SCH 2450 - Molecular Spectroscopy (Notes) PDF

SCH 2450: MOLECULAR SPECTROSCOPY Course objectives - Demonstrate knowledge on the various molecular spectroscopic techniques - Demonstrate competence in the use of molecular spectroscopic techniques in analysis of real samples Course description Absorption of electromagnetic radiation by mo...

SCH 2450: MOLECULAR SPECTROSCOPY Course objectives - Demonstrate knowledge on the various molecular spectroscopic techniques - Demonstrate competence in the use of molecular spectroscopic techniques in analysis of real samples Course description Absorption of electromagnetic radiation by molecules, Molecular electronic energy levels, vibrational energy levels, Raman effects, Lasers. Ultraviolet and visible absorption methods of analysis, Instrumentation. Quantitative methods, differential, difference and derivative spectroscopy, Turbidimetry and Nephelometry. Fluorescence, Phosphorescence and chemiluminescence spectroscopy, instrumentation and applications. Practicals to include applications of the various techniques. Learning outcomes At the end of this course Learner should be able to: - Describe the various molecular energy levels - Explain the principles of UV-Vis, Raman, turbidimetry, nephelometry, fluorescence, phosphorescence and chemiluminescence spectroscopic techniques - Distinguish between differential, difference and derivative spectroscopy - Describe the instrumentation behind the various molecular spectroscopic techniques. Teaching Methodology: Lectures, Tutorials and practicals Instructional Materials Whiteboard/ smart board, white board markers, computer and projector, course notes, Duster Course assessment Written CATS 15%, assignments 5%, Practicals 10%, final written examination 70% 1 Course Journals 1. Journal of Analytical Sciences, Methods and Instrumentation Published/Hosted by Scientific Research Publishing. ISSN (printed): 2164-2745. ISSN (electronic): 2164-2753 2. International Journal of Analytical Chemistry ISSN: 1687-8760 3. Journal of Chromatography A, ISSN 0021-9673 Reference Journals 1. International Journal of Analytical Chemistry ISSN: 1687-8760 2. Talanta Imprint: ELSEVIER ISSN: 0039-9140 3. The Annual Review of Analytical Chemistry ISSN: 19361335, 19361327 4. Trends in Analytical Chemistry ISSN: 0165-9936 REFERENCES 1. Kemp W. (1991). Organic spectroscopy, 3rd edition, McMillan. London 2. Williams, D. H., and Fleming, I (1989) Spectroscopic Methods in Organic Chemistry, 4th Edition, McGraw-Hill Book Company 3. Silverstein, R. M., Webmaster, F.X and Kiemle, D.J. (2005) Spectroscopic Identification of organic compounds. 7th edition. John Wiley REFERENCES (MAIN) 1. Instrumental Methods of Analysis, Willard, H.H., Merritt, L.L., JR; Dean, J.A. and Settle, F.A. (JR), CBS Publishers & Distributors, 6th Edition, 1986. 485 Jain Bhawan, BholaNath Nagar Shahadra Delhi-110032, 1030pp. 2. Modern methods of Chemical Analysis, R.L. Pecsok, L.D. Shields, T. Cairns and I.G. McWilliam. John Wiley and Sons, Inc.,1976. 573pp. 3. Principles of Instrumental Analysis, Douglas A. Skoog, CBS College Publishing (Saunders College Publishing), 3rdEdition, 1985. 4. Internet, e.g. Googles (e.g. Wikipedia, etc.) 5. etc. SCH 2450: MOLECULAR SPECTROSCOPY INTRODUCTION TO ABSORPTION AND EMISSION SPECTROSCOPY Spectroscopy is the measurement and interpretation of electromagnetic radiation absorbed or emitted when the molecules, or atoms, or ions, of a sample move from one energy state to another. Every atom, ion, or molecule has a unique and characteristic relationship with electromagnetic radiation. Areas of spectroscopy may involve changes in the rotational, 2 vibrational, and electronic energies. Other spectroscopic studies may involve energies resulting from energy changes that arise when a sample is placed in a magnetic field (e.g. N.M.R.) or electric field (e.g. E.S.R. study). The Nature of Electromagnetic Radiation A beam of radiation may be regarded as an electromagnetic wave-form disturbance or photon of energy propagated at the speed of light. A photon has the properties of a microscopic particle of definite energy and at the same time has properties of a wave extending over a broad area of space. The Heisenberg uncertainty principle shows that it is not important or even possible to measure both the wave and particle properties of a photon simultaneously. However, it is useful to keep both properties in mind. A photon originating at a point in space radiates from that point in a spherical wave characterized by electric field vectors which have periodic maxima perpendicular to the direction of propagation. The wavelength of the radiation, λ, can be visualized as the distance between these maxima; that is from crest to crest, as shown in the figure below: A λ B 3  P (e.g. WITH AIR AS MEDIUM) (MEDIUM OF HIGHER REFRACTIVE INDEX) Direction of Propagation Fig. 1: Some Characteristics of Electromagnetic radiation In the figure, the distance AB is one wavelength. - The frequency, ν, is the number of waves (or cycles) passing a fixed point, such as P, in a unit length of time. If E is the energy of a photon, then: ∆E = hν = hc/λ Where: h = Planck’s constant and c = the velocity of light in a vacuum, which is about 2.998 x 108 msec-1 or approximately 3.00 x 108 msec-1. The wavelength (λ) and frequency (ν) are related to the velocity of light (c) by the expression: λν = c/n where n = the refractive index (the ratio of the velocity of light in a vacuum to its velocity in the medium in question). Sometimes, it is more convenient to use the number of wavelength units in one centimeter. This number is called the wave-number, ν̅: ν̅ = 1/λ = νn/c (units, cm-1) 4  Fig. 2: Schematic representation of molecular Electronic, Vibrational and Rotational energy levels The Photoelectric Effect When sufficiently energetic radiation impinges on a metallic surface, electrons are emitted. The energy of the emitted electrons is found to be related to the frequency of the incident radiation by the equation: E = hν – w where: w = the work function, i.e. the work required to remove the electron from the metal to a vacuum, E = the energy, and ν = the frequency. N.B. Energy, E is directly dependent upon the frequency, but totally independent of the intensity of the beam. Also note that: - Absorption by polyatomic molecules, particularly in the condensed state, is considerably more complex than atomic absorption. This is because in the former case, the number of energy states is greatly enhanced. Here, the total energy of a molecule (E) is given by: E =Eelectronic + Evibrational + Erotational Where Eelectronic ,Evibrational , and Erotational represent the respective energies of the molecule. 