Problem Set 5 Solutions

Transcription

1 Chemistry 362 Dr Jean M Standard Problem Set 5 Solutions ow many vibrational modes do the following molecules or ions possess? [int: Drawing Lewis structures may be useful in some cases] In all of the cases, there will be 3N 6 vibrations for nonlinear molecules (or ions) and 3N 5 vibrations for linear molecules (or ions) a) C 4 C 4 is a nonlinear molecule; therefore, it has 3*5 6 = 9 vibrational modes b) CN CN is a linear molecule from the Lewis structure shown below C N Therefore, CN possesses 3*3 5 = 4 vibrational modes c) N 3 N 3 is a nonlinear molecule; therefore, it has 3*4 6 = 6 vibrational modes d) N 3 N 3 is a nonlinear molecular ion; therefore, it has 3*4 6 = 6 vibrational modes [Note that N 3 is a planar molecule, not a linear molecule]

2 2 Consider the molecule N 2, with connectivity N-N- 2 a) Is the molecule linear or nonlinear? ow many vibrational modes does N 2 possess? N 2 is a linear molecule (you can predict that from the Lewis dot structure, shown below) N N Since N 2 is linear, the number of vibrational modes is 3N 5, where N is the number of atoms Thus, # modes = 3*3 5 = 4 modes b) Describe the vibrational modes of N 2 () N-N stretch (2) N- stretch (3, 4) degenerate N-N- bends Note here that the two stretching vibrations are not labeled as 'symmetric' and 'antisymmetric' This is because for a molecule such as N 2, the two bonds are not identical and therefore no symmetry combinations exist In contrast, the two bonds in molecules such as 2 or S 2 are identical, and so the stretching vibrations occur as symmetric and antisymmetric combinations c) For N 2, determine which of the vibrational modes are IR active () N-N stretch - IR active (2) N- stretch - IR active (3, 4) degenerate N-N- bends - IR active The dipole moment changes in all three of the vibrations; therefore, all are IR active A gas phase IR spectrum is available from NIST, showing the IR absorption peaks for the three IR active modes (at about 59, 29, and 222 cm, for the bends, N- stretch, and N-N stretch, respectively), as well as several overtone and combination bands The spectrum is reproduced below for your information

3 3 Consider the molecule N 2, with connectivity -N- 3 a) Is the molecule linear or nonlinear? ow many vibrational modes does N 2 possess? N 2 is a nonlinear molecule (you again can predict that from the Lewis dot structure for one resonance contributor, shown below) N Since N 2 is nonlinear, the number of vibrational modes is 3N 6, where N is the number of atoms Thus, # modes = 3*3 6 = 3 modes b) Describe the vibrational modes of N 2 () N- symmetric stretch (2) N- antisymmetric stretch (3) -N- bend c) For N 2, determine which of the vibrational modes are IR active () N- symmetric stretch - IR active (2) N- antisymmetric stretch - IR active (3) -N- bend - IR active As for water (which has a similar bent structure), the dipole moment changes in all three of the vibrations; therefore, all are IR active 4 Consider the formaldehyde molecule, 2 C a) ow many vibrational modes does formaldehyde possess? Since formaldehyde is nonlinear, the number of vibrational modes is 3N 6, where N is the number of atoms Thus, # modes = 3*4 6 = 6 modes b) Describe the vibrational modes of formaldehyde (simple sketches may help) () symmetric C- stretch (2) antisymmetric C- stretch (3) C- stretch (4) -C- bend (also referred to as the C 2 scissors mode) (5) C 2 in-plane rock (6) C 2 out-of-plane bend (also referred to as the C 2 wagging mode)

4 4 b) continued 4 The first three modes listed above are similar to those we have seen before Many times sketches of the vibrational modes use arrows to indicate the direction of motion of the atoms Typical sketches of this type are included below for each of the vibrational modes of formaldehyde C C symmetric C- stretch antisymmetric C- stretch C C C- stretch C 2 scissors C C C 2 in-plane rock C 2 out-of-plane wag Note that for the last sketch, corresponding to the C 2 out-of-plane wag, the plus and minus signs indicate that the atom comes out of the plane of the molecule (+) or goes back into the plane ( ) c) For formaldehyde, list the vibrational modes that are IR active All the modes are IR active Each one leads to a change in the dipole moment (either a change in magnitude or a change in direction)

