where h = 6.62 10-34 J s



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Electromagnetic Spectrum: Refer to Figure 12.1 Molecular Spectroscopy: Absorption of electromagnetic radiation: The absorptions and emissions of electromagnetic radiation are related molecular-level phenomena of compounds. The frequency (ν) of the electromagnetic radiation and the speed with which radiation travels (c = 2.99 10 8 ms -1 in vacuum) can be used to calculate the wavelength λ of the radiation. Thus, the frequency is equal to the number of waves in the radiations of an approaching beam that can arrive at receiver in 1s, and this number is equal to c/ λ. So, we can write or ν = c / λ λ = c / ν The electromagnetic radiation is also treated as stream of packets, called photons. These energies are related to frequency (ν) through Planck relation. ε = h ν where h = 6.62 10-34 J s

Problem 1: What are the frequency and wavelength of electromagnetic radiation for which each quantum has energy of 10-21 J? The energy and frequency are related to each other by Planck s relation ε = h ν where h = 6.62 10-34 J s ν = ε / h = 10-21 J / 6.62 10-34 J s = 1.5 10 12 s -1 The frequency and wavelength are related as: λ = c / ν where c = 2.99 10 8 ms -1 Beer s Law: = 2.99 10 8 ms -1 / 1.5 10 12 s -1 = 2.0 10-4 m When electromagnetic radiation is passed through a sample, some of the radiation is absorbed by the sample. The radiation absorbed by the sample is at particular frequency or of particular frequency range. We will use C for the concentration of the absorbing sample in units of moles per cubic meter.

The common concentration unit is M, molarity i.e. number of moles of solute per liter of solution. Then the concentration in SI units can be obtained by relation: C = M 1000 L m -3 The decrease in the intensity of an incident light beam as it passes through a sample is proportional 1. The path length 2. The concentration of the absorbing component of the solution and 3. Intensity of the beam. Thus, with the proportionality constant k, known as molar absorption coefficient, Refer to Figure 12.2 We have di = - k C I dx di / I = - k C dx Integration over a length l of the sample, di / I = d (ln I) and limits between 0 to l We get I I l di /I = d (ln I) = - k C dx

I 0 I 0 0 ln I / I 0 = - k C l or I = I 0 e -k C l -------- (1) This is known as Beer s law or Beer-Lambert Law In practical analytical use, the equation (1) is converted to base-10 logarithms, concentration is expressed in moles per liter, and length is measured in centimeters. Also, the intensities of outgoing to the incoming radiation i.e. I / I 0 at any wavelength can be directly measured and is called as transmittance and the base-10 logarithmic term, log (I / I 0 ) is called as absorbance. Now, writing ε, molar extinction coefficient for proportionality constant We get, log (I / I 0 ) = ε M l or I = I 0 10 -ε M l -------- (2) The equation (1) and (2) gives us the relation between k and ε as: k = 2.303 10-1 ε For all theoretical purposes, we will use equation (1) and molar absorption coefficient k.

Thus, the extent to which radiations of particular frequencies or wavelengths is absorbed by a sample gives us absorption spectrum. In infrared spectral region, generally amount of absorption is reported as percent transmission i.e. (I/ I 0 ) 100 While in the visible and ultraviolet regions, the absorbance, log (I/ I 0 ) is used. Problem 2: In the infrared absorption spectra of ketones, aldehydes, carboxylic acids, and esters, the carbonyl absorption due to C=O group is prominent feature. For 2-butanone, or methyl ethyl ketone dissolved in carbon tetrachloride this absorption occurs at 5.8µm, or 1724 cm -1. A 0.089 M solution in an absorption cell of thickness, or length, 0.100 mm gives the absorption band shown in following figure. Calculate the value of molar absorption coefficient at the band maximum. Refer to Figure 12.3 From percent transmission values at the center of the absorption band and that estimated for background we obtain I = 49 I 0 = 98 ln (I/ I 0 ) = ln (0.5) = - 0.693 Also C = M 1000 L m -3 = 0.089 mol L -1 1000 L m -3 = 89 mol m -3

l = 0.100 mm = 1.00 10-4 m ln I / I 0 = - k C l -0.693 = - k (89 mol m -3 ) (1.00 10-4 m) k = (0.693) / (89 mol m -3 ) (1.00 10-4 m) = 77.5 mol -1 m 2 Rotational Spectra of Linear Molecules: The absorption of microwave radiation increases the rotational energy of the molecules and gives information about the moment of inertia of the molecules. A rotational spectrum is obtained when a sample of gas absorbs far-infrared or microwave radiation and due to this rotational-energy is changed. Rotational energies: We saw that the angular momentum and rotational energy can be given as follows: Angular momentum = J (J +1) ( h / 2π) Where J = 0,1,2,3 Rotational Energy = J (J + 1) ( h 2 / 8π 2 I ) Where J = 0,1,2,3

In addition, the value of J is quantized and gives us the total rotational states for a particular value of J as 2J +1. i.e. the energy levels specified by the value of J have a degeneracy of 2J + 1 Now, in order to express the rotational states of linear molecules, we use a rotational constant B given as B = ( h 2 / 8π 2 I ) Therefore, we have a rotational energy term as ε rot = J (J+1) ( h 2 / 8π 2 I ) = J (J + 1) B = BJ (J + 1) where B = ( h 2 / 8π 2 I ) and J = 0,1,2,3. Refer to Figure 12.4 Spectroscopic transitions: Most transitions between rotational states of gas phase molecules occur in the microwave region of the electromagnetic spectrum. The microwave radiation is generated in a radiolike tube known as klystron. Microwave beam can be contained in a metal tube called a wave guide into which the gas whose rotational spectrum is to be studied is placed.

The klystron can be tuned to produce various frequencies, and when the frequency that is absorbed by gas molecules, the electrical properties of circuits change. These absorbed frequencies can be measured accurately, thus giving a rotational spectrum of a gas sample. This is also known as farinfrared absorption spectrum. A rotating molecule interacts with electromagnetic radiation through its electric dipole. Thus, in order to get a rotational spectra: 1. The molecule must have a permanent dipole. Homonuclear molecules like N 2 and O 2 and symmetric molecules like CO 2 and CH 4 do not produce rotational spectra. 2. The transition between the adjacent states; i.e. change in the rotational quantum number J must obey a selection rule J = ± 1. Rotational Spectra: The rotational spectra or far-infrared or microwave absorption spectra of gas containing linear polar molecules show regular patterns of nearly equispaced absorption lines. Refer to Figure 12.5 Since, the rotational energies are closely spaced compared to kt, the molecules are distributed through out the lower allowed levels. Therefore, the transition can occur between many levels. The energy differences corresponds to the energies of the quanta of radiation that bring about J = + 1 transition. Therefore, the energy change due to absorption of radiation, due to J = +1 is given as: ε rot = B(J+1) (J+2) - BJ (J + 1) = B (J +1) [ (J + 2) J]

= 2 B (J + 1) where J = 0,1,2,3. Refer to Figure 12.6 The frequency related quantity called wave number with a symbol ν is defined as 1 / λ and therefore proportional to frequency, ν = c / λ is often expressed in terms of constant. The wave-number value is reported in units of cm -1. In this, case a rotational constant expressed in wave numbers with units of cm -1 is represented by B. The value of B with units of cm -1 is related to B with units of joules as: B = h c B Where h = 6.626 10-34 J s and c = 2.99 10 10 cm s -1 Thus, using the overbar to indicate the wave-number quantities, we have ε rot = ν rot = 2 B (J + 1) where J = 0,1,2,3. Thus, we expect a pattern of lines corresponding to the wave-number values 2B, 4B, 6B. and the adjacent spectral lines are spaced by a constant amount that can be identified as 2B. Using the value of B, the spacing between rotational levels can be used to find the value of B using expression B = h c B and this can be used to calculate the moment of inertia. Also, we can use the expression I = µr 2, where µ is the reduced mass, to obtain the internuclear distance, or bond length of the molecules.

Problem 3: The average spacing between the successive rotational absorption lines of diatomic molecules CO(g) is 3.8626 cm -1. Calculate the moment of inertia and the length of CO bond from this spectral result. Since the spacing between the rotational levels = 2B = 3.8626 cm -1 Therefore, B = 1.9313 cm -1 And B = h cb = (6.626 10-34 J s ) (2.99 10 10 cm s -1 ) (1.9313 cm -1 ) = 3.836 10-23 J Now, we have B = ( h 2 / 8π 2 I ) I = ( h 2 / 8π 2 B ) = (6.626 10-34 J s) 2 / [ (8) (3.142) 2 (3.836 10-23 J )] = 14.50 10-47 J s 2 = 14.50 10-47 kg m 2 s -2 s 2

= 14.50 10-47 kg m 2 The reduced mass of C-12 and O-16 of CO molecules can be obtained from the masses in kilograms and Avogadro s number as µ = (m 1 m 2 ) / (m 1 + m 2 ) N A = (0.01200) (0.01600) / (0.02800) (6.022 10 23 ) = 1.139 10-26 kg and I = µ r 2 r 2 = I / µ = 14.50 10-47 kg m 2 / 1.139 10-26 kg = 127.3 10-22 m 2 r = 112.8 10-12 m = 112.8 pm