Physical Chemistry - Problem Drill 15: Vibrational and Rotational Spectroscopy

Transcription

1 Physical Chemistry - Problem Drill 15: Vibrational and Rotational Spectroscopy No. 1 of Internal vibration modes of a molecule containing N atoms is made up of the superposition of 3N-(5 or 6) simple harmonic vibrations called normal modes. How many normal modes of vibration does Carbon Tetrachloride, CCl 4, have? (A) 9 (B) 18 (C) 10 (D) 20 (E) 5 A. Correct! Good job! Since CCl 4 has 5 atoms, and is a nonlinear molecule, N=5 atoms for CCl 4, and N vib =3N-6 = 3 x 5-6 = 9. Good try. Remember to count the atoms in CCl 4, and that this molecule is nonlinear. Good try. Remember to count the atoms in CCl 4, and that this molecule is nonlinear. Good try. Remember to count the atoms in CCl 4, and that this molecule is nonlinear. Good try. Remember to count the atoms in CCl 4, and that this molecule is nonlinear. Correct Answer: A Polyatomic molecules have more possible vibrations. Each fundamental type of vibration of a polyatomic molecule is called a normal mode. A normal mode is a collective motion of all the atoms in the molecule where each atom in the molecule moves in phase with each other at a particular frequency. A nonlinear molecule, like methylene dichloride (CH 2 Cl 2 ), containing N atoms can have 3N-6 fundamental or normal modes of vibration, while a linear molecule like carbon dioxide (CO 2 ) can have 3N-5 normal modes of vibration. 3N-5 for linear molecules (3 translations + 2 rotations) 3N-6 for nonlinear molecules (3 translations + 3 rotations) Since CCl 4 has 5 atoms, and is a nonlinear molecule, N = 5 atoms for CCl 4, and N vib =3N-6 = 3x5-6=9.

2 No. 2 of What is the frequency of light having a wavelength of 6.00 x 10 2 nm? (A) 4.01x10 14 (B) 5.00x10 14 (C) 5.00x10 13 (D) 4.997x10 13 (E) 2.50x10 14 A. Incorrect! Good try. Hint: Apply the formula Frequency = ν=c/λ. B. Correct! Good job! Frequency = ν = c/λ = 2.998x10 8 ms -1 /6.00x10-7 m = 5.00x10 14 s -1. Good try. Good try. Hint: Apply the formula Frequency = ν= c/λ. Good try. Good try. Hint: Apply the formula Frequency = ν = c/λ. Good try. Good try. Hint: Apply the formula Frequency = ν = c/λ. Correct Answer: B Frequency = ν = c/λ = 2.998x10 8 ms -1 /6.00x10-7 m = 5.00x10 14 s -1 ; we have only 3 significant figures in the wavelength data and therefore in the answer.

3 No. 3 of What is the energy of a photon of light having a wavelength of 3.00x10 2 nm? (A) 6.62x10-18 J (B) 1.622x10-29 J (C) 6.62x10-19 J (D) 6.62x10-20 J (E) 3.311x10-19 J A. Incorrect! Good try. Hint: Apply the equation: E photon = hc/λ. Good try. Hint: Apply the equation: E photon = hc/λ. C. Correct! Good job! E photon = hc/λ =6.626x10-34 Js x2.998x10 8 m/s /3.00x10 2 nm=6.62x10-19 J. Good try. Hint: Apply the equation: E photon = hc/λ. Good try. Hint: Apply the equation: E photon = hc/λ. Correct Answer: C E photon = hc/λ = 6.626x10-34 Js x2.998x10 8 m/s /3.00x10 2 nm = 6.62x10-19 J and h = Planck s Constant = 6.626x10-34 Js and c = speed of light = 2.998x10 8 m/s. Remember that we have only 3 significant figures in the given wavelength.

4 No. 4 of For rotational spectra, polyatomic molecules can be treated as rigid rotors to simplify the description of their rotational motions. Such rigid rotors can be classified according to the moments of inertia along the three principal axis (I x, I y, and I z, or also called I a, I b and I c ). Identify the molecules below into one of the three simple types (linear, symmetric and spherical rotors): (1) CH 3 I (2) HCN (3) SF 6. (A) (1) CH 3 I Symmetrical (2) HCN - Symmetrical (3) SF 6 Spherical. (B) (1) CH 3 I Spherical (2) HCN - Linear (3) SF 6 Spherical. (C) (1) CH 3 I Symmetrical (2) HCN - Linear (3) SF 6 Spherical. (D) (1) CH 3 I Symmetrical (2) HCN - Symmetrical (3) SF 6 Symmetrical. (E) (1) CH 3 I Linear (2) HCN - Linear (3) SF 6 Symmetrical. A. Incorrect! Good try. Hint: Use the Lewis Structure model to draw out the geometries and identify the I x, I y and I z. Use the classification definitions to group them. Good try. Hint: Use the Lewis Structure model to draw out the geometries and identify the I x, I y and I z. Use the classification definitions to group them. C. Correct! Good job. Apply the Lewis Structure model to draw out the geometries and identify the I x, I y and I z. Use the classification definitions to group them. Good try. Hint: Use the Lewis Structure model to draw out the geometries and identify the I x, I y and I z. Use the classification definitions to group them. Good try. Hint: Use the Lewis Structure model to draw out the geometries and identify the I x, I y and I z. Use the classification definitions to group them. Correct Answer: C Recall the Lewis Structure method. Draw out the molecular geometries for each molecule and identify the three principal axis for the moments of inertia. Use the rigid rotor s information to classify the molecules. CH 3 I Symmetrical Rotor HCN Linear Rotor SF 6 - Spherical Rotor I a = I b I c I a = I b ; I c = 0 I a = I b = I c Note: I a =I x ; I b =I y ; I c =I z ;

5 No. 5 of Selection rules identify the allowed transitions. A molecule must have a permanent electric dipole moment to have a pure rotational spectrum. Diatomic molecules with a permanent dipole moment can absorb electromagnetic radiation via their rotational motions. Identify the following diatomic molecules with active rotational absorption via their net dipole moments: O 2, CO, NO, N 2, Cl 2 and HCl. (A) O 2, CO, NO, and HCl (B) CO, N 2, Cl 2 and HCl (C) CO, NO, and HCl (D) O 2, CO, NO, N 2, Cl 2 and HCl (E) O 2, N 2, and Cl 2 A. Incorrect! Good try. Hint: Look at the symmetry of each molecule and identify if any net dipole. All homonuclear diatomics have no net dipole (cancelled out). Only heteronuclear diatomic molecules have net dipole moments, hence they are rotational spectrum active. B. Correct! Good job! Look at the symmetry of each molecule and identify if any net dipole. All homonuclear diatomics have no net dipole (cancelled out). Only heteronuclear diatomic molecules have net dipole moments, hence they are rotational spectrum active. Good try. Hint: Look at the symmetry of each molecule and identify if any net dipole. All homonuclear diatomics have no net dipole (cancelled out). Only heteronuclear diatomic molecules have net dipole moments, hence they are rotational spectrum active. Good try. Hint: Look at the symmetry of each molecule and identify if any net dipole. All homonuclear diatomics have no net dipole (cancelled out). Only heteronuclear diatomic molecules have net dipole moments, hence they are rotational spectrum active. Good try. Hint: Look at the symmetry of each molecule and identify if any net dipole. All homonuclear diatomics have no net dipole (cancelled out). Only heteronuclear diatomic molecules have net dipole moments, hence they are rotational spectrum active. Correct Answer: B Draw out the symmetry of each molecule and identify if any net dipole. All homonuclear diatomics have no net dipole (cancelled out). Only heteronuclear diatomic molecules have net dipole moments, hence they are rotational spectrum active. CO, NO and HCl all have net dipole moments due to the differential electronegativities of the different bonding atoms. They will exhibit rotational spectra. All homonuclear diatomic molecules have no rotational spectra.

6 No. 6 of In vibration-rotation spectrum, for infrared, the vibration must cause a change in dipole moment and for Raman, the vibration must change the polarizability. In centro-symmetric molecule, no vibration can be both IR and Raman active. Which molecule pair below is IR active? (A) NH 3 and CH 3 OH (B) O 2 and CH 4 (C) H 2 S and CO 2 (D) CHCl 3 and H 2 (E) HCN and He A. Correct! Good job! Symmetrical linear molecules (homonuclear diatomics) are typically IR inactive. The non-centrosymmetric molecules with polar bonds are usually active since they possess permanent dipole moments. Good try! Hint: Symmetrical linear molecules (homonuclear diatomics) are typically IR inactive. The non-centrosymmetric molecules with polar bonds are usually active since they possess permanent dipole moments. Good try! Hint: Symmetrical linear molecules (homonuclear diatomics) are typically IR inactive. The non-centrosymmetric molecules with polar bonds are usually active since they possess permanent dipole moments. Good try! Hint: Symmetrical linear molecules (homonuclear diatomics) are typically IR inactive. The non-centrosymmetric molecules with polar bonds are usually active since they possess permanent dipole moments. Good try! Hint: Symmetrical linear molecules (homonuclear diatomics) are typically IR inactive. The non-centrosymmetric molecules with polar bonds are usually active since they possess permanent dipole moments. Correct Answer: A In order for a molecule to be IR active, the vibration must produce an oscillating dipole. This usually means that the bond (or bonds) in question are polar so that they have an electric dipole moment. Therefore symmetrical molecules like O 2 and CO 2 are not IR active, but molecules like NH 3 and CH 3 OH with polar bonds are IR active.

7 No. 7 of An electric field applied to a molecule results in its distortion. The distorted molecule acquires a contribution to its dipole moment. The polarizability may be different when the field is applied parallel or perpendicular to the molecular axis. If so, such a molecule has an anisotropic polarizability, which is the selection rule for Raman rotational activity. Use the selection rules to determine if the following molecules have a pure rotational Raman spectrum: HI, HCN, SiH 4, CH 3 Br, SF 6 and O 2. (A) HI, HCN and SF 6 (B) SiH 4, CH 3 Br, SF 6 and O 2 (C) SF 6 and O 2 (D) HI, HCN, SiH 4 and CH 3 Br (E) HI, HCN, CH 3 Br and O 2 A. Incorrect! Good try. Hint: All nonspherical rotors have polarizabilities that depend on the direction of the applied field anisotropical polarizability (Rotational Raman active), while spherical rotors are isotropically polarizable. Determine the rotor type of each molecule and apply the selection rule for its spectral activity. Good try. Hint: All nonspherical rotors have polarizabilities that depend on the direction of the applied field anisotropical polarizability (Rotational Raman active), while spherical rotors are isotropically polarizable. Determine the rotor type of each molecule and apply the selection rule for its spectral activity. Good try. Hint: All nonspherical rotors have polarizabilities that depend on the direction of the applied field anisotropical polarizability (Rotational Raman active), while spherical rotors are isotropically polarizable. Determine the rotor type of each molecule and apply the selection rule for its spectral activity. Good try. Hint: All nonspherical rotors have polarizabilities that depend on the direction of the applied field anisotropical polarizability (Rotational Raman active), while spherical rotors are isotropically polarizable. Determine the rotor type of each molecule and apply the selection rule for its spectral activity. E. Correct! Good job! All nonspherical rotors have polarizabilities that depend on the direction of the applied field anisotropical polarizability (Rotational Raman active), while spherical rotors are isotropically polarizable. Determine the rotor type of each molecule and apply the selection rule for its spectral activity. Correct Answer: E All linear molecules and diatomics (both homonuclear and heteronuclear) have anisotropic polarizabilities (HI, HCN and O 2 ), so as the symmetrical rotors (CH 3 Br). Spherical rotors (SiH 4 and SF 6 ) are isotropically polarizable, therefore they are Raman inactive.

8 No. 8 of What is the degeneracy of rotational energy level J = 3 for Hydrogen Cyanide, HCN? (A) 7 (B) 14 (C) 49 (D) 28 (E) 3 A. Correct! Good job! HCN is a linear molecule and is; therefore, a symmetric rotor molecule. For a symmetric rotor molecule; for J = 3, degeneracy = g(j,k)=(2j+1)=(2x3+1) = 7 Good try. Hint: HCN is a linear molecule and is therefore a symmetric rotor molecule, with degeneracy of 2J+1. Good try. Hint: HCN is a linear molecule and is therefore a symmetric rotor molecule, with degeneracy of 2J+1. Good try. Hint: HCN is a linear molecule and is therefore a symmetric rotor molecule, with degeneracy of 2J+1. Good try. Hint: HCN is a linear molecule and is therefore a symmetric rotor molecule, with degeneracy of 2J+1. Correct Answer: A Degeneracy is the number of states having the same energy level. A rotational energy level in a symmetric rotor molecule has a degeneracy of g(j,k) = 2(2J+1) if K 0. HCN is a linear molecule and is; therefore, a symmetric rotor molecule. For a symmetric rotor molecule; for J = 3, degeneracy = g(j,k)=(2j+1)=(2x3+1) = 7