5 5) The molecule hydrogen peroxide, 2 2, is a nonplanar molecule with an --- torsional angle of about 2 degrees 5 a) ow many vibrational modes does 2 2 have? Since 2 2 is nonlinear, the number of vibrational modes is 3N 6, where N is the number of atoms Thus, # modes = 3*4 6 = 6 modes b) Describe the vibrational modes of 2 2 (simple sketches may help) () symmetric - stretch (2) antisymmetric - stretch (3) - stretch (4) symmetric -- bend (5) antisymmetric -- bend (6) --- torsional vibration Sketches of the vibrational modes are shown below symmetric stretch antisymmetric stretch - stretch symmetric bend antisymmetric bend torsion c) List which vibrational modes are IR active () symmetric - stretch - IR active (2) antisymmetric - stretch - IR active (3) - stretch - IR active (There is only a very small change in the dipole moment, so it is okay if you stated that this mode was not IR active) (4) symmetric -- bend - IR active (5) antisymmetric -- bend - IR active (6) --- torsional vibration - IR active

6 6 The ground state wavefunction of the harmonic oscillator is ψ ( x) = N e α x 2 / 2 Determine the normalization constant N 6 The normalization constant N is defined by the equation Substituting the form of the wavefunction, N = & ( '( ψ * ( x) ψ( x) dx ) + * + ψ * ( x)ψ( x) dx = e α x 2 dx From the handout on integrals, e bx 2 dx = 2 & π ) ( + ' b * 2 This is for only half the range Since the function is an even function, e bx 2 dx = & π ) ( + ' b * Then, making the substitution that b = α, the integral becomes ψ * ( x) ψ( x) dx = e α x 2 dx = ( π + * - ) α, Substitution into the normalization equation becomes N = N = & ( '( α 3 / π 2 ψ * ( x) ψ( x) dx / 4 ) + * + = & π ( 3 '( / α 2 ) + * +

7 7 The first excited state wavefunction of the harmonic oscillator is given by ψ x where α = # µk & % ( $ '! 2 ( ) = N 2 α x e α x 2 / 2, Show that this function is an eigenfunction of the harmonic oscillator amiltonian operator and determine the energy eigenvalue ere, N is the normalization constant and µ is the reduced mass 7 In order to show that the function is an eigenfunction, we start by operating the amiltonian operator on it, Evaluating the first derivative, $ ˆ ψ (x) =!2 d 2 2µ dx k x ' % 2 ( & ) N 2 α x e α x = d $ dx % Evaluating the second derivative, d 2 dx 2 2N α!2 2µ x e α x 2 / 2 x e α x 2 / 2 ( ) = d Substituting into the amiltonian equation, ˆ ψ (x) = = = 2N α!2 2µ + -%! 2 α (,' * 3 α x 2 - & 2µ ) d 2 dx 2 2 / 2 x e α x 2 / 2 ( ) + k 2 2N α x 3 e α x 2 / 2 & ' = e α x 2 / 2 α x 2 e α x 2 / 2 = e α x 2 / 2 ( α x 2 ) { ( )} dx e α x 2 / 2 α x 2 = 2α x e α x 2 / 2 α x ( α x 2 )e α x 2 / 2 = α x e α x 2 / 2 ( 3 α x 2 ) α x e α x 2 / { 2 ( 3 α x 2 )} + k 2 2N α x 3 e α x 2 / 2 ( ) + k 2 x 2 + -%! 2 α (,' * 3 α x 2 - & 2µ ) ( ) + k 2 x 2 /- x 2N α x 2 / 2 e α - /- - ψ ( x)

8 7 continued 8 Using the definition of α, ˆ ψ (x) = + - $! 2 α ',& ) 3 α x 2 - % 2µ ( + - $ = 3 &!2 ' $ ) µ k ', & % 2µ (%! 2 ) - ( = ˆ ψ (x) = ! $ k ', - % µ ( = 3 2! $ k ' % µ ( h 2π ( ) + k 2 x 2 $ k ' % µ ( /- - ψ x ( ) $! 2 ' $ µ k ' & % 2µ (%! 2 ) x 2 + k / ( 2 x ψ x k 2 x 2 + k / 2 x ψ x ψ ( x) ψ ( x) Next, the definition of the harmonic frequency can be used, ( ) ( ) ν = 2π $ k ' % µ ( Substituting, or ˆ ψ (x) = 3 2 h 2π $ k ' % µ ( ˆ ψ (x) = 3 2 hν ψ x ( ) ψ ( x), Thus, we see that the function ψ x ˆ Since ψ ( x) = E ψ ( x), the energy eigenvalue must be given by ( ) is an eigenfunction of the Schrödinger equation for the harmonic oscillator E = 3 2 hν