5 The Electromagnetic Spectrum Ionizatio- Energy n of Chang- atoms es and Involv- Inner shell molecul- Valence Spin orientations ed Nuclear Electrons es electrons Molecular vibrations: (in magnetic field) X- Near Visib- Electro- Nuclei ray UV le ns NMR s Stretching Bending (ESR) "Soft" Near IR Infrared Far Micro- Radio- Gamm- X- Vacuum (Overtones) (Fundam- IR waves waves a rays rays UV ental) Region in Electr- omag- 1Å 10Å 100Å 200nm 400nm 800nm 0.04cm 25cm netic 0.8μm 2.5μm 25μm 400μm Spectr- um Wavelength Figure 3: Schematic Diagram of the Electromagnetic Spectrum. Note that the wavelength scale is non-linear MOLECULAR ELECTRONIC ENERGY LEVELS SINGLET TRIPLET SINGLET Inter-atomic distance along critical coordinate N.B. The curve tails on the right represent the following multiplicities: Topmost (Singlet), Middle (Triplet) and Bottom (Singlet) as shown. 6 Fig. 4: Schematic diagram for a diatomic molecule Energy levels Fig. 5: Partial Energy Diagram for a Photo-luminescent System N.B. Fluorescence lifetime is 10-9 to 10-7 sec., while Phosphorescence lifetime is 10-4 to 10 sec. or longer. (To be discussed in details later) 7 VIBRATIONAL ENERGY LEVELS - Vibrational energies are usually an order of magnitude smaller than the electronic energy. - For a diatomic molecule, the allowed energies as given by the quantum-mechanical vibrational energy level, are: ℎ 𝑘 Evib = (v+½) √µ (v = 0, 1, 2, …………….) 2𝜋 where k = the force constant, i.e. the stiffness of the chemical bond and µ = the reduced mass for the two atoms, m1 and m2; m1m2 µ= m1+m2 RAMAN EFFECT In Raman spectroscopy, the interaction stems from the electric field and an oscillating polarizability within the molecule. Quasi-excited State Anti-Stokes Stokes hν0 h(ν0+νm) hν0 h(ν0-νm) hνm Fig. 6: Quantum representation of energy interchange involved in the Raman effect Laser Sources - A laser provides an almost ideal monochromatic source of narrow line width. - It emits radiant energy that is coherent, parallel, and polarized. - Laser operation involves three principles of physics: stimulated emission, population inversion, and optical resonance. - Examples are: He-Ne laser and Ar-Kr laser. 8  Energy transfer 160,000 1s 2s 1s22s22p54s F via He-Ne 8673 cm-1 (1153 nm) r collisions 1s22s22p53p e 15,803 cm-1 (632.8 nm) q 1s22s22p53s u EED e n CD c y (cm-1) 0 1s2 1s22s22p6 Helium Neon Key: EED =Excitation by electrical discharge; CD = Collisional deactivation Fig. 7: Energy levels involved in the He-Ne Laser INTRODUCTION TO UV-VISIBLE SPECTROSCOPY, AAS, etc. Spectroscopy This is the measurement and interpretation of electromagnetic radiation that is absorbed, emitted or scattered by atoms, molecules or ions. - Absorption or emission is associated with changes in the energy states of the interacting species. - Since each species has characteristic energy states, it may be used to identify and quantify species in a sample. Examples Radiation in UV-Visible region: - Atomic Absorption Spectrometry - UV-Visible Spectrophotometry - Flame Photometry 9  - Atomic Emission Spectrometry - Atomic fluorescence, phosphorescence Luminescence Radiation scattering – Turbidimetry N.B. Photometer- An instrument that measures the ratio or some function of the two radiant power(s) of the two beams over a small wavelength range. - It uses a filter to isolate wavelengths and an inexpensive detector, e.g. Photocell. Spectrophotometer - An instrument that measures the ratio or some function of the two radiant power(s) of the two beams over a large wavelength range. - It uses a monochromator and a very sensitive detector, e.g. Photomultiplier Tube (PMT). Theory of Beer-Lambert’s Law - The (practical) UV region generally represents: ≅ 200 - < 400 nm, while Visible region represents: 400 – 800 nm - When light is incident upon a