Microwave Spectroscopy
Dr. P. ELANGO Associate Professor & Head Department of Physics Government Arts College (Autonomous) Coimbatore 641018
Spectroscopy
Using electromagnetic radiation as a probe to obtain information about atoms and molecules that are too small to see. Electromagnetic radiation is propagated at the speed of light through a vacuum as an oscillating wave. Seeing the unseeable” Classification of Molecules
Molecules are classified on the basis of • Type of bonding: Covalent, Ionic, Metallic, etc. • On the basis of atomicity (total number of atoms that constitute a molecule), molecules can be classified as : • Monoatomic-composed of 1 atom e.g. He, Ne, Ar (all noble gases are monoatomic)
• Diatomic-composed of 2 atoms e.g. H2 , N2 , O2 • Triatomic-composed of 3 atoms e.g. O3
• Polyatomic-composed of 3 or more atoms e.g. P4 , S8 • Reactivity: Acidic or alkaline • Number of bonds: single, double, triple • Moment of Inertia The rotation of a three dimensional molecule is quite complex, but it can be described in terms of [c] THREE PRINCIPAL MOMENTS OF INERTIA [b] [a] 푰풂, 푰풃, & 푰풄 푎푙표푛푔 푎, 푏 & 푐 푎푥푒푠 Type of Rotator Moment of Inertia Example [a] c Linear Ia = 0 HCl, OCS b
Ib = Ic I a Symmetric Top
Prolate Ia=Ib>Ic CH3F
Oblate Ia=Ib Spherical Top Ib=Ic=Ia CH4 , SF6 Asymmetric Top Ib≠Ic≠Ia H20 Interaction of Radiation with Rotating Molecule: Coupling Mechanism An e.m.wave is an oscillating electric field and interacts only with molecules that can undergo a change in dipole moment Rotation of a polar diatomic molecule showing the fluctuation in the dipole moment MICROWAVE SPECTROSCOPY (Rotational Spectroscopy) • Determination of the structure of molecules which do not give good results by IR or Raman spectroscopy • The better results in microwave spectroscopy may by attributed to the more precise measurement of the frequencies in this region • Klystron and Magnetron oscillators • Microwave Region: • Wavelength : 100 cm to 1 mm • Frequency : 3 x 1013Hz to 3 x 1010Hz • Lies between far IR and conventional rf region • Mostly absorption spectra • Rotational energy of a molecule is quantized • Permitted energy values are called rotational energy levels • These energy values can be obtained by solving Schrodinger equation for the system represented by that molecule A molecule interacts with the oscillating electric field of the incident radiation to absorb rotational energy and produce absorption spectra Only such molecule which have resultant dipole moment on rotation can generate an electric field which is capable of interacting with the electric component of the incident microwave radiation Homonuclear & symmetric molecules are microwave inactive Hetronuclear and assymmetric molecules are microwave ACTIVE since they posses permanent dipole moment ROTATIONAL SPECTRA OF LINEAR DIATOMIC MOLECULES • Rotating diatomic molecule whose nuclei are separated by a definite distance • The bond between atoms is stiff and does not change in length (Rigid Rotator) • A diatomic molecule can rotate around a vertical axis. The rotational energy is quantized. • Rotational energy level, frequency of spectral lines due to transition, selection rule, absorption spectra are studied Inter nuclear distance = r Molecule rotates about G 2 2 2 Moment of inertia of ‘i’ particles is = I = m1r1 + m2r2 +……..+miri 2 = ∑miri 2 2 I = m1r1 + m2r2 (In case of two particles) As the system is balanced about G, we can write , m1r1 = m2r2 CM m1 m2 r1 r2 r = r1+ r2 m1r1 m2 r2 m1 (r r2 ) m2 r2 m1r m1r2 m2 r2 m1r m1r2 m2 r2 m1r r2 (m1 m2 ) m1 r2 r m1 m2 m2 r1 r Derive m1 m2 2 2 I m1r1 m2 r2 2 2 m1m2 2 m2 m1 2 I 2 r 2 r (m1 m2 ) (m1 m2 ) m m I 1 2 r 2 (m1 m2 ) I .r 2 m1m2 where...... Reduced mass (m1 m2 ) 1 Energy of a rotating molecule = 퐼휔2 2 Angular momentum of a rotating molecule = 퐼휔 푛ℎ 퐽ℎ Bohr quantum condition 퐼휔 = = 2휋 2휋 ퟏ (푰흎)ퟐ 푳ퟐ 풉ퟐ 푬 = 푰흎ퟐ = = = 푱ퟐ Classical Expression 푱 ퟐ ퟐ푰 ퟐ푰 ퟖ흅ퟐ푰 But, quantum restriction with which angular momentum 퐿 is quantized is given by ℎ 퐿 = 퐽(퐽 + 1) ℏ = 퐽(퐽 + 1) 2휋 This value of L will convert chemical expression into quantum mechanical expression. ℎ ퟐ 2 ퟐ 퐽 퐽 + 1 ퟐ h 푳 2휋 풉 E J (J 1)...... joules 푬 = = = 푱(푱 + ퟏ) J 2 푱 ퟐ푰 ퟐ푰 ퟖ흅ퟐ푰 8 I • Frequency 휈 = Number of vibrations/ second (Unit: 푠−1or Hz) • For Molecular vibrations, this number is very large (of the order of 1013푠−1) and therefore INCONVINIENT. • More convenient parameter is Wavenumber 휈ҧ which is defined as ퟏ the reciprocal of wavelength 흀 1 퐹푟푒푞푢푒푛푐푦 훾 3푥1013푠−1 • 휈ҧ = = = = = 1000 푐푚−1 휆 푣푒푙표푐𝑖푡푦 푐 3푥1010푐푚.