ChemWiki - User contributions [en]

MRD:ny517

2019-05-24T11:22:04Z

Ny517:

== Molecular Reaction Dynamics ==
=== Exercise 1: H + H2 ===

''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''

The transition state is defined as the maximum energy point on the minimum energy path and the point on the potential energy surface when the gradient (partial first derivative) of the potential is equal to zero (∂V(ri)/∂ri=0). The transition state can be identified as a saddle point which means it has a partial second derivative greater than and less than zero (dependant on the variables and direction of axis). A local minimum is the point where it is solely a minimum and so the second derivative is always a positive value.

''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The best estimate of the transition state to 3 decimal places was:
rts=r1=r2= 0.908 Å

This was estimated through looking at the contour plot of the initial conditions set for a H + H2 reaction and following the reactive trajectory to the point where r1 was approximate to r2 - which was in the region of 0.9 Å (as seen on the contour plot below). This is because for a system with identical atoms, the equidistant point is equal to the transition state position. After trialling values in this region, 0.908 Å produced a trajectory with a constant internuclear distance between both atoms (as seen on the plot below) indicating that the trajectory will not fall off the ridge. This is a transition state since the atoms are stationary over time when they have no initial momenta.

[[File:Ny517_q2.JPG|300px|thumb|left|Contour plot from initial conditions]] [[File:Ny517_q2_Internuclear_distance.JPG|600px|thumb|center|Internuclear Distance vs Time plot where rts = 0.908 Å]]

''Comment on how the mep and the trajectory you just calculated differ.''

The mep produces a trajectory that follows the valley floor and is also known as the reaction path. It shows a straight-line path along this trajectory because the velocity is always reset to zero after each time step. This does not provide a realistic account of the motion of atoms during the reaction since it assumes the atoms have no mass. The “Dynamics” calculation provides a more realistic trajectory since it accounts for the mass of the system and so the motion of the atoms in the gas phase will be inertial. In a MEP, kinetic energy is assumed to be 0 which explains why there are no oscillations along the path. As the potential energy decreases towards its minimum from the transition state, there is no kinetic energy producing a momentum and causing the molecule to travel up the potential energy curve at small bond distances so it remains at the potential energy minimum resulting in no oscillations.

[[File:Ny517_q3_mep.JPG|500px|thumb|left|mep surface plot]] [[File:Ny517_q3_dynamics.JPG|500px|thumb|center|Dynamics surface plot]]

If the initial conditions were reversed, the A-B and B-C plots would switch for the “Distances vs Time” and “Momenta vs Time” graphs. This is because the magnitude of the forces in the reaction do not change but the direction does. As the time increases, the A-B distance increases whilst the B-C bond distance stays constant (bar slight oscillations) at 0.75 because atom A is travelling further away from the BC diatomic molecule. At 5 seconds the A-B bond distance is 18 Å. As the time increases, the A-B momentum plateaus at 2.5 kg ms-1 whilst the B-C momentum oscillates about an average of 1.25 kg ms-1. If the values were reversed, the B-C bond distance would increase whilst the A-B bond distance would remain constant.

[[File:Ny517_q3_larget_bond.JPG|500px|thumb|left|Distance vs Time]] [[File:Ny517_q3_larget_momentum.JPG|500px|thumb|center|Momenta vs Time]]

''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''

Initial positions '''r1''' = 0.74 Å and '''r2''' = 2.0 Å

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0 ||Yes || Reactants travel through transition state then start to vibrate ||Plot 1
|-
| -1.5 || -2.0 ||-100.4 ||No || Reactants do not overcome activation energy and rebound off the transition state. Atoms vibrate on approach and after reaching transition state ||Plot 2
|-
| -1.5 || -2.5 ||-98.9 ||Yes || Reactants vibrate whilst travelling through transition state then vibrate more vigorously ||Plot 3
|-
| -2.5 || -5.0 ||-85.0 ||No || Reactants pass through transition state then pass back through the transition state to reform the reactants ||Plot 4
|-
| -2.5 || -5.2 ||-83.4 ||Yes || Reactants travel through transition state, then reform reactants and pass through the transition state again. This occurs with strong vibrations throughout reaction ||Plot 5
|}

[[File:Ny517_plot1.JPG|300px]]

[[File:Ny517_plot2.JPG|300px]]

[[File:Ny517_plot3.JPG|300px]]

[[File:Ny517_plot4.JPG|300px]]

[[File:Ny517_plot5.JPG|300px]]

''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

Transition state theory (TST) is used to qualitatively understand the reaction rates of elementary chemical reactions with the idea that all reactions occur through a transition state. One assumption is that once the reagents gain an energy sufficient enough to overcome the activation energy, they will react to form the product. In reality this is not the case as is seen in the calculated examples above where some reactants reached the transition state but reformed the reagents and not the products. This is because of another assumption of TST which is that the energy is looked at classically rather than quantum mechanically so fails to account for the quantization of molecular vibrations. Failing to factor this is why TST predictions for reaction rates will typically be higher than experimental values.

=== EXERCISE 2: F - H - H system ===

''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic).''

 F + H2 

[[File:Ny517_h2_plus_f.JPG|300px|]]

H-H distance: 0.74 Å

F-H distance: 2.00 Å

H-H momentum: -1.5 kg ms-1

F-H momentum: -2.5 kg ms-1

The reactants are higher in energy than the products therefore the reaction is exothermic.

 H + HF 

[[File:Ny517_f_h_plus_h.JPG|300px|]]

F-H distance: 0.74 Å

H-H distance: 2.00 Å

F-H momentum: -1.5 kg ms-1

H-H momentum: -2.5 kg ms-1

The reactants are lower in energy than the products therefore the reaction is endothermic.

''How does this relate to the bond strength of the chemical species involved?''

Through inspection of the potential energy surfaces, it is clear that the bond strength of H-F is greater than the bond strength of H-H since more energy is needed to break the H-F bond in the second reaction.

''Locate the approximate position of the transition state.''

[[File:Ny517_hf_transition_state_distance.JPG|400px|thumb|left|Internuclear Distance vs Time]][[File:Ny517_hf_transition_state_contour.JPG|400px|thumb|center|Contour plot]]

H-H distance: 0.745 Å

H-F distance: 1.815 Å

To find the transition state, both momenta were set to equal zero. Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. The transition state will be closer in energy to the reactants of the exothermic (first) reaction and the products of the endothermic (second) reaction. The contour plot shows the atoms do not fall off the ridge and the Internuclear Distances plot shows the distances remained constant (bar slight oscillations).

''Report the activation energy for both reactions.''

The activation energy is defined as the difference in energy between the reactants and transition state. The energy of the transition state is known, which is -103.752 kcal mol-1. 0.3 A was added to and subtracted from the HF bond length to calculate the minimum energy difference between the reactants and transition state for both reactions. The reactant energies were extracted using an mep calculation. The calculated activation energies are stated below:

[[File:Ny517_activation_exo.JPG|300px|]] [[File:Ny517_activation_endo.JPG|300px|]]

*For the F + H2 system: 0.236 kcal mol-1
*For the H + HF system: 30.256 kcal mol-1

=== Reaction Dynamics ===

''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ''

For a F + H2 reaction trajectory:

H-H distance: 0.744832 Å

H-F distance: 2.00 Å

H-H momentum: -0.7 kg ms-1

H-F momentum: -1.6 kg ms-1

[[File:Ny517_reaction_dynamics_momenta.JPG|400px|thumb|left|Momenta vs Time]][[File:Ny517_reaction_dynamics_energy.JPG|400px|thumb|center|Energy vs Time]]

It is seen that the B-C momentum plateaus over time due to the H2 bond being split and the H atoms becoming further apart so the bond vibrations begin to stop. The A-B momentum oscillates as the B-C momentum stops oscillating which is due to the H-F atom being formed. The H-F bond vibrations are larger than the H-H bond vibrations due to the strength of the bond being greater and hence a have a higher enthalpy of formation. If the H-F bond is stronger than the H-H bond then this will result in an exothermic reaction. This could be confirmed experimentally through a reaction calorimeter. The plot displaying energy over time indicates that the total energy is constant and hence conserved. It also shows that the the potential and kinetic energy convert between each other as the atoms vibrate between their bonds. Notice that the energy oscillations become larger after the transition state- similar to that seen in the momentum plot.

Changing momenta:

*Keeping the same initial conditions but changing the H-H momentum:

[[File:Ny517_momentum_-3.JPG|500px|thumb|left|pHH=-3 kg ms-1]]
[[File:Ny517_momentum_-2.JPG|500px|thumb|center|pHH=-2 kg ms-1]]
[[File:Ny517_momentum_-1.JPG|500px|thumb|left|pHH=-1 kg ms-1]]
[[File:Ny517_momentum_0.JPG|500px|thumb|center|pHH=0 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=1 kg ms-1]]
[[File:Ny517_momentum_2.JPG|500px|thumb|center|pHH=2 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=3 kg ms-1]]

Only when pHH was equal to -1 and -2 kg ms-1 was there a reaction trajectory. When the time step was doubled for pHH=0 a reaction trajectory was produced indicating that the H2 bond was broken and a HF bond was formed. For the other reactions, the reactants passed the transition state but went back and reformed the reactants.

When pFH = -0.8 kg ms-1 and pHH = 0.1 kg ms-1, a reaction trajectory occurs:

[[File:Ny517_momentum_change.JPG||300px|]]

For the reverse reaction, H + HF:

The initial conditions are set with :

H-H distance: 2.00 Å

H-F distance: 0.74 Å

H-H momentum: -0.59 kg ms-1

H-F momentum: -6.55 kg ms-1

[[File:Ny517_reverse_1.JPG||300px|]]

Keeping the atom positions constant, the momenta were varied to have a low vibrational motion on on the H-F bond and an arbitrarily high value of pHH above the activation energy and obtain a reactive trajectory:

[[File:Ny517_reverse_2.JPG||300px|]]

H-H distance: 2.00 Å

F-H distance: 0.74 Å

H-H momentum: -7.55 kg ms-1

F-H momentum: -0.35 kg ms-1

''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.''

The efficiency of a reaction is dependant on the position of the transition state. If there is a late transition state, the vibrational energy is the most influential on the efficiency of the reaction and hence its ability to overcome the reaction energy. If there is an early transition state, the translational energy is the most influencial on the efficiency. Using this, it can be deduced that for the endothermic H + HF system, the reaction is affected more by the vibrational energy and that for the exothermic F + H2 system, the reaction is affected more by the translational energy.

MRD:ny517

2019-05-23T19:04:53Z

Ny517: /* Exercise 1: H + H2 */

== Molecular Reaction Dynamics ==
=== Exercise 1: H + H2 ===

''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''

The transition state is defined as the maximum energy point on the minimum energy path and the point on the potential energy surface when the gradient (partial first derivative) of the potential is equal to zero (∂V(ri)/∂ri=0). The transition state can be identified as a saddle point which means it has a partial second derivative greater than and less than zero (dependant on the variables and direction of axis). A local minimum is the point where it is solely a minimum and so the second derivative is always a positive value. The figure below shows a surface plot of a reaction between molecule A-B and atom C where the transition state is highlighted.

''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The best estimate of the transition state to 3 decimal places was:
rts=r1=r2= 0.908 Å

This was estimated through looking at the contour plot of the initial conditions set for a H + H2 reaction and following the reactive trajectory to the point where r1 was approximate to r2 - which was in the region of 0.9 Å (as seen on the surface plot below). This is because for a system with identical atoms, the equidistant point is equal to the transition state position. After trialling values in this region, 0.908 Å produced a trajectory with a constant internuclear distance between both atoms indicating that the trajectory will not fall off the ridge. This is a transition state since the atoms are stationary over time when they have no initial momenta.

[[File:Ny517_q2.JPG|300px|thumb|left|Surface plot from initial conditions]] [[File:Ny517_q2_Internuclear_distance.JPG|600px|thumb|center|Internuclear Distance vs Time plot where rts = 0.908 Å]]

''Comment on how the mep and the trajectory you just calculated differ.''

The mep produces a trajectory that follows the valley floor and is also known as the reaction path. It shows a straight-line path along this trajectory because the velocity is always reset to zero after each time step. This does not provide a realistic account of the motion of atoms during the reaction since it assumes the atoms have no mass. The “Dynamics” calculation provides a more realistic trajectory since it accounts for the mass of the system and so the motion of the atoms in the gas phase will be inertial. In a MEP, kinetic energy is assumed to be 0 which explains why there are no oscillations along the path. As the potential energy decreases towards its minimum from the transition state, there is no kinetic energy producing a momentum and causing the molecule to travel up the potential energy curve at small bond distances so it remains at the potential energy minimum resulting in no oscillations.

[[File:Ny517_q3_mep.JPG|500px|thumb|left|mep surface plot]] [[File:Ny517_q3_dynamics.JPG|500px|thumb|center|Dynamics surface plot]]

If the initial conditions were reversed, the A-B and B-C plots would switch for the “Distances vs Time” and “Momenta vs Time” graphs. This is because the magnitude of the forces in the reaction do not change but the direction does. As the time increases, the A-B distance increases whilst the B-C bond distance stays constant (bar slight oscillations) at 0.75 because atom A is travelling further away from the BC diatomic molecule. At 5 seconds the A-B bond distance is 18 Å. As the time increases, the A-B momentum plateaus at 2.5 kg ms-1 whilst the B-C momentum oscillates about an average of 1.25 kg ms-1. If the values were reversed, the B-C bond distance would increase whilst the A-B bond distance would remain constant.

[[File:Ny517_q3_larget_bond.JPG|500px|thumb|left|Distance vs Time]] [[File:Ny517_q3_larget_momentum.JPG|500px|thumb|center|Momenta vs Time]]

''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''

Initial positions '''r1''' = 0.74 Å and '''r2''' = 2.0 Å

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0 ||Yes || Reactants travel through transition state then start to vibrate ||Plot 1
|-
| -1.5 || -2.0 ||-100.4 ||No || Reactants do not overcome activation energy and rebound off the transition state. Atoms vibrate on approach and after reaching transition state ||Plot 2
|-
| -1.5 || -2.5 ||-98.9 ||Yes || Reactants vibrate whilst travelling through transition state then vibrate vigorously ||Plot 3
|-
| -2.5 || -5.0 ||-85.0 ||No || Reactants pass through transition state then vibrate vigorously leading to the reactants being reformed ||Plot 4
|-
| -2.5 || -5.2 ||-83.4 ||Yes || Reactants travel through transition state, then reform reactants and pass through the transition state again. The atoms vibrating strongly leads to constantly passing through the transition state ||Plot 5
|}

[[File:Ny517_plot1.JPG|300px]]

[[File:Ny517_plot2.JPG|300px]]

[[File:Ny517_plot3.JPG|300px]]

[[File:Ny517_plot4.JPG|300px]]

[[File:Ny517_plot5.JPG|300px]]

''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

Transition state theory (TST) is used to qualitatively understand the reaction rates of elementary chemical reactions by through the belief that all reactions occur through a transition state. One assumption is that once the reagents gain an energy sufficient enough to overcome the activation energy, they will react to form the product. In reality this is not the case as is seen in the calculated examples above where some reactants reached the transition state and reformed the reagents but not the products. This is because of another assumption of TST which is that the energy is looked at classically rather than quantum mechanically so fails to account for the quantization of molecular vibrations. Failing to factor this is why TST predictions for reaction rates will typically be higher than experimental values.

=== EXERCISE 2: F - H - H system ===

''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic).''

 F + H2 

[[File:Ny517_h2_plus_f.JPG|300px|]]

H-H distance: 0.74 Å

F-H distance: 2.00 Å

H-H momentum: -1.5 kg ms-1

F-H momentum: -2.5 kg ms-1

The reactants are higher in energy than the products therefore the reaction is exothermic.

 H + HF 

[[File:Ny517_f_h_plus_h.JPG|300px|]]

F-H distance: 0.74 Å

H-H distance: 2.00 Å

F-H momentum: -1.5 kg ms-1

H-H momentum: -2.5 kg ms-1

The reactants are lower in energy than the products therefore the reaction is endothermic.

''How does this relate to the bond strength of the chemical species involved?''

Through inspection of the potential energy surfaces, it is clear that the bond strength of H-F is greater than the bond strength of H-H since more energy is needed to break the H-F bond in the second reaction.

''Locate the approximate position of the transition state.''

[[File:Ny517_hf_transition_state_distance.JPG|400px|thumb|left|Internuclear Distance vs Time]][[File:Ny517_hf_transition_state_contour.JPG|400px|thumb|center|Contour plot]]

H-H distance: 0.745 Å

H-F distance: 1.815 Å

To find the transition state, both momenta were set to equal zero. Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. The transition state will be closer in energy to the reactants of the exothermic (first) reaction and the products of the endothermic (second) reaction. The contour plot shows the atoms do not fall off the ridge and the Internuclear Distances plot shows the distances remained constant (bar slight oscillations).

''Report the activation energy for both reactions.''

The activation energy is defined as the difference in energy between the reactants and transition state. The energy of the transition state is known which is -103.752 kcal mol-1. 0.3 A was added and subtracted to the HF bond length to calculate the minimum energy difference between the reactants and transition state for both reactions. The reactant energies were extracted using an mep calculation. The calculated activation energies are stated below:

[[File:Ny517_activation_exo.JPG|300px|]] [[File:Ny517_activation_endo.JPG|300px|]]

*For the F + H2 system: 0.236 kcal mol-1
*For the H + HF system: 30.256 kcal mol-1

=== Reaction Dynamics ===

''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ''

For a F + H2 reaction trajectory:

H-H distance: 0.744832 Å

H-F distance: 2.00 Å

H-H momentum: -0.7 kg ms-1

H-F momentum: -1.6 kg ms-1

[[File:Ny517_reaction_dynamics_momenta.JPG|400px|thumb|left|Momenta vs Time]][[File:Ny517_reaction_dynamics_energy.JPG|400px|thumb|center|Energy vs Time]]

It is seen that the B-C momentum plateaus over time due to the H2 bond being split and the H atoms becoming further apart so the bond vibrations begin to stop. The A-B momentum oscillates as the B-C momentum stops oscillating which is due to the H-F atom being formed. The H-F bond vibrations are larger than the H-H bond vibrations due to the strength of the bond being greater and hence a have a higher enthalpy of formation. If the H-F bond is stronger than the H-H bond then this will result in an exothermic reaction. This could be confirmed experimentally through a reaction calorimeter. The plot displaying energy over time indicates that the total energy is constant and hence conserved. It also shows that the the potential and kinetic energy convert between each other as the atoms vibrate between their bonds. Notice that the energy oscillations become larger after the transition state- similar to that seen in the momentum plot.

Changing momenta:

*Keeping the same initial conditions but changing the H-H momentum:

[[File:Ny517_momentum_-3.JPG|500px|thumb|left|pHH=-3 kg ms-1]]
[[File:Ny517_momentum_-2.JPG|500px|thumb|center|pHH=-2 kg ms-1]]
[[File:Ny517_momentum_-1.JPG|500px|thumb|left|pHH=-1 kg ms-1]]
[[File:Ny517_momentum_0.JPG|500px|thumb|center|pHH=0 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=1 kg ms-1]]
[[File:Ny517_momentum_2.JPG|500px|thumb|center|pHH=2 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=3 kg ms-1]]

Only when pHH was equal to -1 and -2 kg ms-1 was there a reaction trajectory. When the time step was doubled for pHH=0 a reaction trajectory was produced indicating that the H2 bond was broken and a HF bond was formed. For the other reactions, the reactants passed the transition state but went back and reformed the reactants.

When pFH = -0.8 kg ms-1 and pHH = 0.1 kg ms-1, a reaction trajectory occurs:

[[File:Ny517_momentum_change.JPG||300px|]]

For the reverse reaction, H + HF:

The initial conditions are set with :

H-H distance: 2.00

H-F distance: 0.74

H-H momentum: -0.59 kg ms-1

H-F momentum: -6.55 kg ms-1

[[File:Ny517_reverse_1.JPG||300px|]]

Keeping the atom positions constant, the momenta were varied to have a low vibrational motion on on the H-F bond and an arbitrarily high value of pHH above the activation energy and obtain a reactive trajectory:

[[File:Ny517_reverse_2.JPG||300px|]]

H-H distance: 2.00 Å

F-H distance: 0.74 Å

H-H momentum: -7.55 kg ms-1

F-H momentum: -0.35 kg ms-1

''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.''

The efficiency of a reaction is dependant on the position of the transition state. If there is a late transition state, the vibrational energy is the most influential on the efficiency of the reaction and hence its ability to overcome the reaction energy. If there is an early transition state, the translational energy is the most influencial on the efficiency. Using this, it can be deduced that for the endothermic H + HF system, the reaction is affected more by the vibrational energy and that for the exothermic F + H2 system, the reaction is affected more by the translational energy.

MRD:ny517

2019-05-23T18:55:32Z

2019-05-23T18:49:01Z

Ny517: /* Exercise 1: H + H2 */

== Molecular Reaction Dynamics ==
=== Exercise 1: H + H2 ===

''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''

The transition state is defined as the maximum energy point on the minimum energy path and the point on the potential energy surface when the gradient (partial first derivative) of the potential is equal to zero (∂V(ri)/∂ri=0). The transition state can be identified as a saddle point which means it has a partial second derivative greater than and less than zero (dependant on the variables and direction of axis). A local minimum is the point where it is solely a minimum and so the second derivative is always a positive value. The figure below shows a surface plot of a reaction between molecule A-B and atom C where the transition state is highlighted.

''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The best estimate of the transition state to 3 decimal places was:
rts=r1=r2= 0.908 Å

This was estimated through looking at the contour plot of the initial conditions set for a H + H2 reaction and following the reactive trajectory to the point where r1 was approximate to r2 - which was in the region of 0.9 Å (as seen on the surface plot below). This is because for a system with identical atoms, the equidistant point is equal to the transition state position. After trialling values in this region, 0.908 Å produced a trajectory with a constant internuclear distance between both atoms indicating that the trajectory will not fall off the ridge. This is a transition state since the atoms are stationary over time when they have no initial momenta.

[[File:Ny517_q2.JPG|300px|thumb|left|Surface plot from initial conditions]] [[File:Ny517_q2_Internuclear_distance.JPG|600px|thumb|center|Internuclear Distance vs Time plot where rts = 0.908 Å]]

''Comment on how the mep and the trajectory you just calculated differ.''

The mep produces a trajectory that follows the valley floor and is also known as the reaction path. It shows a straight-line path along this trajectory because the velocity is always reset to zero after each time step. This does not provide a realistic account of the motion of atoms during the reaction since it assumes the atoms have no mass. The “Dynamics” calculation provides a more realistic trajectory since it accounts for the mass of the system and so the motion of the atoms in the gas phase will be inertial.

[[File:Ny517_q3_mep.JPG|500px|thumb|left|mep surface plot]] [[File:Ny517_q3_dynamics.JPG|500px|thumb|center|Dynamics surface plot]]

'''Why is mep used?'''

If the initial conditions were reversed, the A-B and B-C plots would switch for the “Distances vs Time” and “Momenta vs Time” graphs. This is because the magnitude of the forces in the reaction do not change but the direction does. As the time increases, the A-B distance increases whilst the B-C bond distance stays constant (bar slight oscillations) at 0.75 because atom A is travelling further away from the BC diatomic molecule. At 5 seconds the A-B bond distance is 18 Å. As the time increases, the A-B momentum plateaus at 2.5 kg ms-1 whilst the B-C momentum oscillates about an average of 1.25 kg ms-1. If the values were reversed, the B-C bond distance would increase whilst the A-B bond distance would remain constant.

[[File:Ny517_q3_larget_bond.JPG|500px|thumb|left|Distance vs Time]] [[File:Ny517_q3_larget_momentum.JPG|500px|thumb|center|Momenta vs Time]]

''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''

Initial positions '''r1''' = 0.74 Å and '''r2''' = 2.0 Å

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0 ||Yes || Reactants travel through transition state then start to vibrate ||Plot 1
|-
| -1.5 || -2.0 ||-100.4 ||No || Reactants do not overcome activation energy and rebound off the transition state. Atoms vibrate on approach and after reaching transition state ||Plot 2
|-
| -1.5 || -2.5 ||-98.9 ||Yes || Reactants vibrate whilst travelling through transition state then vibrate vigorously ||Plot 3
|-
| -2.5 || -5.0 ||-85.0 ||No || Reactants pass through transition state then vibrate vigorously leading to the reactants being reformed ||Plot 4
|-
| -2.5 || -5.2 ||-83.4 ||Yes || Reactants travel through transition state, then reform reactants and pass through the transition state again. The atoms vibrating strongly leads to constantly passing through the transition state ||Plot 5
|}

''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

Transition state theory (TST) is used to qualitatively understand the reaction rates of elementary chemical reactions by through the belief that all reactions occur through a transition state. One assumption is that once the reagents gain an energy sufficient enough to overcome the activation energy, they will react to form the product. In reality this is not the case as is seen in the calculated examples above where some reactants reached the transition state and reformed the reagents but not the products. This is because of another assumption of TST which is that the energy is looked at classically rather than quantum mechanically so fails to account for the quantization of molecular vibrations. Failing to factor this is why TST predictions for reaction rates will typically be higher than experimental values.

=== EXERCISE 2: F - H - H system ===

''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic).''

 F + H2 

[[File:Ny517_h2_plus_f.JPG|300px|]]

H-H distance: 0.74 Å

F-H distance: 2.00 Å

H-H momentum: -1.5 kg ms-1

F-H momentum: -2.5 kg ms-1

The reactants are higher in energy than the products therefore the reaction is exothermic.

 H + HF 

[[File:Ny517_f_h_plus_h.JPG|300px|]]

F-H distance: 0.74 Å

H-H distance: 2.00 Å

F-H momentum: -1.5 kg ms-1

H-H momentum: -2.5 kg ms-1

The reactants are lower in energy than the products therefore the reaction is endothermic.

''How does this relate to the bond strength of the chemical species involved?''

Through inspection of the potential energy surfaces, it is clear that the bond strength of H-F is greater than the bond strength of H-H since more energy is needed to break the H-F bond in the second reaction.

''Locate the approximate position of the transition state.''

[[File:Ny517_hf_transition_state_distance.JPG|400px|thumb|left|Internuclear Distance vs Time]][[File:Ny517_hf_transition_state_contour.JPG|400px|thumb|center|Contour plot]]

H-H distance: 0.745 Å

H-F distance: 1.815 Å

To find the transition state, both momenta were set to equal zero. Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. The transition state will be closer in energy to the reactants of the exothermic (first) reaction and the products of the endothermic (second) reaction. The contour plot shows the atoms do not fall off the ridge and the Internuclear Distances plot shows the distances remained constant (bar slight oscillations).

''Report the activation energy for both reactions.''

The activation energy is defined as the difference in energy between the reactants and transition state. The energy of the transition state is known which is -103.752 kcal mol-1. 0.3 A was added and subtracted to the HF bond length to calculate the minimum energy difference between the reactants and transition state for both reactions. The reactant energies were extracted using an mep calculation. The calculated activation energies are stated below:

[[File:Ny517_activation_exo.JPG|300px|]] [[File:Ny517_activation_endo.JPG|300px|]]

*For the F + H2 system: 0.236 kcal mol-1
*For the H + HF system: 30.256 kcal mol-1

=== Reaction Dynamics ===

''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ''

For a F + H2 reaction trajectory:

H-H distance: 0.744832 Å

H-F distance: 2.00 Å

H-H momentum: -0.7 kg ms-1

H-F momentum: -1.6 kg ms-1

[[File:Ny517_reaction_dynamics_momenta.JPG|400px|thumb|left|Momenta vs Time]][[File:Ny517_reaction_dynamics_energy.JPG|400px|thumb|center|Energy vs Time]]

It is seen that the B-C momentum plateaus over time due to the H2 bond being split and the H atoms becoming further apart so the bond vibrations begin to stop. The A-B momentum oscillates as the B-C momentum stops oscillating which is due to the H-F atom being formed. The H-F bond vibrations are larger than the H-H bond vibrations due to the strength of the bond being greater and hence a have a higher enthalpy of formation. If the H-F bond is stronger than the H-H bond then this will result in an exothermic reaction. This could be confirmed experimentally through a reaction calorimeter. The plot displaying energy over time indicates that the total energy is constant and hence conserved. It also shows that the the potential and kinetic energy convert between each other as the atoms vibrate between their bonds. Notice that the energy oscillations become larger after the transition state- similar to that seen in the momentum plot.

Changing momenta:

*Keeping the same initial conditions but changing the H-H momentum:

[[File:Ny517_momentum_-3.JPG|500px|thumb|left|pHH=-3 kg ms-1]]
[[File:Ny517_momentum_-2.JPG|500px|thumb|center|pHH=-2 kg ms-1]]
[[File:Ny517_momentum_-1.JPG|500px|thumb|left|pHH=-1 kg ms-1]]
[[File:Ny517_momentum_0.JPG|500px|thumb|center|pHH=0 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=1 kg ms-1]]
[[File:Ny517_momentum_2.JPG|500px|thumb|center|pHH=2 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=3 kg ms-1]]

Only when pHH was equal to -1 and -2 kg ms-1 was there a reaction trajectory. When the time step was doubled for pHH=0 a reaction trajectory was produced indicating that the H2 bond was broken and a HF bond was formed. For the other reactions, the reactants passed the transition state but went back and reformed the reactants.

When pFH = -0.8 kg ms-1 and pHH = 0.1 kg ms-1, a reaction trajectory occurs:

[[File:Ny517_momentum_change.JPG||300px|]]

For the reverse reaction, H + HF:

The initial conditions are set with :

H-H distance: 2.00

H-F distance: 0.74

H-H momentum: -0.59 kg ms-1

H-F momentum: -6.55 kg ms-1

[[File:Ny517_reverse_1.JPG||300px|]]

Keeping the atom positions constant, the momenta were varied to have a low vibrational motion on on the H-F bond and an arbitrarily high value of pHH above the activation energy and obtain a reactive trajectory:

[[File:Ny517_reverse_2.JPG||300px|]]

H-H distance: 2.00 Å

F-H distance: 0.74 Å

H-H momentum: -7.55 kg ms-1

F-H momentum: -0.35 kg ms-1

''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.''

The efficiency of a reaction is dependant on the position of the transition state. If there is a late transition state, the vibrational energy is the most influential on the efficiency of the reaction and hence its ability to overcome the reaction energy. If there is an early transition state, the translational energy is the most influencial on the efficiency. Using this, it can be deduced that for the endothermic H + HF system, the reaction is affected more by the vibrational energy and that for the exothermic F + H2 system, the reaction is affected more by the translational energy.

MRD:ny517

2019-05-23T18:28:28Z

Ny517: /* Exercise 1: H + H2 */

== Molecular Reaction Dynamics ==
=== Exercise 1: H + H2 ===

''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''

The transition state is defined as the maximum energy point on the minimum energy path and the point on the potential energy surface when the gradient (partial first derivative) of the potential is equal to zero (∂V(ri)/∂ri=0). The transition state can be identified as a saddle point which means it has a partial second derivative greater than and less than zero (dependant on the variables and direction of axis). A local minimum is the point where it is solely a minimum and so the second derivative is always a positive value. The figure below shows a surface plot of a reaction between molecule A-B and atom C where the transition state is highlighted.

''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The best estimate of the transition state to 3 decimal places was:
rts=r1=r2= 0.908 Å

This was estimated through looking at the contour plot of the initial conditions set for a H + H2 reaction and following the reactive trajectory to the point where r1 was approximate to r2 - which was in the region of 0.9 Å (as seen on the surface plot below). This is because for a system with identical atoms, the equidistant point is equal to the transition state position. After trialling values in this region, 0.908 Å produced a trajectory with a constant internuclear distance between both atoms indicating that the trajectory will not fall off the ridge. This is a transition state since the atoms are stationary over time when they have no initial momenta.

[[File:Ny517_q2.JPG|300px|thumb|left|Surface plot from initial conditions]] [[File:Ny517_q2_Internuclear_distance.JPG|600px|thumb|center|Internuclear Distance vs Time plot where rts = 0.908 Å]]

''Comment on how the mep and the trajectory you just calculated differ.''

The mep produces a trajectory that follows the valley floor and is also known as the reaction path. It shows a straight-line path along this trajectory because the velocity is always reset to zero after each time step. This does not provide a realistic account of the motion of atoms during the reaction since it assumes the atoms have no mass. The “Dynamics” calculation provides a more realistic trajectory since it accounts for the mass of the system and so the motion of the atoms in the gas phase will be inertial.

[[File:Ny517_q3_mep.JPG|500px|thumb|left|mep surface plot]] [[File:Ny517_q3_dynamics.JPG|500px|thumb|center|Dynamics surface plot]]

'''Why is mep used?'''

If the initial conditions were reversed, the A-B and B-C plots would switch for the “Distances vs Time” and “Momenta vs Time” graphs. This is because the magnitude of the forces in the reaction do not change but the direction does. As the time increases, the A-B distance increases whilst the B-C bond distance stays constant (bar slight oscillations) at 0.75 because atom A is travelling further away from the BC diatomic molecule. At 5 seconds the A-B bond distance is 18 Å. As the time increases, the A-B momentum plateaus at 2.5 kg ms-1 whilst the B-C momentum oscillates about an average of 1.25 kg ms-1. If the values were reversed, the B-C bond distance would increase whilst the A-B bond distance would remain constant.

[[File:Ny517_q3_larget_bond.JPG|500px|thumb|left|Distance vs Time]] [[File:Ny517_q3_larget_momentum.JPG|500px|thumb|center|Momenta vs Time]]

''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''

Initial positions '''r1''' = 0.74 Å and '''r2''' = 2.0 Å

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0 ||Yes || ||Plot 1
|-
| -1.5 || -2.0 ||-100.4 ||No || ||Plot 2
|-
| -1.5 || -2.5 ||-98.9 ||Yes || ||Plot 3
|-
| -2.5 || -5.0 ||-85.0 ||No || ||Plot 4
|-
| -2.5 || -5.2 ||-83.4 ||Yes || ||Plot 5
|}

''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

Transition state theory (TST) is used to qualitatively understand the reaction rates of elementary chemical reactions by through the belief that all reactions occur through a transition state. One assumption is that once the reagents gain an energy sufficient enough to overcome the activation energy, they will react to form the product. In reality this is not the case as is seen in the calculated examples above where some reactants reached the transition state and reformed the reagents but not the products. This is because of another assumption of TST which is that the energy is looked at classically rather than quantum mechanically so fails to account for the quantization of molecular vibrations. Failing to factor this is why TST predictions for reaction rates will typically be higher than experimental values.

=== EXERCISE 2: F - H - H system ===

''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic).''

 F + H2 

[[File:Ny517_h2_plus_f.JPG|300px|]]

H-H distance: 0.74 Å

F-H distance: 2.00 Å

H-H momentum: -1.5 kg ms-1

F-H momentum: -2.5 kg ms-1

The reactants are higher in energy than the products therefore the reaction is exothermic.

 H + HF 

[[File:Ny517_f_h_plus_h.JPG|300px|]]

F-H distance: 0.74 Å

H-H distance: 2.00 Å

F-H momentum: -1.5 kg ms-1

H-H momentum: -2.5 kg ms-1

The reactants are lower in energy than the products therefore the reaction is endothermic.

''How does this relate to the bond strength of the chemical species involved?''

Through inspection of the potential energy surfaces, it is clear that the bond strength of H-F is greater than the bond strength of H-H since more energy is needed to break the H-F bond in the second reaction.

''Locate the approximate position of the transition state.''

[[File:Ny517_hf_transition_state_distance.JPG|400px|thumb|left|Internuclear Distance vs Time]][[File:Ny517_hf_transition_state_contour.JPG|400px|thumb|center|Contour plot]]

H-H distance: 0.745 Å

H-F distance: 1.815 Å

To find the transition state, both momenta were set to equal zero. Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. The transition state will be closer in energy to the reactants of the exothermic (first) reaction and the products of the endothermic (second) reaction. The contour plot shows the atoms do not fall off the ridge and the Internuclear Distances plot shows the distances remained constant (bar slight oscillations).

''Report the activation energy for both reactions.''

The activation energy is defined as the difference in energy between the reactants and transition state. The energy of the transition state is known which is -103.752 kcal mol-1. 0.3 A was added and subtracted to the HF bond length to calculate the minimum energy difference between the reactants and transition state for both reactions. The reactant energies were extracted using an mep calculation. The calculated activation energies are stated below:

[[File:Ny517_activation_exo.JPG|300px|]] [[File:Ny517_activation_endo.JPG|300px|]]

*For the F + H2 system: 0.236 kcal mol-1
*For the H + HF system: 30.256 kcal mol-1

=== Reaction Dynamics ===

''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ''

For a F + H2 reaction trajectory:

H-H distance: 0.744832 Å

H-F distance: 2.00 Å

H-H momentum: -0.7 kg ms-1

H-F momentum: -1.6 kg ms-1

[[File:Ny517_reaction_dynamics_momenta.JPG|400px|thumb|left|Momenta vs Time]][[File:Ny517_reaction_dynamics_energy.JPG|400px|thumb|center|Energy vs Time]]

It is seen that the B-C momentum plateaus over time due to the H2 bond being split and the H atoms becoming further apart so the bond vibrations begin to stop. The A-B momentum oscillates as the B-C momentum stops oscillating which is due to the H-F atom being formed. The H-F bond vibrations are larger than the H-H bond vibrations due to the strength of the bond being greater and hence a have a higher enthalpy of formation. If the H-F bond is stronger than the H-H bond then this will result in an exothermic reaction. This could be confirmed experimentally through a reaction calorimeter. The plot displaying energy over time indicates that the total energy is constant and hence conserved. It also shows that the the potential and kinetic energy convert between each other as the atoms vibrate between their bonds. Notice that the energy oscillations become larger after the transition state- similar to that seen in the momentum plot.

Changing momenta:

*Keeping the same initial conditions but changing the H-H momentum:

[[File:Ny517_momentum_-3.JPG|500px|thumb|left|pHH=-3 kg ms-1]]
[[File:Ny517_momentum_-2.JPG|500px|thumb|center|pHH=-2 kg ms-1]]
[[File:Ny517_momentum_-1.JPG|500px|thumb|left|pHH=-1 kg ms-1]]
[[File:Ny517_momentum_0.JPG|500px|thumb|center|pHH=0 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=1 kg ms-1]]
[[File:Ny517_momentum_2.JPG|500px|thumb|center|pHH=2 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=3 kg ms-1]]

Only when pHH was equal to -1 and -2 kg ms-1 was there a reaction trajectory. When the time step was doubled for pHH=0 a reaction trajectory was produced indicating that the H2 bond was broken and a HF bond was formed. For the other reactions, the reactants passed the transition state but went back and reformed the reactants.

When pFH = -0.8 kg ms-1 and pHH = 0.1 kg ms-1, a reaction trajectory occurs:

[[File:Ny517_momentum_change.JPG||300px|]]

For the reverse reaction, H + HF:

The initial conditions are set with :

H-H distance: 2.00

H-F distance: 0.74

H-H momentum: -0.59 kg ms-1

H-F momentum: -6.55 kg ms-1

[[File:Ny517_reverse_1.JPG||300px|]]

Keeping the atom positions constant, the momenta were varied to have a low vibrational motion on on the H-F bond and an arbitrarily high value of pHH above the activation energy and obtain a reactive trajectory:

[[File:Ny517_reverse_2.JPG||300px|]]

H-H distance: 2.00 Å

F-H distance: 0.74 Å

H-H momentum: -7.55 kg ms-1

F-H momentum: -0.35 kg ms-1

''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.''

The efficiency of a reaction is dependant on the position of the transition state. If there is a late transition state, the vibrational energy is the most influential on the efficiency of the reaction and hence its ability to overcome the reaction energy. If there is an early transition state, the translational energy is the most influencial on the efficiency. Using this, it can be deduced that for the endothermic H + HF system, the reaction is affected more by the vibrational energy and that for the exothermic F + H2 system, the reaction is affected more by the translational energy.

MRD:ny517

2019-05-23T18:25:22Z

2019-05-23T17:00:43Z

Hello

NY517 Mod

2019-05-20T12:55:33Z

Ny517: /* MO Analysis of [Cr(CO)6] */

== '''BH3''' ==

Method/Basis Set: '''B3LYP/6-31G(d,p)'''

[[File:Ny517_BH3_Summary.jpeg|250px]]

<pre>
Item Value Threshold Converged?
Maximum Force 0.000189 0.000450 YES
RMS Force 0.000095 0.000300 YES
Maximum Displacement 0.000746 0.001800 YES
RMS Displacement 0.000373 0.001200 YES
</pre>

Frequency analysis log file: [[NY_BH3_FREQ.LOG]]

<pre>
Low frequencies --- -0.2263 -0.1037 -0.0054 47.9770 49.0378 49.0383
Low frequencies --- 1163.7209 1213.6704 1213.6731
</pre>

<jmol><jmolApplet>
<title>Optimised borane molecule</title>
<color>black</color>
<size>250</size>
<uploadedFileContents>NY_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Vibrations and IR ===

{| class="wikitable" border="1"
|+ Vibrations
|-
|Frequency/ cm-1|| