wagging twisting scissoring rocking

Bending:

http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared%3A_Theory

How do you distinguish whether the structure of transition metal complex molecule M(CO)4L2 is cis or trans by inspection of the CO stretching region of the IR spectra?

-> determine the symmetry group:

CO substitution pattern

-> Check the character tables:

c2v

d4h

Reducing stretching motions

4 stretching vectors

4 0 0 2

4 0 0 0 0 0 0 4 2 0

Experimental

ATR Method – attenuated total reflection

RAMAN

https://www.youtube.com/watch?v=TMLnUmbLwUI

Figliola, et. al. Organometallics 2014, 33, 4449

Compound 1A = S

20612021197619551878

Compound 2A = Se

20542014197019501875

We should decide which structure is more likely based on IR:5 peaks, higher frequencies for A = Sulfur

① or ② ?

Find the point group of both molecules:

①

②

(Practise with: http://symmetry.otterbein.edu/challenge)

C2v E C (z) s(xz) s(yz) Linear f, rotations

Quadratic f

A1 1 1 1 1 z x, y, zA2 1 1 -1 -1 R xyB1 1 -1 1 -1 x, R xzB2 1 -1 -1 1 y, R yz

G 8 0 4 0

Molecule ① : 8 CO groups

Reduce G: ¼ (8x1x1 + 0x1x1 + 4x1x1 + 0x1x1) = 3 A1

¼ (8x1x1 + 0x1x1 + 4x-1x1 + 0x-1x1) = 1 A2

¼ (8x1x1 + 0x-1x1 + 4x1x1 + 0x-1x1) = 3 B1

¼ (8x1x1 + 0x-1x1 + 4x-1x1 + 0x1x1) = 1 B2

The A2 can be ignored since it does not contain x, y or z and is therefore not IR active.This gives 7 IR active CO vibrations.

C2v E C (z) s(xz) s(yz)G 6 0 2 0

Molecule ②: also Point Group C2v

Reduce G: ¼ (6x1x1 + 0x1x1 + 2x1x1 + 0x1x1) = 2 A1

¼ (6x1x1 + 0x1x1 + 2x-1x1 + 0x-1x1) = 1 A2

¼ (6x1x1 + 0x-1x1 + 2x1x1 + 0x-1x1) = 2 B1

¼ (6x1x1 + 0x-1x1 + 2x-1x1 + 0x1x1) = 1 B2

The A2 can be ignored since it does not contain x, y or z and is therefore not IR active.This gives 5 IR active CO vibrations.

A1 1 1 1 1 z x, y, zA2 1 1 -1 -1 R xyB1 1 -1 1 -1 x, R xzB2 1 -1 -1 1 y, R yz