푠−1 • Therefore frequency in terms of wavenumber is obtained by simply dividing frequency by velocity of light. • Similarly, ENERGY can be expressed in terms of wavenumber by dividing ENERGY by hc ℎ 푐 퐸 • 퐸 = ℎ휈 = = ℎ푐 휈ҧ 푊푎푣푒푛푢푚푏푒푟 휈ҧ = 휆 ℎ푐 In spectroscopy, it is convenient to use wavenumber units 2 E h 20B J J (J 1) J=4 J hc 8 2 Ihc h 12B J=3 J (J 1) J 8 2 Ic 6B J=2 BJ (J 1)...... cm1 J 2B J=1 0 J=0 where ‘B’ is called Rotational constant NOTE: Rotational Constant B is inversely proportional to Moment of Inertia I BJ (J 1) • Selection rule: ∆J = ±1 J • Other transitions are spectroscopically forbidden J J(J 1) [B(J 1)(J 2)][BJ (J 1)] -1 Frequency of the absorption line J J 2B(J 1) cm • If a molecule absorbs incident radiation and raise from J=0 state to J=1 state, then εJ=1 – εJ=0 = 2B – 0 = 2B Similarly, εJ=2 – εJ=1 = 6B – 2B = 4B εJ=3 – εJ=2 = 12B – 6B = 6B For J=1 Molecule has lowest angular momentum Molecule does not rotate for J=0 Energy difference (∆휺푱) between adjacent lines in a rotation spectrum of a diatomic molecule is CONSTANT (ie) FREQUENCY LINES ARE EQUALLY SPACED Energy levels are NOT equally spaced J BJ (J 1) 2B 0 B0(0 1) 0 1 o 1 B1(11) 2B 2 1 4B 2 B2(2 1) 6B 3 2 6B 3 B3(31) 12B 4 B4(4 1) 20B Calculate the reduced mass and the moment of inertia of 퐷2퐶푙35, given that the bond length of the molecule is 127.5 p.m. The masses of 퐷2 and 퐶푙35 are 2.01410 and 34.96885 amu. 1 푝푖푐푎 푚푒푡푒푟 = 10−12푚 푚 × 푚 2.01410 × 34.96885 휇 = 퐷 퐶푙 = = 1.90441 푎푚푢 푚퐷 + 푚퐶푙 2.01410+34.96885 1 푎푚푢 = 1.661 × 10−27 푘푔 휇 = 1.90441 × 1.661 × 10−27 푘푔 = 3.16 × 10−27 푘푔 퐼 = 휇푟2 = 3.16 × 10−27 푘푔 × 127.5 × 10−12 푚 2 = 5.14 × 10−47푘푔. 푚2 The B value estimated for 퐻1퐶푙35 is 10.59342 cm-1. The masses of 퐻1 and 퐶푙35 are 1.0078250 and 34.9688527 amu. What is the bond length of the molecule? 푚 × 푚 1.0078250 × 34.9688527 휇 = 퐻 퐶푙 = = 0.97959 푎푚푢 푚퐻 + 푚퐶푙 1.0078250+34.9688527 1 푎푚푢 = 1.661 × 10−27 푘푔 휇 = 0.97959 × 1.661 × 10−27 푘푔 = 1.627 × 10−27 푘푔 ℎ ℎ 퐵 = = 8휋2퐼푐 8휋2휇푟2푐 ℎ 6.626 × 10−34퐽. 푠 푟2 = = 8휋2휇퐵푐 10.59342 푐푚−1 8 × 휋2 × 1.627 × 10−27 × 푚−2 × 3 × 108푚. 푠−2 10−2 푟2 = 1.62464 × 10−20푚2 푟 = 1.274 × 10−10푚 = 1.274 Å Isotopic Effect in Rotational Spectra • Isotopes are atoms of same element having same number of protons (or electrons), but different number of neutrons • Same chemical property but different physical property • Same atomic number (Z) but different mass number (A) • An atom when replaced by one of its isotopes, the inter-bond distance remains same as the electron charge distribution does not change • But the MASS of the nucleus changes leading to a change in the moment of inertia I • Hence the value of rotational constant ‘B’ changes On going from 12C16O 13C16O, there is a mass increase and hence decrease in the B value. Let us write the rotational constant B of 13C molecule with a prime, and we have 퐵 > 퐵′ This will result in the lowering of energy levels of 13C molecule Spectrum of heavier species will show a smaller separation between the lines (2퐵′) than that of the lighter one (2퐵) This decreased separation is used to calculate the From 12C16O 13C16O, mass increases, B atomic weights decreases and so Energy levels lower For 12C16O & 13C16O 퐵 = 1.92118 푐푚−1 푎푛푑 퐵′ = 1.83669 푐푚−1 ℎ 퐵 2 퐼′ 휇′ 1.92118 = 8휋 퐼푐 = = = = 1.046 퐵′ ℎ 퐼 휇 1.83669 8휋2퐼′푐 Taking mass of 16O= 15.9994 & that of 12C to be 12.00 휇′ 15.9994 푚′ 12 + 15.994 = 1.046 = × 휇 15.9994 + 푚′ 12 × 15.9994 We can find 푚′=13.0007, which is within 0.02% of the From 12C16O 13C16O, mass increases, B best value decreases and so Energy levels lower transmission % % 15 20 25 30 35 40 (cm-1) 12C16O (major species) 13C16O and 12C18O lines Intensity of Spectral Lines Depends on the number of molecules present in each level. This depends on the Boltzmann Factor & Degeneracy of the level (degeneracy = existence of 2 or more energy states having exactly the same energy) Each allowed energy of rigid rotor is (2J+1)-fold degenerate. Hence, there exists (2J+1) different wave functions with that energy Energy in joule 퐸퐽 = 휀퐽. ℎ푐 = 퐵 퐽 퐽 + 1 . ℎ푐 EJ BhcJ (J 1) N J No exp No exp kT kT Taking degeneracyinto account BhcJ (J 1) N J No (2J 1)exp kT To find the value of J for which the population is maximum, 푑푁 we have 퐽 = 0. This leads to 푑퐽 kT 1 J 2Bhc 2 ie., the level nearest to this J value will have the maximum population. Hence transitions originating from this level have maximum intensity Non-rigid Rotator • Our assumption of rigid bond is only an approximation • All the bonds are elastic to some extend • Due to elasticity of the bond, following changes may occur: • The elastic bond may have vibrational energy • There is a change in the value of B and r during vibrations • When the molecule rotates with more energy the bond length increases due to centrifugal distortion • Does this centrifugal distortion affect the rotational spectrum? Non-rigid Rotator Rotational data of HF molecule 푱 흂ഥ ퟐ푩 = 휟흂ത푱 휟흂ഥ푱 Bond 푱 푩 = • Theory of Rigid diatomic molecule ퟐ Length 풓 (Å) concludes that the frequency lines 0 41.08 are equally spaced = 2B 1 82.19 41.11 20.56 0.929 • But in experimental data. It is 2 123.15 40.96 20.48 0.931 found that B decreases with 3 164.00 40.85 20.43 0.932 increasing J. 4 204.62 40.62 20.31 0.935 • As J increases rotational energy 5 244.93 40.31 20.16 0.938 increases the molecule rotates 6 285.01 40.08 20.04 0.941 with higher energy Centrifugal 7 324.65 39.64 19.82 0.946 distortional force increases 8 363.93 39.28 19.64 0.951 Bond length increases Non-rigid Rotator This effect of increase in bond length due to centrifugal force is called centrifugal effect. Bond length depends on which rotational energy level the molecule is in. To improve our model let us consider the molecule as a Non-rigid Rotator. A diatomic molecule with masses m1 & m2 is rotating about an axis passing through center of mass (CM) with an angular velocity 휔. Such rotating molecule with two masses can be visualized as a molecule with single mass (휇) rotating with angular velocity 휔. 휔 휔 푚 × 푚 휇 = 1 2 = 푅푒푑푢푐푒푑 푀푎푠푠 푚1 + 푚2 Non-rigid Rotator 휇푣2 Centrifugal force experienced by the rotating molecule with single mass is = 퐹 = 푐 푟 푣 휇푣2 Since 휔 = , we can write 퐹 = = 휇휔2푟 푟 푐 푟 This centrifugal force is balanced by restoring force 퐹푟 = 푘(푟 − 푟표) where 푘 is the force constant. 휇휔2푟 퐹 counterbalances 퐹 . ∴ 퐹 = 퐹 푟 − 푟 = 푐 푟 푐 푟 표 푘 2 (or) 휇휔 푟 = 푘(푟 − 푟 ) We shall 표 2 4 2 휇 휔 푟 use this 2 푟 − 푟 2 = 휇휔 푟 = 푘푟 − 푘푟표 표 푘2 later 2 푘푟 − 휇휔 푟 = 푘푟표 푘푟 푟 = 표 As angular velocity 휔 increases, the denominator decreases. 푘−휇휔2 Therefore the bond length 푟 increases. Non-rigid Rotator 푇퐸 = 퐾퐸 + 푃퐸 This eq. can be written in terms of angular momentum 퐿 = 퐼휔 1 1 퐾퐸 = 퐼휔2 & 푃퐸 = − 푘(푟 − 푟 ) 2 2 2 표 퐿2 퐿4 푇퐸 = − 1 1 2퐼 2퐼2푘푟2 푇퐸 = 퐼휔2 − 푘 푟 − 푟 2 2 2 표 The quantum restriction with which 퐿 is quantized is ℎ 퐿 = 퐽(퐽 + 1) ℏ = 퐽(퐽 + 1) 2휋 1 1 휇2휔4푟2 = 퐼휔2 − 푘 This quantum restriction will convert a classical result into 2 2 푘2 a quantum mechanical result. 2 1 1 퐼휔2 ℎ2 ℎ3 = 퐼휔2 − Since 퐼 = 휇푟2 ∴ 푇퐸 = 퐸 = 퐽 퐽 + 1 − 퐽(퐽 + 1) 2 2 2 푘2 퐽 8휋2퐼 32휋4퐼2푘푟2푐 ℎ2 ℎ4 퐸 = 퐽 퐽 + 1 − 퐽(퐽 + 1) 2 퐽 8휋2퐼 32휋4퐼2푘푟2 In spectroscopically convenient wavenumber units, 퐸 ℎ ℎ3 휀 = 휈ҧ = 퐽 = 퐽 퐽 + 1 − 퐽(퐽 + 1) 2 퐽 퐽 ℎ푐 8휋2퐼푐 32휋4퐼2푘푟2푐 2 휀퐽 = 휈퐽ҧ = 퐵 퐽 퐽 + 1 − 퐷[퐽 퐽 + 1 ] Assignment: 4퐵3 3 Show that 퐷 = 2 ℎ ℎ 휈휐 푤ℎ푒푟푒 퐵 = & 퐷 = 8휋2퐼푐 32휋4퐼2푘푟2푐 where 1 푘 휈ഥ = vibrational frequency 휐 2휋푐 휇 = Non-rigid Rotator Selection Rule : ∆J=±1: J J 1 J 1 J 2 2 J J 1 B(J 1)(J 2) D(J 1) (J 2) BJ (J 1) DJ 2 (J 1) 2 2 2 2 J J 1 B(J 1)[J 2 J ] D(J 1) [(J 2) J ] 2 2 2 J J 1 2B(J 1) D(J 1) [J 4 4J J ] 2 J J 1 2B(J 1) 4D(J 1) (J 1) 3 J J 1 2B(J 1) 4D(J 1) This term is responsible for shift of lines (Frequency separation decreases with increasing J value) SAME AS RIGID MOLECULE Non-rigid Rotator Decrease in separation between successive lines with increasing J Consequently there is a decrease in the value of B Thus, Bond length increases with increasing J Polyatomic molecules Things get much more complicated, but the general principles are the same. LINEAR MOLECULES OCS Carbon oxysulphide HCCCl Chloroacetylene Moment of Inertia Condition IB = IC; IA = 0 All atoms lie on a straight line – Only non-cyclic molecule considered. Energy levels for Diatomic molecules are given by ퟐ ퟐ −ퟏ 휺푱 = 흂ഥ푱 = 푩 푱 푱 + ퟏ − 푫푱 푱 + ퟏ 풄풎 Spectral lines will show 2B separation modified by the distortion constant • Discussion on diatomic molecule applies equally to linear molecules • IA is along the length of the molecule and it is zero. So there is only one independent moment of inertia, either IB or IC , Say simply I. • Moment of Inertia of end-over-end rotation of polyatomic molecule, I, is greater than that of a diatomic molecule. • Therefore, B smaller and spectral lines more closely spaced. • ‘B’ for diatomic molecule is 10 cm-1 and for triatomic molecule 1 cm-1 • Molecule must possess dipole moment if it is to be microwave active • OCS – microwave active, but, OCO – inactive • Linear polyatomic molecule with N atoms has (N-1) bonds. • In case of OCS molecule there are two bonds [C—O bond and C—S bond] • Therefore the values of TWO bond lengths and ONE moment of inertia have to be found • OCS in detail: 16 16 풓푪 MI of OCS = 퐼 Mass of O = 푚푂 18 18 ′ MI of OCS = 퐼′ Mass of O = 푚푂 풓푶 풓푺 CM Considering moments about CM 푚푂푟푂 + 푚퐶푟퐶 = 푚푆푟푆 −−−−−−−− − 2 풓푪푶 풓푪푺 Considering MI 풓푶 = 풓푪푶 + 풓푪 2 2 2 (1) 퐼 = 푚푂푟푂 + 푚퐶푟퐶 + 푚푆푟푆 −−−−− −(3) 풓푺 = 풓푪푺 − 풓푪 Substituting (1) in (2), 푚푂 푟퐶푂 + 푟퐶 + 푚퐶푟퐶 = 푚푆(푟퐶푆−푟푐) 푚푂푟퐶푂 + 푚푂푟퐶 + 푚퐶푟퐶 = 푚푆푟퐶푆 − 푚푆푟퐶 푚푂푟퐶 + 푚퐶푟퐶 + 푚푆푟퐶 = 푚푆푟퐶푆 − 푚푂푟퐶푂 (푚푂+푚퐶 + 푚푆)푟퐶 = 푚푆푟퐶푆 − 푚푂푟퐶푂 푀푟퐶 = 푚푆푟퐶푆 − 푚푂푟퐶푂 where 푀 = (푚푂+푚퐶 + 푚푆) = 푇표푡푎푙 푚푎푠푠 표푓 푡ℎ푒 푀표푙푒푐푢푙푒 푚 푟 − 푚 푟 푟 = 푆 퐶푆 푂 퐶푂 −−−−−−−−−−−−−−−−−−−−−−−−−−−− −(4) 퐶 푀 Substituting (1) in (3) 2 2 2 퐼 = 푚푂 푟퐶푂 + 푟퐶 + 푚퐶푟퐶 + 푚푆 푟퐶푆 − 푟푐 2 2 2 2 2 퐼 = 푚푂푟퐶푂 + 푚푂푟퐶 + 2푚푂푟퐶푂푟퐶 + 푚퐶푟퐶 + 푚푆푟퐶푆 + 푚푆푟푐 − 2푚푆푟퐶푆푟푐 2 2 2 퐼 = (푚푂+푚퐶 + 푚푆)푟퐶 + 2푟퐶 푚푂푟퐶푂 − 푚푆푟퐶푆 + 푚푂푟퐶푂 + 푚푆푟퐶푆 2 2 2 퐼 = 푀푟퐶 + 2푟퐶 푚푂푟퐶푂 − 푚푆푟퐶푆 + 푚푂푟퐶푂 + 푚푆푟퐶푆 −−−−−−−−−−− −(5) Finally substituting (4) in (5) 푚 푟 − 푚 푟 2 푚 푟 − 푚 푟 퐼 = 푀 푆 퐶푆 푂 퐶푂 + 2 푆 퐶푆 푂 퐶푂 푚 푟 − 푚 푟 + 푚 푟2 + 푚 푟2 푀 푀 푂 퐶푂 푆 퐶푆 푂 퐶푂 푆 퐶푆 푀 푚 푟 − 푚 푟 2 2 푚 푟 − 푚 푟 2 퐼 = 푂 퐶푂 푆 퐶푆 − 푂 퐶푂 푆 퐶푆 + 푚 푟2 + 푚 푟2 푀2 푀 푂 퐶푂 푆 퐶푆 푚 푟 − 푚 푟 2 퐼 = 푚 푟2 + 푚 푟2 − 푂 퐶푂 푆 퐶푆 −−−−−−−−−−−−−−−−−−−− −(6) 푂 퐶푂 푆 퐶푆 푀 푚 푟 − 푚 푟 2 퐼 = 푚 푟2 + 푚 푟2 − 푂 퐶푂 푆 퐶푆 −−−−−−−−−−−−−−−− −(6) 푂 퐶푂 푆 퐶푆 푀 Similarly, for isotopically substituted 18OCS molecule, we can write, 푚′ 푟 − 푚 푟 2 퐼′ = 푚′ 푟2 + 푚 푟2 − 푂 퐶푂 푆 퐶푆 −−−−−−−−−−−−−−−− −(7) 푂 퐶푂 푆 퐶푆 푀′ Equations (6) and (7) are two simultaneous equations. We can solve these equations ′ to find the values of 푟퐶푂& 푟퐶푆 provided we have the values of 퐼 & 퐼 of the molecule. Similar approach can be employed to study the linear molecules having more than 3 atoms by constructing more equations. Symmetric Top Molecules Has two independent moments of inertia. IA is due to rotation about TOP axis (or) molecular symmetry axis IB & IC is due to PROLATE CH3Cl, CH3F, IA < IB = IC rotation about an SYMMETRIC CH3CN, NH3, etc axis perpendicular to TOP top axis IC is due to rotation about TOP axis (or) molecular symmetry OBLATE BF , BCl , etc axis IA = IB < IC 3 3 SYMMETRIC IB & IA is due to TOP rotation about an axis perpendicular to top axis In Symmetric top molecules, there are two independent moments of inertia, one about molecular symmetry axis (Top axis – C-F Bond) and the other about an axis perpendicular to the molecular symmetry axis. We need two quantum numbers to describe these two moments of inertia. 1) 퐽 Total angular Sum of separate angular Note: In linear molecules, there was momentum momenta about two only one independent MI and so the total angular momentum was taken as 퐽 quantum number different axes 2) K Angular Represents the projection momentum about of total angular molecular momentum 퐽 on the symmetry axis molecular symmetry axis Now let us discuss the allowed values for 퐽 & 퐾. Both 퐽 & 퐾 must be integral values or zero. Theoretically, 퐽 can be as large as we like, that is, 퐽 can be 0, 1, 2, 3, 4, ……. ∞. But at high rotational energies, molecule dissociates. Therefore, dissociation energy restricts the upper limit of the 퐽 value. Rotational energy represented by 퐽 can be divided in several ways between 1) The motion about the main symmetry axis and 2) Motion about the axis perpendicular to the main symmetry axis When J=2; K= -2, -1, 0, +1, +2 When 퐽 = 3 K=±3 Represents all rotation about the main symmetry axis K=±2 Represents two third of rotation about symmetry axis and one third about axis perpendicular to it K=±1 Represents one third of rotation about symmetry axis and two third about axis perpendicular to it K=0 Represents rotation about perpendicular axis K can have negative values which represent anticlockwise rotation (Positive for clockwise) But K cannot be greater than J. To be precise 퐾 cannot be greater than J. To sum up: J = 0, 1, 2, 3, …… and K=0, ±1, ±2, ±3, …..±J Thus K has (2J+1) values for each J. Let us now consider a rigid symmetric top molecule in which bonds are rigid and do not stretch under centrifugal force. Solution of Schrodinger equation for rigid PROLATE symmetric top molecules gives the energy levels. 2 2 2 ℎ ℎ ℎ 2 퐸퐽,퐾 = 2 퐽 퐽 + 1 + 2 − 2 퐾 푗표푢푙푒 8휋 퐼퐵 8휋 퐼퐴 8휋 퐼퐵 퐸 휀 = 퐽,퐾 = 퐵퐽 퐽 + 1 + 퐴 − 퐵 퐾2 푐푚−1 −−−−−−−−− −(1) 퐽,퐾 ℎ푐 ℎ ℎ Where 퐴 = 2 & 퐵 = 2 8휋 퐼퐴퐶 8휋 퐼퐵퐶 In case of rigid OBLATE symmetric molecules assuming 퐼퐶 as the top axis, 퐸 휀 = 퐽,퐾 = 퐵퐽 퐽 + 1 + 퐶 − 퐵 퐾2 푐푚−1 −−−−−−−−− −(2) 퐽,퐾 ℎ푐 ℎ ℎ Where 퐶 = 2 & 퐵 = 2 8휋 퐼퐶퐶 8휋 퐼퐵퐶 In both (1) and (2), we see 퐾2 term. It means, it is immaterial whether the top spins clockwise or anticlockwise, the energy is same for the given angular momentum. For all K>0, the rotational energy levels are DOUBLY DEGENERTATE. 2 For PROLATE molecules since IB > IA, B < A and hence (A - B)K Energy increases as K increases. is always POSITIVE. 2 For OBLATE molecules since IB < IC, B > C and hence (C - B)K Energy decreases as K increases. is always NEGATIVE. For PROLATE molecules Energy increases as K since IB > IA, B < A and increases. hence (A - B)K2 is always POSITIVE. For OBLATE molecules Energy decreases as K since IB < IC, B > C and increases. hence (C - B)K2 is always NEGATIVE. PROLATE OBLATE Selection Rule: ∆퐽 = ±1 as before 푎푛푑 ∆퐾 = 0 2 2 ∴ 휀 퐽+1 , 퐾 − 휀퐽, 퐾 = 휈퐽ҧ ,퐾 = 퐵 퐽 + 1 퐽 + 2 + 퐴 − 퐵 퐾 − [퐵퐽 퐽 + 1 + 퐴 − 퐵 퐾 ] = 2퐵 퐽 + 1 푐푚−1 −−−−−−−−−−− −(3) The frequency spectrum is independent of K, indicating that the rotational changes about the symmetry axis do not give rise to rotational spectrum. The reason is that the rotation about the symmetry axis causes NO CHANGE in the dipole moment since the dipole moment of the symmetric top is directed along the symmetry axis. Hence the energy of this type of rotation will not interact with the e.m. radiation. −1 The frequencies of the transition is given by 휈퐽ҧ ,퐾 = 2퐵 퐽 + 1 푐푚 . This spectrum is same as that of linear molecules and that only one MI can be measured (end-over-end rotation). For non-rigid symmetric top molecules, we have to the centrifugal stretching into account. The Energy levels will be 2 2 2 2 2 −1 휀퐽,퐾 = 퐵퐽 퐽 + 1 + 퐴 − 퐵 퐾 − 퐷퐽퐽 퐽 + 1 − 퐷퐽퐾퐽 퐽 + 1 퐾 − 퐷퐾퐾 푐푚 3 2 휈퐽ҧ ,퐾 = 2퐵 퐽 + 1 − 4퐷퐽 퐽 + 1 − 2퐷퐽퐾 퐽 + 1 퐾 This spectrum depends on K. That means rotation about symmetry axis causes change in the dipole moment since the molecule undergoes stretching and bending. Again, like linear molecules, only one MI can be determined. However, using isotopic substitution, other MIs can be found. All the bond lengths and angles of symmetric top molecules can be found. 1. Source 2. Measurement of frequency 3. Guidance of the radiation to the absorbing substance (Wave guides) 4. Sample cell 5. Detector and recorder Source • Klystron • Varied mechanically over a wide rang or electrically over a small range • Very high stability • Backward wave oscillators • Tuned electronically over a wide range • Gunn Diodes • Low Power consumption • Mechanical tuning over whole range Frequency Measuring device • Cavity Wavemeters used for rough estimates • Accuracy 1 to 5 MHz • For accurate frequency measurements done by frequency counters or by the use of a beat technique Guidance of radiation to cell • Guidance of microwave radiation form the source to sample cell is done by using rectangular waveguides • In addition to this, waveguide bends, connectors, tapers, impedance matching devices, attenuators etc., are also needed Sample Cell • Stark cell is commonly used which Stark modulation techniques • Long rectangular waveguide (3 to 4 m), ends of which are sealed off by means of thin mica windows • Can be evacuated and then sample can be admitted in Detectors • A silicon crystal mounted in a coaxial cartridge is mostly used • Incoming radiation gives rise to a dc current • Transmitted radiation is modulated only when molecular resonance occurs • Due to resonance small ac component overlaps dc • The ac component is amplified and detected by phase sensitive detector • This detector is connected to CRO IR Spectroscopy Frequency, n in Hz ~1019 ~1017 ~1015 ~1013 ~1010 ~105 Wavelength, l ~.0001 nm ~0.01 nm 10 nm 1000 nm 0.01 cm 100 m Energy (kcal/mol) > 300 300-30 300-30 ~10-4 ~10-6 g-rays X-rays UV IR Microwave Radio nuclear core electronic molecular molecular Nuclear Magnetic excitation electron excitation vibration rotation Resonance NMR (PET) excitation (p to p*) (MRI) (X-ray cryst.) Visible What Happens When a Molecule Absorbs Infrared Radiation? • Bonds may stretch back and forth • Bonds may rotate • Bonds may bend • Groups on atoms may “wag” and “scissor” • The molecule may rotate and vibrate in other ways Electromagnetic Radiation Causes a Vibration of the Electric and Magnetic Fields of the medium through which they pass This Vibration will affect the Polar portions of a Molecule For a molecule to absorb IR, the vibrations or rotations within a molecule must cause a net change in the dipole moment of the molecule. Symmetrical Antisymmetrical Scissoring stretching stretching Rocking Wagging Twisting Various Modes of Vibration Will Require Different Wavelengths of IR Radiation to be Absorbed With So Many Possible Modes of Vibration, You Might Expect a Huge Number of Absorptions for a Given Compound But You Get Fewer Absorptions Because • The modes of vibration are quantized • Not all motion results in a change in the dipole moment of the