a.electronic spectra (chapter 11 pp. 380-381)

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A. Electronic Spectra (Chapter 11 pp. 380-381) 1) A characteristic of transition metal complexes is color arising from electronic transitions between d-orbitals of different energies a) Everyday Examples of colored transition metal complexes b) The UV-Vis Experiment and the spectral result “Both rubies and sapphires owe their intense colors to impurities, ruby to the presence of chromium, and blue sapphire to both titanium and iron. In rubies, the color can be explained by crystal field theory, but in sapphire, a slightly different process, known as charge transfer, produces the color blue.” http:// www.webexhibits.org/ causesofcolor/8.html

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A.Electronic Spectra (Chapter 11 pp. 380-381) A characteristic of transition metal complexes is color arising from electronic transitions between d-orbitals of different energies Everyday Examples of colored transition metal complexes The UV-Vis Experiment and the spectral result. - PowerPoint PPT Presentation

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Page 1: A.Electronic Spectra (Chapter 11 pp. 380-381)

A. Electronic Spectra (Chapter 11 pp. 380-381)

1) A characteristic of transition metal complexes is color arising from electronic transitions between d-orbitals of different energies

a) Everyday Examples of colored transition metal complexes

b) The UV-Vis Experiment and the spectral result

“Both rubies and sapphires owe their intense colors to impurities, ruby to the presence of chromium, and blue sapphire to both titanium and iron. In rubies, the color can be explained by crystal field theory, but in sapphire, a slightly different process, known as charge transfer, produces the color blue.”http://www.webexhibits.org/causesofcolor/8.html

Page 2: A.Electronic Spectra (Chapter 11 pp. 380-381)

2) The Beer-Lambert Law

a) Absorbance = A = log (Io/I) Most modern systems give readings in A directly, so you don’t have to do the math. A has no unit.

b) A = lc describes the absorption of light in a solution

i. l = the length of the cell containing the solution, usual 1 cm

ii. c = concentration in mol/L = M

iii. = Molar extinction coefficient = constant for a given molecule at a given wavelength of light = how well the molecule absorbs light

3) Plotting Spectra

a) E = h c = = 3 x 108 m/s h = 6.626 x 10-34 Js = Planck’s constant

b) Wavenumber =

i. higher energy = larger wavenumbers

ii. Higher energy = smaller wavelength

1-1- cmM (cm)(M)

1

c

A ε l

λh wavelengtoflight ofenergy λ

hc E

cm λ

1 ν 1-

νhc λ

hc E

1-

7-

1- cm 000,20

nmcm10

(500nm)

1ν cm ?? nm 500

Page 3: A.Electronic Spectra (Chapter 11 pp. 380-381)

4) Color

a) What we see as the color of a compound is the complementary color to what the compound absorbs

b) Example: Absorbs red, we see green

c) Example: Absorbs yellow/orange, we see blue/purple

d) Many complexes have multiple absorption band, so prediction is hard

Page 4: A.Electronic Spectra (Chapter 11 pp. 380-381)

B. Interpreting the Electronic Spectrum of d1 Ti3+

1. Single d-electron species are relatively easy to understand

2. o can be calculated directly from the energy of light absorbed

17

300,20101

1

493

1

cm

cmx

nm

nmo

Page 5: A.Electronic Spectra (Chapter 11 pp. 380-381)

C. Interpreting the Spectra of Multi-Electron Species

1. Electron-Electron Interactions complicate the spectra of multi-electron species

2. Example: [V(H2O)6]3+ is an octahedral d2 metal ion

286nm 333nm 400nm 500nm 667nm

[V(H2O)6]3+

Page 6: A.Electronic Spectra (Chapter 11 pp. 380-381)

3. Because of the multiple electrons, now we have multiple excited states

a) dxy1dxz

1 is not the only possible ground state, but is convenient to use

b) We can promote one e- from dxy or dxz to dz2 or dx2-y2 (we show 2 above)

c) Excited states produced: dxy1dz2

1 and dxy1dx2-y2

1

d) The energy needed for this excitation is not the same for each: Ea ≠ Eb

dxy1dx2-y2

1 will be higher energy because more repulsion; both in xy plane

Page 7: A.Electronic Spectra (Chapter 11 pp. 380-381)

e) Excitation dxy1dxz

1 ----> dxy2 = Ec

i. Energy change is for pairing the electrons

ii. Absorption is extremely week (“spin forbidden” to spin flip)

4. Example: [Mn(H2O)6]2+

a) This spectrum has many very

weak; solution is very pale

b) Weak field ligand makes this a

high spin octahedral d5 metal ion

c) All possible transitions include

a “spin forbidden” spin flip, so all absorptions are very weak

[Mn(H2O)6]2+

Page 8: A.Electronic Spectra (Chapter 11 pp. 380-381)

5. Example: [Cr(NH3)6]3+

a. There are two absorptions in its visible spectrum, one strong and one weak

b. Explanation

i. Octahedral d3 metal ion has two types of excitations available

ii. One is “allowed”, the other spin forbidden

iii. Relatively strong field NH3 makes o (450nm) > P (650nm)

Page 9: A.Electronic Spectra (Chapter 11 pp. 380-381)

6. The Spectrochemical Series and Electronic Spectra of Metal Complexes

a) Ligand field strength affects o, which effects the of light absorbed

b) It is difficult to correlate color with ligand field strength due to multiple absorbances of the multi-electron metal ions

c) A particular metal ion’s color may correlate with related ligands

I-

Cl-

NO2-

Page 10: A.Electronic Spectra (Chapter 11 pp. 380-381)

d) Example: A sample of one Cr3+ complex is green and another is yellow. One is [Cr(NH3)6]3+ and the other is [CrF6]3-. Which is which?

i. Determine the color of light absorbed to determine energy

ii. Green complex absorbs red (less energetic)

iii. Yellow complex absorbs blue (more energetic)

iv. NH3 complex must be yellow and F- complex must be green

e) Example: Assign the following two spectra to [Ni(H2O)6]2+ and [Ni(NH3)6]2+

and determine the color of their solutions.

i. NH3 is stronger field, so it should have higher energy peaks = (a)

ii. The NH3 complex absorbs all but ~750nm, so solution is red

iii. The H2O complex absorbs all but ~500nm, so solution is green = (b)

f) Example: A sample of one Co3+ complex is green and another is yellow. One is [Co(F)6]3- and the other is [Cr(CN)6]3-. Which is which?

i. Green complex absorbs red (low energy) = F- complex (weak field)

ii. Yellow complex absorbs violet (high energy) = CN- complex (strong)