homogeneous medium, part of it is absorbed, reflected or transmitted. Po(Incident) Pa Pt (transmitted) Pr (Reflected) (Absorbed) Po = Pa +Pt +Pr - For an all-glass interface, Pr is very small (i.e. ≅ 4%) and may be neglected. Thus: Po = Pa +Pt - If a beam of radiant power, Po, traverses an infinitesimally small distance, δx, of an absorber, the decrease in radiant power, -δP, is given by: 10  -δP = k’Pδx, since the number of absorbing species is proportional to the thickness of the medium. Thus -δP = k’δx …………………………………………………………………………Equation ① P or -δP = -δ(ln P) = k’δx P Thus, radiant power absorbed is proportional to the thickness traversed. - If Po is the radiant power at x = 0 and P is the radiant power transmitted at distance x = l, the equation above can be integrated to give: 𝑃 𝑙 -∫𝑃0 𝑑𝑙𝑛 𝑃 = k’∫0 𝑑𝑥 𝑃0 Thus, giving ln P0 – ln P = ln( ) = k’l ……………………………………….. Equation ② 𝑃 This gives the Lambert’s Law. - The number of absorbing species that collide with photons is proportional to the concentration, C. -dP = k’’PdC 𝑃0 and as above, ln( ) = k”C ……………………………………………………….. Equation ③ 𝑃 which gives the Beer’s Law. - Combining equations ② and ③ gives: 𝑃0 ln ( ) = klC 𝑃 Replacing natural logarithms with base 10 logarithms, gives: 𝑃0 log( ) = abC, where b is the path length and a is a proportionality constant. 𝑃 𝑃0 Absorbance, A, is defined as log ( ). Thus 𝑃 𝑃0 A = log( ) = abC and 𝑃 𝑃 Transmittance is T = 𝑃0 1 i.e. A = log ( ) = -log (T) 𝑇 A = abC where a is the specific absorptivity. - The proportionality constant, a, is designated as ε, when C is expressed as molar concentration and path length,l, is in centimeters. Thus giving: A = εlC The above is referred to as the Beer-Lambert’s Law. The units of ε are Mole-1 L cm-1 or mole-1 dm3 cm-1. ε is the molar absorptivity or molar extinction coefficient. - Value of molar absorptivity will be different at different wavelengths. Example A 0.02 mM solution of standard potassium permanganate has an absorbance of 0.511 when measured in a 2.0 cm cell at 520 nm. Calculate the concentration of a sample whose 11 absorbance is 0.120 when measured in a 1.0 cm cell at the same wavelength (K = 39, Mn = 55, O = 16). Solution A = εCl For a 2cm cell: 0.511 = ε x 0.02 x 10-3M x 2cm => Molar absorptivity, ε = 0.511 2 x 10-5 M x 2cm Thus ε = 12775 mol-1 L cm-1 For 1.0 cm cell: 0.120 = 12775 x C x 1cm 0.120 => C = = 9.393 x 10-6 moles/litre 12775 𝑥 1 = 9.393 µM In ppm, Rfm for KMnO4 = 39+55+4(16) =39+55+64 = 158 g/mol 1 litre of solution contains 9.393 x 10-6 moles. Thus 1 litre of solution contains 9.393 x 10-6 moles x 158g/mol = 1.4841 x 10-3 g i.e. 1000 mLof solution contains 1.4841 x 10-3 g ∴ 106 ml of solution contains 1.4841 x 10-3 x 106/103 g = 1.4841 x 10-3 x 103 g ≅1.484 ppm (m/v) or 1.484 mg/L, or 1.484 µg/ml, etc. Deviation from Beer-Lambert’s Law - A plot of absorbance versus concentration should give a straight line. - However, in practice, there are positive and negative deviations, as shown below in the following graph of Absorbance versus Concentration. 12 Absorbance Expected Linear relation Concentration Reasons for deviation (i) Real deviations - The Law applies only at low concentrations. At high concentrations (> 10-3 M), refractive index changes considerably, and so does absorptivity, leading to deviations. (ii) Instrumental deviations - Beer’s Law assumes monochromatic light. However, truly monochromatic light may not be attained. (iii) Chemical deviations - Caused by a shift in the position of a chemical or physical equilibrium involving the absorbing species, e.g. Dissociation or Association: 4C6H5CH2OH ⇌ (C6H5CH2OH)4 Cr2O72- + H2O ⇌ 2HCrO4- (Orange) (Yellow) (iv) Presence of impurities that fluoresce or absorb at the particular wavelength. REGRESSION AND CORRELATION ANALYSIS Classical methods of Analytical Chemistry are also sometimes referred to as “Wet Chemistry”. Examples of such methods are Titrimetry and Gravimetry. These can sometimes have high precision but mainly for major constituents and can handle a few samples. Classical methods are usually absolute methods, i.e. there are no reference standards required. They are also tedious, but less costly. On the other hand, instrumental methods can go to low 13 concentrations and so are relatively highly sensitive. They are ideal where many samples have to be analyzed, i.e. they have high sample throughput. They also display multi-component analysis capability. They also enable interfacing with computers, which affords high degree of instrumental control and data handling. Over 90% of all analytical work is currently done by instrumental methods. All instrumental methods require a calibration step in which standards are treated in the same way as the samples. 0.35 HYPOTHETICAL INSTRUMENTAL CALIBRATION GRAPH Instrumental Response e.g. 0.3 e.g. y = 0.713x + 0.001 0.25 R² = 0.999 Absorbance 0.2 0.15 0.1 0.05 Series1 0 0 0.1 0.2 0.3 0.4 0.5 Concentration, in Units (e.g. mg/L) - The concentration of the sample is obtained by interpolation (or cross referencing, as shown in the hypothetical curve drawn above). This is referred to as the Analytical curve method. - The calibration curve has to be linear, and given that every point is subject to errors, we need a best line of fit through the points. - Statistics provides a mechanism for objectively obtaining the best line of fit and also specifying errors associated with it. This refers to as Regression analysis. Assumptions (1) Errors in calibration experiment occur only in the y-values. (2) The magnitude of errors in y-values is independent of analyte concentration. (3) A linear relationship exists between the y- and x- values. The Product-moment Correlation coefficient (r) - A straight line has the general equation: y = bx + a, where b = slope of the curve and a = y-intercept - For n points on the curve: (x1, y1), (x2, y2), …………………., (xn, yn) 14  There exists a centroid point (x̅, y̅) - The product-moment correlation coefficient, r, is used to test how well the points fit a straight line. r = ∑𝑖(𝑥𝑖 − 𝑥)(𝑦𝑖 − 𝑦) 2 2 √∑(xi − x̅) ∑(yi– y̅) 𝑖 i where r takes values: -1 ≤ r ≤ +1 (as illustrated below): y r =+1 (Perfect positive correlation) r could be 0.90 y r = 0, i.e. for other curves No correlation x r = -1 (Perfect negative corre- x lation) Example Consider the data: Conc. (pg/ml) 0.00 2.00 4.00 6.00 8.00 10.00 12.00 (x): Fluorescence 2.10 5.00 9.00 12.6 17.3 21.0 24.7 intensity (units) (y): Calculate the correlation coefficient of the data. Solution x̅ = 0.00 + 2.00 + 4.00 + 6.00 + 8.00 + 10.00 + 12.00 = 42.00 = 6.00 7 7 y̅ = 2.10 + 5.00 +9.00 + 12.6 + 17.3 + 21.0 + 24.7 = 91.7 = 13.1 7 7 xi yi (xi-x̅) (xi-x̅)2 (yi-y̅) (yi-y̅)2 (xi-x̅) (yi-y̅) 15 0.00 2.10 -6 +36.0000 -11 121.0000 +66.0000 2.00 5.00 -4 +16.0000 -8.1 65.6100 32.4000 4.00 9.00 -2 +4.0000 -4.1 16.8100 8.2000 6.00 12.6 0 0.0000 -0.5 0.2500 0.0000 8.00 17.3 2 4.0000 +4.2 17.6400 8.4000 10.00 21.0 4 16.0000 +7.9 62.4100 31.6000 12.00 24.7 6 36.0000 +11.6 134.5600 69.6000 TOTALS: 112.0000 418.2800 