molecule • Different modes of vibration are caused by the absorption of radiation of similar wavelength The Bottom Line Is That a Given Compound Will Only Have specific Set of Wavelengths That Will Be Absorbed Having a Specific Set of Absorptions Caused By Specific Functional Groups Allows Us to Identify the Presence or Absence of Specific Groups in a Given Compound INFRA RED SPECTROSCOPY Diatomic molecule – atoms A and B separated by equilibrium distance re. Minimum of the curve correspond to r = re Squeezing the atoms Cl results in increase of repulsive force Cl Pulling the atoms apart results in increase of attractive force Potential Energy Potential PE increases in both the cases 0 Internuclear distance VIBRATING DIATOMIC MOLECULE AS A HARMONIC OSCILLATOR In a diatomic molecule, two atoms of mass m1 and m2 are joined by a spring Molecule is distorted from its equilibrium length re to a new length r The restoring forces on each atom are d 2 r m 1 k(r r ) k is the force constant. 1 dt 2 e It measures the stiffness d 2 r m 2 k(r r ) of the bond. 2 dt 2 e m2 r m1r r1 and r m m 2 1 2 m1 m2 d 2 r m 1 k (r r ) 1 dt 2 e 2 m1m2 d r 2 k(r re ) m1 m2 dt 2 m1m2 d (r re ) Since r is k(r r ) e 2 e a constant m1 m2 dt Put...... (r re ) x m m and...... 1 2 m1 m2 2 d x 2 k k.x where dt 2 d 2 x k Frequency of vibration is given by x 0 dt 2 1 k n 2p d 2 x 2 x 0 1 k In terms of dt 2 n Wave number 2pc Unit: cm-1 This is the equation of simple harmonic Vibrational energies are quantized. motion The permitted values of vibrational energies can be found by solving Schrodinger equation. The eigen values for energy of a linear Harmonic oscillator are given by is the classical is called Here, fundamental TERM VALUE & frequency Lowest vibrational Energy When v=0, ZERO POINT ENERGY Atoms can never have ZERO vibrational energy Atoms can never be at REST relative to other Depends only on classical vibrational energy Classical Mechanics could find no objection to molecule possessing zero vibrational energy But Quantum Mechanics insists that molecule must vibrate to some extent even at lowest point Results of Quantum Mechanics are well supported by experiments Selection rule = ∆v=±1 Vibrational energy changes will give rise to an observable spectrum only if it interacts with the incident radiation That is, if the vibration involves a change in dipole moment of the molecule Homonuclear Molecule – No dipole moment – No Vibrational Spectrum Hetro nuclear molecule – IR Spectrum is observed (Vibrational Spectrum) Since, vibrational levels are equally spaced, the transition between any two neighboring states will give rise to SAME ENERGY CHANGE Anharmonic Oscillator Real molecules do not exactly obey the laws of SHM When the bonds are stretched beyond a certain limit, (10% of bond length), a much more complicated behavior is seen Bond breaks at this point An empirical expression (using curve fitting method) has been derived by P.M. Morse (Morse function) 2 E Deq 1 expa(req r) where ‘a’ is constant for a particular molecule Deq is called dissociation Energy Energy values are obtained by solving the Schrodinger equation The pattern of allowed vibrational energy levels are found to be Calculation of Ground State Energy at 0 = Zero Point Energy = Slightly differs from harmonic oscillator Selection Rule ∆v=±1, ±2, ±3…… Same as harmonic oscillator with additional possibility of larger jumps But, in Practice ∆v=±1, ±2 and ±3 have observable intensity Spacing between vibrational levels is nearly 10+3 cm-1 at room temperature According to Boltzmann distribution, at room temperature, the population of v=1 level is nearly 0.01% to 1% of ground state population Therefore, transitions originating at v=1 can be ignored Three spectral lines lie very close to SECOND OVERTONE (8347 cm-1) FIRST OVERTONE (5668 cm-1) FUNDAMENTAL ABSORPTION (2886 cm-1) 1 2. 1 very weak 2 2 1 1 1 1 01 2 e e 2 e 1 e e 1 e 2 2 2 2 1 12 e (1 4 e ) cm Slightly lower wave number than fundamental absorption Such weak bands are called HOT BANDS Since high temperature is necessary for their occurrence Such bands are observed by raising the temperature of the sample DIATOMIC VIBRATING ROTATOR Rotational Energy Separation 1 – 10 cm-1 -----FINE STRUCTURE Vibrational Energy Separation 10+3 cm-1 -----COARSE STRUCTURE Vibrational Energy Separation for HCl molecule = 3000 cm-1. Energies of two motions (vibrational and rotational) are so different Molecule can execute these two motions INDEPENDENTLY The combined rotational and vibrational energy is simply the sum of the separate energies Etotal = Erot + Evib joule -1 Εtotal = εrot + εvib cm This is called BORN-OPPENHEIMER approximation (Ignoring the small centrifugal distortion constant D) Selection rule ∆ = ±1, ±2, …. ∆J = ±1 ∆ = 0 corresponds to PURE rotational spectrum Usually, the vibrational changes in a molecule will be accompanied by a simultaneous rotational changes Rotational quantum numbers in the =0 state are denoted as J” (Double prime for lower vibrational state) Rotational quantum numbers in the =1 state are denoted as J’ (Single prime for lower vibrational state) Considering = 0 = 1 transition only Lines arising from ∆J = -2 -1 0 +1 +2 are called O P Q R S BRANCH J 1, that is, J 'J" 1 & J ' J"1 BRANCH - ‘R’ ‘R’ J 1, that is, J 'J" 1 & J" J '1 J ,v o B(J 'J")(J 'J"1) BRANCH - J ,v o B( 1)(J 'J '11) ‘P’ ‘P’ J ,v o 2B(J '1) Here, J ' 0,1,2..... Combining the equations for P branch and R branch, we get J ,v o 2B( J '1) J ,v o 2B.m where m 1, 2...... The value of ‘m’ CANNOT be ZERO because this will mean that the values of J’ or J” be negative (-1), which is not possible This equation represents the vibration – rotation spectrum Such a spectrum contain equally spaced lines with spacing 2B on either sides of the band origin (ωo) As m = 0, no lines will appear at ωo and thus Q branch lines will be absent LINEAR MOLECULES The rotation-vibration spectra of DIATOMIC molecules give rise to equally spaced lines for P and R branches For vibrations to be IR active, there must be a change in dipole moment during the vibration. For complex molecules, vibrations can be divided into two categories i. Vibrations causing dipole change parallel to the major axis of rotational symmetry (PARALLEL vibrations) ii. Vibrations causing a dipole change perpendicular to the major axis of rotational symmetry (PERPENDICULAR Vibrations) SELECTION RULES are (i) Different for parallel and perpendicular vibrations and (ii)Depend on the shape of the molecule All LINEAR molecules (those with or without any permanent dipole moment) are found to be IR ACTIVE if some of their vibrations can produce an OSCILLATING DIPOLE PARALLEL BAND The oscillating dipole moment is parallel to the molecular axis The spectra will have the P and R branches with almost equally spaced lines (similar to diatomic molecules) The spacing between lines is less in linear molecules compared to diatomic molecules as the moment of inertia is large (B value less) For a still larger molecule (with large MI and less B value), separate lines can no longer be resolved Individual lines COALESCE resulting in two lines one for P Branch and other for R branch PERPENDICULAR BAND SELECTION RULE: ∆J = 0, ±1, ∆n = ±1 ∆J=0 Transition implies that a vibrational change can take place with no simultaneous rotational change This transition gives rise to a new group of lines called ‘Q’ Branch Fourier transform infrared spectroscopy If the radiant power is recorded as a function of frequency, it is called frequency domain spectroscopy (FDS) If it is recorded as a function of time, it is called as Time Domain Spectroscopy (TDS). In FTIR spectroscopy, a time domain plot is converted into a frequency domain spectrum. Fourier Transform of f(t) is defined by Then the inverse relation is These two equations form a Fourier transform pair Fourier transform infrared spectroscopy Superposition of two sine waves of same amplitude but slightly different frequency The third wave from the top represents the superposed wave And it represents the time domain spectra Fourier transform infrared spectroscopy Several frequencies can be sounded off at once to create a beat frequency. A graph usually displays amplitude vs. time, making it difficult to identify the different frequencies. A Fourier Transform changes this graph into amplitude vs. frequency graph. This allows us to clearly see what frequencies are being sounded off together. Fourier transform infrared spectroscopy Fourier transform infrared spectroscopy The multiplex or Fellgett's advantage. This arises from the fact that information from ADVANTAGES all wavelengths is collected simultaneously. It results in a higher Rapid Scanning Signal-to-noise ratio for a given scan-time or a shorter scan-time for a given S/N ratio resolution. High sensitivity High resolution The throughput or Jacquinot's advantage. This results from the fact that, in a Data processing dispersive instrument, the monochromator has entrance and exit slits which restrict the amount of light that passes through it. The interferometer throughput is determined only by the diameter of the collimated beam coming