9 No. 9 of The Doppler is the shift in frequency of a wave for an observer moving relative to its source, such as train whistles, police and fire sirens, race car engines. The received frequency is higher when approaching to the source with a formula ν = ν[1/(1-v/c)], where v is the speed of the speed of the moving object and c is the speed of the light or sound. What is the Doppler-shifted frequency for a 750 Hz siren sound from a police car approaching at the high speed of 80 MPH (Speed of sound is m/s)? (A) 838 Hz (B) 750 Hz (C) 625 Hz (D) 340 Hz (E) 1230 Hz A. Correct! Good job! Apply the Doppler shift frequency equation given and calculate the ν with the same unit in Hz. It will be in higher frequency. Good try. Hint: Apply the Doppler shift frequency equation given and calculate the ν with the same unit in Hz. It should be in higher frequency. Good try. Hint: Apply the Doppler shift frequency equation given and calculate the ν with the same unit in Hz. It should be in higher frequency. Good try. Hint: Apply the Doppler shift frequency equation given and calculate the ν with the same unit in Hz. It should be in higher frequency. Good try. Hint: Apply the Doppler shift frequency equation given and calculate the ν with the same unit in Hz. It should be in higher frequency. Correct Answer: A The Doppler effect was named after Christian Doppler in He thought that the frequency of sound waves would change if either the source or the observer was moving. If they were approaching, the frequency would be higher; if they were diverging, the frequency could be lower. There are many everyday examples of the Doppler effect: train whistles, police and fire sirens, race car engines. When a train is approaching, the whistle has a higher pitch than normal. You can hear the change in pitch as the train passes. After the train has passed, the pitch should be lower than normal. The same is true with sirens on police cars and the engines of race cars. In this case, the v = 80 MPH = m/s and sound wave speed = m/s. ν = ν[1/(1-v/c)] = 750[1/( /340.29)] = Hz Therefore, when the police car is approaching with the siren on, you will hear higher pitch sound (higher frequency).

10 No. 10 of The spectral data can be used to provide structural information of a molecule. The quantum energy states are allowed with F(J) = BJ(J+1). For J=1 rotational transitions, the lines of energy levels can be given as E J = F(J+1) - F(J) = B(J+1)(J+2)-BJ(J+1) = 2B(J+1). The lines in rotational spectrum are therefore expected to occur at 2B(J=0), 4B(J=1), 6B(J=2) etc. with the constant line spacing of 2B. If the rotational line spacing for H 35 Cl is found to be 21.18cm -1. Calculate its bond length. (A) 1.28Å (B) 1.15Å (C) 2.28Å (D) 0.97Å (E) Å A. Correct! Good job! Apply the inertia formula in terms of B (I = ħ/4 CB).and the original definition of inertia (I = μr 2 ). With all the known data, you can calculate the bond length R. Good try. Hint: Use the inertia formula in terms of B (I = ħ/4 CB).and the original definition of inertia (I = μr 2 ). With all the known data, you can calculate the bond length R. π Good try. Hint: Use the inertia formula in terms of B (I = ħ/4 CB).and the original definition of inertia (I = μr 2 ). With all the known data, you can calculate the bond length R. Good try. Hint: Use the inertia formula in terms of B (I = ħ/4 CB).and the original definition of inertia (I = μr 2 ). With all the known data, you can calculate the bond length R. Good try. Hint: Use the inertia formula in terms of B (I = ħ/4 CB).and the original definition of inertia (I = μr 2 ). With all the known data, you can calculate the bond length R. Correct Answer: A Apply two equations below to connect line spacing to bond length. By equating these two, we can computer R with the known B. I = ħ/4 CB and I = [m a m b /(m a +m b )]R 2 The line spacing is cm -1, then B = 21.18/2 = cm -1. I = ħ/4 CB = x10-34 Js/(4 x x10 10 cm*s -1 x10.59cm -1 ) = 2.649x10-47 kg*m 2 μ = m H m Cl /(m H +m Cl ) = 1.008x35/( ) = g/mol = x10-27 kg/molecule I = μr 2 ; R = (I/μ) = x10-10 m = Å Therefore, the bond length of HCl is Å