216.2000 r = ∑𝑖(𝑥𝑖 − 𝑥)(𝑦𝑖 − 𝑦) 2 2 √∑(xi − x̅) ∑(yi– y̅) 𝑖 i = 216.20000 = 0.9989 (which is > 0.9900, hence very close to +1.0000) √(112)𝑥(418.28) This represents good correlation. The line of Regression of y on x - Linear relationship is assumed. Draw best line of fit through the points. - Since errors are assumed to be in y, we seek the line that minimizes the deviations in the y-direction between experimental points and the calculated line (i.e. fitted values). These deviations are referred to as y-residuals. y - y - residuals x 16  - Since some deviations will be positive and some negative, we try to minimize the sum of the squares of the residuals, hence the method of Least squares. - The best line of fit is calculated on this principle. - The best line of fit must pass through the centroid, i.e. (x̅, y̅). - It can be shown that for the regression line y = a + bx 𝑏 = ∑(𝑥𝑖 − 𝑥)(𝑦𝑖 − 𝑦) 𝑖 ∑𝑖(𝑥𝑖 − 𝑥)2 and since y̅ = a + bx̅, then a = y̅ – bx̅ which gives the line of regression of y on x. Using the earlier example: b = 216.2/112 = 1.93035714285 and a = 13.1 –(6.00 x 1.93) = 13.1 – 11.58 = 1.52 => y = 1.93x + 1.52 is the regression equation Thus for a sample that gives fluorescent intensity of 3.00, then: 3.0 = 1.93x + 1.52 => x = (3.00 – 1.52)/1.93 = 0.7668394 ≅ 0.767 pg/ml. Errors in Slope and Intercept - Random errors in y and x are of importance. From Statistics: ∑𝑖(𝑦𝑖−ŷ𝑖)2 Sy/x = √ (𝑛−2) where yi = experimental value ŷ = fitted value = corresponding value on the regression line n-2 = degrees of freedom, since only two points are needed for a straight line. - Thus, the standard deviation for slope and intercept are calculated as follows: 𝑆𝑦/𝑥 Sb = 2 √∑𝑖(xi−x̅) ∑𝑖 xi2 and Sa = Sy/x 𝑥√ 𝑛 ∑𝑖(xi−x̅)2 17  - Thus, confidence limits on slope and intercept are: b±tSb and a ± tSa COMPUTATION RULES (ROUNDING) (i) MULTIPLICATION AND DIVISION Example: (a) To Work out the following calculation and round the final answer appropriately: 56𝑥0.003462𝑥43.22 1.684 “Exact Answer” = 4.9757409976….. The above is rounded as ≅ 5.0 But why is the final answer rounded as above? Answer: We round to 2 significant figures, which is the least number of significant figures displayed in the equation (expression) above, i.e. 56 [2 significant figures]. General Rule: In multiplication or division, the result should contain the same number of significant figures as that term with the fewest number of significant figures. Other examples: (b) A = 𝜋r2 (where A = Area of circle, r = its radius). Calculate A when r = 3.82 units. Solution: 22 A = ( ) x 3.822 = 3.1428571…. x 3.822 = 3.1428571…. x 14.5924 = 45.86182857…. 7 ≅ 45.9 𝑎𝑟𝑒𝑎 𝑢𝑛𝑖𝑡𝑠 (i.e. 3 s.f.) 𝐴 (c) C = (Recall A = εbC, where A = absorbance, ε = Extinction coefficient, b = path length and 𝜀𝑏 C = concentration) 1.20 C= = 3.613459123466 x 10-5≅3.61 x 10-5 moles/L (i.e. to 3 s.f.) 33176 𝑥 1.001 Assignment (For marking) – contained in Main Assigment I – Answer all the questions: 35.63 𝑥 0.5481 𝑥 0.05300 Calculate and round appropriately: x 100% 1.1689 (ii) ADDITION AND SUBTRACTION 18 General Rule: In addition or subtraction of a set of numbers, all are rounded to the same number of decimal places. We generally use the lowest number of decimal places. Examples Addition (4 (a) 1.6375 (4 d.p.) (b) 0.0072 (2 s.f.) d.p.) (2 75.2 (1 d.p.) 12.02 (4 s.f.) d.p.) (4 6.002 (3 d.p.) 4.0078 (5 s.f.) d.p.) (1 82.8 (1 d.p.) 25.9 (3 s.f.) d.p.) (0 4886 (4 s.f.) d.p.) (i.e. 0 d.p. or whole N.B. 4927.9278 4928 number) 1.6375+75.2+6.00 2 = 82.8395 82.8 (1 d.p.) Subtraction (c) 1.32 x 10-2 ≡ 13.2x10-3 (1 d.p.) 6.72x10-3 6.7 x 10-3 (1 d.p.) 6.5 x 10-3 (1 d.p.) J.K.U.A.T. ASSIGNMENT I : COMPUTATION RULES AND ANALYTICAL CHEMISTRY CALCULATIONS (1ST SEMESTER) 19 B. Sc (Analytical Chemistry) – 4th Years SCH 2450 (MOLECULAR SPECTROSCOPY) ANSWER ALL THE QUESTIONS Q1. Express the results of the following calculations using the appropriate number of significant figures or decimal places. (i) 15.41 + 14.7 (ii) 58.13 + 5.20 + 20.67 – 12.00 (iii) 71.4 + 0.0074 + 50.35 (iv) (1.412 x 10-5) x (6.1 x 10-6) (v) 35.63 x 0.5481 x 0.05300 x 100% 1.1689 (5 marks) Q2. In an analysis of nitrite ions in water supplies, it was found necessary to make up 500 ml of a solution containing 1000 μg/mL of nitrite ions. The purity of the nitrite used was 95.4 %. (N = 14, O = 16, Na = 23) (i) Calculate the amount of sodium nitrite that would be required. (ii) Calculate the concentration (ppm) of sodium ions in the resulting solution. (5 marks) Q3 (i) A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 μg of zinc. What is the concentration of zinc in the plant in (a) ppm (b) ppb? (2 marks) (ii) Calculate the molar concentrations of 1.00 ppm solutions each of Li+ and Pb2+. (Pb = 207, Li = 6.94) (2 marks) Q4. Explain briefly how you would round the final answer if the calculation involved all the four major operations, i.e. addition, subtraction, multiplication and division together. (1 mark) TOTAL No. OF MARKS = 15 MOLECULAR ULTRAVIOLET AND VISIBLE ABSORPTION SPECTROSCOPY Absorbing Species - Absorption of UV/Visible radiation by atomic or molecular species M is a two-step process. (i) M + hν M* i.e. excitation (by absorption) The lifetime of excited particle is brief (in the range of 10-9 to 10-8 sec.) (ii) M* M + Relaxation e.g. heat energy, decomposition of M*, etc. Such a process is called a photochemical reaction. Alternatively, relaxation may involve fluorescent or phosphorescent reemission of radiation. Absorbing species can be categorized into three types of electronic transitions: (a) 𝜋, σ, and n electrons, (b) d and f electrons, and (c) Charge-transfer electrons, i.e. transitions involving these. 20 ABSORBING SPECIES CONTAINING 𝜋, σ, AND n ELECTRONS Absorbing species containing 𝜋, σ, and n electrons include organic molecules and ions as well as a number of inorganic anions. The following diagrams show how σ and 𝜋 bonds are formed. THEORY + + + s s σ (Bonding molecular orbital) + - + - s s σ* (Antibonding molecular orbital) π Orbital (Bonding M.O.) π* Orbital (Antibonding M.O.) Figure 9(a): Bonding and antibonding molecular orbitals (M.O.) from 2px atomic orbitalsSimilarly, when a 2py orbital of one atom overlaps with a 2py orbital of another atom, they also overlap sidewise forming two molecular orbitals p2py and p*2py. These are exactly similar to p2px and p*2px molecular orbitals. 21 Energy σ* σ → σ* n → σ* Antibonding 𝜋* 𝜋→𝜋* n → 𝜋* Antibonding n Non bonding 𝜋 Bonding σ Bonding Figure 9(b): Electronic Molecular Orbital Energy Levels N.B. - Sigma (σ) bonds are formed when there is “head-on” atomic orbital overlap and pi (𝜋) bonds result when there is “parallel” atomic orbital overlap, e.g. parallel overlap of two atomic orbitals. - In Organic Chemistry, n (non-bonding) electrons are located principally in the atomic orbitals of nitrogen, oxygen, sulphur and the halogens. - UV and Visible radiation absorption promotes electronic transitions as shown above. - The∆E values for transitions are in the order: n→𝜋* < 𝜋→𝜋* < n→σ*