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CHM 321:PHYSICAL CHEMISTRY II SPECTROSCOPY. WHAT IS SPECTROSCOPY? ORIGINATED FROM THE STUDY OF VISSIBLE LIGHT DISPERSED ACCORDING TO WAVELENGTH OR INTERACTION.

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Presentation on theme: "CHM 321:PHYSICAL CHEMISTRY II SPECTROSCOPY. WHAT IS SPECTROSCOPY? ORIGINATED FROM THE STUDY OF VISSIBLE LIGHT DISPERSED ACCORDING TO WAVELENGTH OR INTERACTION."— Presentation transcript:

1 CHM 321:PHYSICAL CHEMISTRY II SPECTROSCOPY

2 WHAT IS SPECTROSCOPY? ORIGINATED FROM THE STUDY OF VISSIBLE LIGHT DISPERSED ACCORDING TO WAVELENGTH OR INTERACTION WITH RELATIVE ENERGY AS A FUNCTION TO FREQUENCY OR WAVELENGTH. STUDY OF INTERACTION BETWEEN MATTER AND RADIATED ENERGY. STUDY OF INTERACTTION ELECTROMAGNETIC WAVES AND MATTER.

3 BORN-OPPENHEIMER APPROXIMATION TOTAL ENERGY OF A SYSTEM IS GIVEN THE BORN-OPPENHEIMER. IS THE SUM OF THAT DUE TO ROTATIONAL, VIBRATIONAL AND ELECTRONIC ENERGY. THIS ENERGY IS: E T =E ROT +E VIB +E ELE CAN BE CALCULATED USING THE SCHRODINGER EQUATION.

4 WHAT IS AN ELECTROMAFNETIC WAVE? SIMPLE HARMINIC WAVE PROPAGATED FROM A SOURCE TRAVELS IN A STRAIGHT LINE EXCEPT WHEN REFRACTED OR REFLECTED FOR A SIMPLE HARNONIC MOTION: Y=AsinƟ. Here y is the displacement with A being the maximum, Ɵ is the angle varying from 0 to 360 o ( 0 to 2π)

5 PRESENTING AN EQUATION OF A WAVE IF A WAVE TRAVELS WITH A UNIFORM VELOCITY ω radians s -1 IN A CIRCULAR PATH WITH RADIUS A, AFTER TIME t SECONDS WILL HAVE GONE THROUGH AN ANGLE Ɵ=ωt RADIANS. FOR THE VERTICAL DISPLACEMENT, IT WOULD HAVE GONE THROUGH y=A sinƟ=Asinωt AFTER TIME t= 2 π/ω, THE PARICLE WOULD HAVE TRAVELED THROGH ONE CYCLE.

6 PRESENTING AN EQUATION OF A WAVE CONT’D FOR 1 SECOND THE WAVE WILL REPEAT ITSELF ω/2π TIMES. THIS IS CALLED FREQUENCY UNITS OF FREQUENCY(SI):HERTZ (s -1 ). A WAVE EQUATION CAN THEREFORE BE REPRESENTED AS: y-Asinωt=Asin2πᴠt

7 OTHER PROPERTIES OF A WAVE IF THE VEOLCITY OF LIGHT IS c AND THE WAVE TRAVELS OR t SECONDS, IT WOULD HAVE TRAVELED; x=ct COMBING WITH THE EARLIER EQUATION; Y=Asin2πγt =Asin 2πγx/c ANOTHER PROPERTY OF THE WAVE IS THE WAVELENGTH (ʎ) AND IF THE VELOCITY IS c AND FRE, ITS FREQUENCY (ᴠ) i.e THE NUMBER OF ᴠ CYCLES PER SECOND, THEN THERE ARE ᴠWAVES IN c METERS 0R c=ᴠʎ OR ʎ =c/ᴠ METERS

8 OTHER PROPERTIES CONT’D FINALLY THE EQUATION FOR AN ELECTROMAGNETIC WAVE IS GIVEN BY: y=Asin2πx/ƛ WAVELENGTHS ARE EXPRESSED IN A NUMBER OF UNITS USUALLY CHOSEN TO SUIT THE RANGE FOR THE CONVENIENCE OF THE PERSON TO AVOID HIGH POWERS OF TEN