9 8 Recall that two functions f(x) and g(x) are said to be orthogonal if 9 f * ( x) g( x) dx = Show that the ground state (v=) and first excited state (v=) wavefunctions of the harmonic oscillator are orthogonal For the two functions to be orthogonal, we must show that ψ * ( x) ψ ( x) dx = The harmonic oscillator functions for the ground and first excited states are ψ (x) = N e α x 2 / 2 and ψ (x) = 2 N α x e α x 2 / 2 Substituting, ψ * ( x) ψ ( x) dx = 2 N N α x e αx 2 dx From the handout on integrals, x e bx 2 dx = 2b This is for only half the range Since the function is an odd function, Then, the orthogonality integral is ψ * x e bx 2 dx = ( x) ψ ( x) dx = 2 N N α x e α x 2 dx = Therefore, the two functions are orthogonal

10 9 Determine the average value of the position for the ground state of the harmonic oscillator The expectation value of position is defined as x = ψ * ( x) x ˆ ψ ( x ) dx The wavefunction for the ground state of the harmonic oscillator is ψ (x) = N e αx 2 / 2, where the normalization constant N is given by $ N = α ' % π ( Substituting into the expression for the expectation value, / 4 x = = x = ( α + * - ) π, ( α + * - ) π, ψ * ( x) x ˆ ψ( x) dx e α x 2 / 2 x e α x 2 / 2 dx x e α x 2 dx This integral is the same one evaluated in Problem 8, and it was shown to be Therefore, x =

11 Determine the average value of the momentum for the ground state of the harmonic oscillator The expectation value of momentum for normalized wavefunctions is defined as p x = ψ * ( x) p ˆ ψ ( x ) dx x The wavefunction for the ground state of the harmonic oscillator is ψ (x) = N e α x 2 / 2, where the normalization constant N is given by $ N = α ' % π ( Substituting into the expression for the expectation value, p x = ψ * x ( ) ˆ ( = i! α + * - ) π, p x ψ x / 4 ( ) dx e α x 2 / 2 d dx e α x 2 / 2 dx In the last expression above, the definition of the momentum operator, ˆ p x = i! d dx, has been used Now, evaluating the derivative, d dx e α x 2 / 2 = α x e α x 2 / 2 Substituting, the expectation value becomes $ p x = iα! α ' % π ( x e α x 2 dx This integral has already been shown to be Therefore, p x =

12 What are the most probable positions for the ground and first excited states of the harmonic oscillator? 2 To determine the most probable value, the maximum in the probability density must be determined A maximum in a function can be found by setting the derivative equal to zero Ground State The wavefunction for the ground state of the harmonic oscillator is ψ (x) = % α ( ' * & π ) / 4 e α x 2 / 2, and so the probability density is ψ * ( x)ψ ( x) = ψ 2 ( x) = % α ( ' * & π ) e α x 2 Setting the derivative equal to zero, d dx ψ * x ( )ψ x ( ) = % α ( d ' * & π ) dx e α x 2 = % α ( ' * 2α x e α x 2 & π ) ( ) = Dividing both sides by the constant terms, x e α x 2 = Since the exponential term is never zero except if x, the solution is x = Thus, the most probable position for the ground state of the harmonic oscillator occurs at x = First Excited State The wavefunction for the first excited state of the harmonic oscillator is The probability density for this function is ψ (x) = 2N α x e α x 2 / 2 ψ * ( x) ψ ( x) = ψ 2 ( x) = 4 N 2 α x 2 e α x 2

13 continued 3 Setting the derivative equal to zero, d dx { ψ * ( x)ψ ( x) } = 4 N 2 α d { x 2 e α x 2 } dx = ( ) = 4 N 2 α 2x e α x 2 2α x 3 e α x 2 8 N 2 α x ( α x 2 )e α x 2 = Dividing both sides by the constant term 8 N 2 α, the equation becomes x ( α x 2 )e α x 2 = Since the exponential is never zero unless x, the equation can be simplified to This equation has solutions when x ( α x 2 ) = x = or α x 2 = It is easy to see from a plot of the probability density for the first excited state that the solution x = corresponds to a node in the wavefunction and therefore a minimum in the probability distribution The other solution is α x 2 = α x 2 = x 2 = α x = ± α Thus, there are two most probable positions for the first excited state of the harmonic oscillator The most probable positions are x = α and x = α