from the source.. INFRARED SPECTRA OF POLYATOMIC MOLECULES We know that the selection rules for simultaneous rotation and vibration of a DIATOMIC molecule are: = ±1, ±2, ±3 … J= ±1 J 0 . These selection rules give rise to a spectrum consisting of approximately equally spaced line series on each side of central minimum (Band Centre). . The vibrations of complex molecules are subdivided into two categories Viz., those causing a dipole change either (i) Parallel or (ii) Perpendicular to the major axis of rotational symmetry. Selection rules for these two cases are different. Also the selection rule and energies depend on the shape of the molecule. . The rotational transitions of such complex molecules depend on whether the vibrations are parallel or perpendicular. LINEAR MOLECULES: Parallel vibrations: . The selection rules for linear molecules executing parallel vibrations are: J= ±1 = ±1 For Simple Harmonic Motion J= ±1 = ±1, ±2, ±3 … For Anharmonic Motion . This spectrum consists of P and R Branches with equally spaced lines on each side and no line occurring at the band centre. . In case of linear molecules, the Moment of Inertia is considerably large and so the rotational constant B will be small. Therefore P and R line spacing will be less. . For larger molecules the value of B is much smaller and the lines cannot be resolved. Perpendicular vibrations: . The selection rules for linear molecules executing perpendicular vibrations are: J= 0, ±1 = ±1 . J= 0 means that the vibrational changes can take place with no simultaneous rotational transition. . The spectrum obtained from the perpendicular vibrations of linear molecule will give P, Q and R branches. . If the B value of upper and lower vibrational states are equal, all J= 0 transitions give line at Band centre. (Q Branch) 퐹표푟 ∆퐽 = 0, ∆휀 = 휀퐽,푣+1 – 휀퐽,푣 1 1 1 1 = {1 훾̅ − 1 휒 훾̅ + 퐵퐽(퐽 + 1)} − { 훾̅ − 휒 훾̅ + 퐵퐽(퐽 + 1)} 2 푒 4 푒 푒 2 푒 4 푒 푒 = 훾표̅ (FOR EQUAL ‘B’ VALUE) . This Q Branch line is broad if B value slightly differ between lower and upper vibrational states, ∆휀 = 휀퐽,푣+1 – 휀퐽,푣 1 1 1 1 = {1 훾̅ − 1 휒 훾̅ + 퐵′퐽(퐽 + 1)} − { 훾̅ − 휒 훾̅ + 퐵"퐽(퐽 + 1)} 2 푒 4 푒 푒 2 푒 4 푒 푒 ′ = 훾표̅ + 푱(푱 + ퟏ)(푩 − 푩") . Case (i): If B’ < B”, Q branch lines will split into a series of lines on low frequency side since (B’-B”) is negative . Case (ii): If B’ > B”, Q branch lines will split into a series of lines on high frequency side since (B’-B”) is positive . Case (i): If B’ < B”, Q branch lines will split into a series of lines on low frequency side since (B’-B”) is negative . Case (iii): If B’ = B”, Q branch will have a single line at the band centre and no broadening seen. INFRA RED SPECTRA OF SYMMETRIC TOP MOLECULES Vibrational and rotational energy levels of Symmetric Top Molecule are given by: 휀푣,퐽 = 휀푣𝑖푏푟푎푡𝑖표푛푎푙 + 휀푟표푡푎푡𝑖표푛푎푙 1 1 2 = (푣 + ) 훾̅ − (푣 + ) 휒 훾̅ + 퐵퐽(퐽 + 1) + (퐴 − 퐵)퐾2 푐푚−1 2 푒 4 푒 푒 ℎ ℎ Where, 퐴 = 2 푎푛푑 퐵 = 2 8휋 퐼푎푐 8휋 퐼푏푐 Ia and Ib are principal moments of inertial of the symmetric top molecules. K represents the projection of total angular momentum on molecular symmetry axis. In case of symmetric top molecules, the molecular symmetry axis will be the axis about which the top rotates. Vibrations of this type of molecules are divided into two categories: i. Those which change the dipole parallel to the main symmetry axis and ii. Those which change the dipole perpendicular to the main symmetry axis. The rotational selection rules differ for the two types. Parallel Vibrations: The Selection rues are J= 0, ±1 = ±1 K = 0 Since K = 0, terms in K will be identical in the upper and lower states and so the spectral frequencies will be independent of K. The spectrum consists of P, Q and R Branches, with strong Q Branch. The 퐼푎 intensity of Q Branch varies with the ratio . When 퐼푎 ⟶ 0, the symmetric top 퐼푏 molecule becomes linear molecule and the Q branch has Zero intensity. Perpendicular Vibrations: The Selection rues are J= 0, ±1 = ±1 K = ±1 Since K = ±1, terms in K will not be identical in the upper and lower states. These rules will lead to following frequencies for three branches: (i) R BRANCH: J= +1, = ±1 and K = ±1 Δ휀 = 휈푠푝푒푐푡̅ = 훾표̅ + 2퐵 (퐽 + 1) + (퐴 − 퐵)(1 ± 2퐾) (ii) P BRANCH: J= –1 , = ±1 and K = ±1 Δ휀 = 휈푠푝푒푐푡̅ = 훾표̅ − 2퐵 (퐽 + 1) + (퐴 − 퐵)(1 ± 2퐾) (iii) Q BRANCH: J= 0, = ±1 and K = ±1 Δ휀 = 휈푠푝푒푐푡̅ = 훾표̅ + (퐴 − 퐵)(1 ± 2퐾) Since for each J value, K has the allowed values of 0, ±1, ±2, ±3 … ± J, we can expect many sets of P and R branches. The Q–Branch also consists of a series of lines on both sides of band origin 휔̅표, with 2(A-B) separation between lines.