9 OTHER PROPERTIES CONT’D A WAVE CAN ALSO BE DESCRIBED BY WAVE NUMBERS (ṽ) DEFINED AS THE RECIPROCAL OF CENTIMETRES WAVELENTH EXPRESSED γ =1/ʎ HENCE y=Asin2πxṽ

10 QUANTIZATION OF ENERGY BEFORE THE 2OTH CENTURY, ENERGY WAS THOUGHT TO BE EMITTED OR ABSORBED CONTINUOUSLY BUT EARLY IN THE 20 TH CENTURY AROUND 1900 MAX PLANCK HYPOTHESIZED THAT ENERGY EMITTED OR ABSORBED IN DISCRED AMOUNTS BY AN OSCILLATOR. AN OSCILLATOR WOULD MOVE FROM ONE ENERGY LEVEL TO ANOTHER BY EMITTING OR ABSORBING A CERTAIN DISCREED AMOUNTS

11 QUANTIZATION OF ENERGY CONT’D FROM NIELS BOHR’S DISCOVERY OF THE ENERGY LEVELS OF THE ATOM, EACH ENERGY STATE HAD A DISCREED AMOUNT OF ENERGY. AN ELECTRON WOULD THEREFORE MOVE FROM A LOWER ENERGY LEVEL TO A HIGHER ONE BY ABSORBING A DEFINITE AMOUNT OF ENERGY CHARACTERISTIC OF THE ATOM OR EMIT THE AMOUNT WHEN IT IS FROM HIGH TO LOW.

12 QUANTIZATION OF ENERGY CONT’D THE LEVELS IN WHICH THE ELECTRON MOVES WITHIN THE ATOM ARE CALLED QUANTUM LEVELS THE ENERGY ABSORBED OR EMITTED IS OBTAINED AS; E=hᴠ, WHERE h IS THE PLANCKS CONSTANT AND ᴠ IS THE FREQUENCY. IF PARTICLE MOVES FROM E 1 TO E 2 ENERGY LEVELS, THE AMOUNT OF ENERGY ABSORBED OR EMITTED IS E 2 -E 1 =ΔE=hᴠ (joules) h HAS A VALUE OF 6.63 x 10-34 S -1

13 GENERAL FEATURES OF SPECTROSCOPY SPECTRA CAN BE OBTAINED IN THREE WAYS: – (i) EMISSION SPECTROSCOPY WHERE A MOLECULE OR ATOM DROPS FROM HIGHER TO LOWER ENERGY LEVEL WHERE THE EXCESS ENERGY IS EMITTED – (ii) ABSORPTION SPECTROSCOPY: IT IS WHEN AN ATOM OR MOLECULE MOVES FROM A LOWER ENERGY LEVEL TO HIGHER ONE WHERE ENERGY IS ABSORBED

14 GENERAL FEATURES (iii) RAMAN SPECTROSCOPY: THIS EXPLORES THE ENERGY LEVELS OF MOLECULES BY EXAMINING SCATTERED INCIDENT LIGHT ON MOLECULE. NEW FREQUENCIES ARE ADDED SINCE THE PHOTONS CAN ABSORB OR LOSE ENERGY DURING COLLISION. (IV) MOLECULES EXCITED BY PHOTONS ABSORB SOME OF THE ENERGY LEAVING THE INCIDENT LIGHT WITH LOWER ENERGY AND THE REMAINING LIGHT EMERGES WITH LOWER FREQUENCY. ON THE HAND IF THE MOLECULES LOSE ENERGY TO THE INCIDENT LIIGHT, THE INCIDENT LIGHT BECOMES HIGHER IN ENERGY AND EMERGED LIGHT HAS HIGHER FREQUENCIES.

15 INTENSITY OF SPECTRAL LINES SPECTRA ARE REPRESENTED BY LINES, SHOWING A VARIETY OF INTENSITIES; WEAK OR STRONG. SOME DO NOT APPEAR AT ALL. SOME MAY HAVE A FREQUENCY γ= ΔE/h AND WILL NOT BE DETECTED. THIS IS ACCOUNTED FOR BY THE NUMBER OF MOLECULES IN THE VARIOUS STATES WHEN THE SPECTRUM IS TAKEN.

16 INTENSITY OF SPECTRAL LINES SPECTRAL LINE INTENSITY OF MOLECULES FROM ENERGY LEVEL E i TO A FINAL LEVEL E f DEPENDS ON THE NUMBER OF MOLECULES THAT HAVE THE INITIAL ENERGY. IT ALSO DEPENDS ON THE PROBABILITY OR THE LIKELIHOOD OF THE TRANSITION AND, THE CONCENTRATION OR THE PATH LENGTH OF THE SAMPLE (BEER-LAMBERT LAW; I/I o =exp(-kcl) OR IN TERMS OF, I/I o =10 -ԑcl =T WHICH IS ALSO DEFINED AS TRANSMITTANCE, ԑ IS THE MOLAR ABSORPTION COEFFICIENT. OR LOG 10 (I/Io)= ԑcl=A WHERE A IS THE ABSORBANCE.

17 INTENSITY OF SPECTRAL LINES THIS NUMBER IS CALLED THE POPULATION. FOR A SAMPLE OF N MOLECULES AT TEMPERATURE T THE NUMBER WITH Ei IS GIVEN N(E i )α N exp(-E i /kT). ANOTHER LINE IN THE TRANSITION ON THE SPECTRUM WITH A STATE WITH ENERGY E i ’.

18 INTENSITY OF SPECTRAL LINES THE RATIO OF THE INTENSITIES OF THE TWO ENERGY LEVELS WILL BE I(E i )/I(E i ’)=N(E i )/N(E i ’)=exp [(-E i -E i ’)/kT] IF E i -E i ’>>kT THE NUMBER OF MOLECULES WITH ENERGY E i ARE FAR LESS THAN THE NUMBER WITHENERGY E i ’. HENCE THE INTENSITY OF THE LINE AT E i WILL BE LESS THAN THAT IN E’ i.

19 SELECTION RULES THESE ARE STATEMENTS THAT GOVERN WHETHER A TRANSITION IS ALLOWED OR FORBIDDEN. ALLOWED TRANSITIONS APPEAR AS LINES ON A SPECTRUM WHILE FORBIDDEN TRANSITIONS DO NOT APPEAR AT ALL EVEN THOUGH ENERGY LEVELS OF THE APPROPRIATE SEPARATION ARE PRESENT. SELECTION RULES DIFFER FROM THE VARIOUS TYPES OF TRANSITIONS; e.g. ROTATIONAL, VIBRATIONAL OR ELECTRONIC.

20 REGIONS OF THE SPECTRUM RADIO FREQUENCY; 3x10 6 -3x10 10 Hz OR 10m TO 1 cm WAVELENGTH (nmr/esr spectroscopy) MICROWAVE REGION;3x10 10 -3x10 12 Hz OR 1 cm TO 100 µm WAVELENGTH (Rotational spectroscopy). INFRARED REGION; 3x10 12 -3x10 14 Hz OR 100 µm- 1 µm WAVELENGTH. (Vibrational spectroscopy). VISSIBLE AND ULTRA VIOLET REGIONS; 3x10 14 - 3x10 16 Hz OR 1 µm – 10 nm WAVELENGTH.

21 REGIONS OF THE SPECTRUM CONT’D X-RAY REGION; 3x10 16 -3x10 18 Hz OR 10 nm - 100 pm. ENERGY CHANGES INVOLVING INNER ELECTRONS OF AN ATOM OR MOLECULES ϒ-ray REGION; 3x10 18 -3x10 20 Hz OR 100 pm TO 1 pm

22 REQUIREMENT CHARACTERISTICS RADIOFREQUENCY: THERE IS REVERSAL OF SPIN TO INTERACT WITH THE MAGNETIC FIELD OF ELECTROMAGNETIC RADIATION FOR MICROWAVE, THE SPECIES MUST POSSESS A PERMANENT DIPOLE MOMENT e.g. HCl, CO IN INFRARED, SPECIES SHOULD BE ABLE TO HAVE A CHANGE IN ITS DIPOLE MOMENT DURING INTERACTION, e.g O---C---O

23 MICROWAVE SPECTROSCOPY DEALS WITH RADIATION OF ʎ-RANGE FROM 1.0 cm TO 100 µm OR ᴠ OF 3 x10 10 - 3x 10 12. THIS REGION IS WHERE THE ENERGY IS ADEQUATE TO ROTATE A MOLECULE. THE ABILITY OF A MOLECULE TO ROTATE IN THE MICROWAVE REGION DEPENDS ON A NUMBER OF FACTORS; – STRUCTURE OF THE MOLECULES – SIZE OF THE MOLECULE

24 STRUCTURE OF MOLECULES MOLECULES CAN BE DIVIDED INTO FOUR GROUPS IN TERMS OF STRUCTURE – 1. LINEAR MOLECULES – 2. SYMMETRIC TOPS – 3. SPHERICAL TOPS – 4. ASYMMETRIC TOPS

25 RESOLUTION OF AXES A BODY HAS THREE MOMENTS OF INERTIA RESOLVED ACCORDING TO THE CARTESSIAN COORDINATES; I x, I y AND I z. THE KINETIC ENERGY OF A MOVING BODY IS GIVEN BY E= MV 2 FOR A ROTATING BODY, THE ENERGY IS GIVEN AS FOLLOWS; E= Iω 2 WHERE I IS THE MOMENTS OF INERTIA, ω IS THE ANGULAR MOMENTUM. A FREE BODY ROTATING IN THREE AXES HAS ENERGY E= I x ω 2 + I y ω 2 + I z ω 2

26 MOMENTS OF INERTIA ALL THE DIFFERENT MOLECULAR STRUCTURES ARE DESCRIBED BY I x, I y OR I z NOT ALL MOLECULES HAVE ALL THE I x, I y AND I z EQUAL. HENCE ANY MOLECULE THAT HAS I x = I y = I z WILL NOT BE MICROWAVE ACTIVE. THIS WILL BE ILLUSTRATED IN SPECIFIC STRUCTURES.

27 LINEAR MOLECULES MOLECULES IN WHICH ALL THE ATOMS ARE IN A STRAIGHT LINE. AN EXAMPLE HCl AND OCS HCl H-----------Cl OCS O----------C----------S FOR THESE MOLECULES, I x,= I y AND I z =0, THAT IS TWO HAVE VALUES AND THE THIRD IS ZERO.

28 SYMMETRIC TOPS FOR SYMMETRIC TOPS, ALL THE THREE MOMENTS OF INERTIA ARE EQUAL i.e. I y =I z ≠I x THERE IS PERMANENT DIPOLE MOMENT

29 SPHERICAL TOPS HAS ALL THREE MOMENTS OF INERTIA EQUAL I x =I y =I z MOLECULES HAVE NO DIPOLE OR CHANGE IN DIPOLE MOMENT AND NOT MICROWAVE ACTIVE

30 ASYMMETRIC TOPS THESE HAVE ALL THE MOMENTS OF INERTIA DIFFERENT. I x ≠I y ≠I z MAJORITY OF MOLECULES BELONG TO THIS GROUP e.g. H 2 O, CH 2 CH 2

31 RIGID DIATOMIC MOLECULES CALCULATION OF THE MOMENT INERTIA

32 MOMENT OF INERTIA FROM THE DIAGRAM; r 0 =r 1 +r 2................(1) if the molecule rotates about a point C, then by the principle of moments, m 1 r 1 =m 2 r 2..............................(2) and the moment of inertia about the same point C is; I=m 1 r 1 2 + m 2 r 2 2................................(3) =m 2 r 2 r 1 +m 1 r 1 r 2 =r 1 r 2 (m 1 +m 2 )

33 MOMENT OF INERTIA From the above equations; m 1 r 1 =m 2 r 2 =m 2 (r 0 -r 1 ) and therefore r 1 =r 2 =..............(4) Replacing (4) into (3) then the moment of inertia I= =µr 0 2.....................................(5) Where µ= IS CALLED THE REDUCED MASS

34 MOMENT OF INERTIA r 0 is the bond length THE ENERGY OF A ROTATING RIGID MOLECULE CAN BE OBTAINED USING THE FAMOUS SCHRODINGER’S EQUATION AS; E J = JOULES WHERE J=0,1,2,3,........ h is the Planck’s constant, I is the moment of inertia either, I y or I z since they are both equal. J TAKES VALUES FROM ZERO UPWARDS IS CALLED THE ROTATIONAL QUANTUM NUMBER

35 MOMENT OF INERTIA THE ROTATIONAL QUANTUM NUMBER IS RESTRICTED TO INTEGERS IT ARISES FROM THE SOLUTION OF THE SCHRODINGER’S EQUATION. THE FREQUENCY FROM THE ABOVE EQUATION IS; γ= Hz OR WAVE NUMBER; ṽ= cm -1 cm-1 (J=0.1,2,3......) WHERE c IS THE SPEED OF LIGHT WITH UNITS IN cm s -1

36 ROTATIONAL CONSTANT THE WAVE NUMBER UNITS ARE EXPRESSED IN RECIPROCAL cm WHICH IS cm-1 IT IS USUALLY ABBREVIATED AS; cm-1(J=0, 1, 2, 3.......) WHERE B IS THE ROTATIONAL CONSTANT WHICH IS cm -1

37 ROTATIONAL STATES ALLOWED ROTATIONAL ENERGIES OF DIATOMIC MOLECULES

38 EXAMPLES TRANSITION FROM ԑ J=0 to ԑ J=1 MEANS THAT ԑ J=1 - ԑ J=0 2B- 0 = 2B cm -1 and therefore ṽ= 2B cm -1 GIVE MORE EXAMPLES. IF THE MOLECULE IS RAISED FROM STATE J TO J+1, THEN ṽ = 2B(J+1) cm -1 IN ROTATIONAL SPECTRA, TRANSITIONS ARE STEPWISE i.e. J=O→J=1 J=2→J=3.............. BUT NOT J=0→ J=2 OR J=1 → J=3......

39 SELECTION RULE THIS GIVES RISE A SELECTION RULE FOR TRANSITIONS IN ROTATIONAL SPECTRA. THE SELECTION IS: ΔJ= ±1 THEMOLECULE MUST POSSESS A PERMANENT DIPOLE MOMENT. ENERGY DIFFERENCE BETWEEN ROTATIONAL LEVELS IS USUALLY VERY SMALL WHICH ALLOWS HIGHER LEVELS TO BE OCCUPIED EVEN AT ROOM TEMPERATURES

40 BOND DISTANCE CAN BE CALCULATED IF BIS KNOWN. IT IS GIVEN BY: r=m CALCULATION THE ROTATIONAL SPECTRUM OF HCl MOLECULES SHOWS THAT THE ROTATIONAL LINES ARE WQUALLY SEPARATED BY 20.70cm-1 CALCULATE THE INTERNUCLEAR DISTANCE.

41 SOLUTION STEP 1: OBTAIN THE VALUE OF B. SINCE 2B IS EQUAL TO SEPARATION DISTANCE, THEN; 2B=20.70 WHICH IMPLIES B=10.37 cm-1 STEP 2 CALUCLATE REDUCED MASS µ= µ= 1.627 x 10 -24 g OR 1.627 x 10 -27 kg

42 SOLUTION STEP 3:CALCULATE r USING r= r= =0.129 nm

43 USES OF ROTATIONAL SPECTRA ROTATIONAL OF A DIATOMIC MOLECULES CONSISTS OF EQUALLY SPACE LINES EACH SPACE IS EQUAL TO 2B WHICH ALSO A VALUE OF h/8π 2 Ic cm -1 FROM THIS VALUE, THE MOMENT OF INERTIA AND BOND DISTANCE CAN BE OBTAINED.


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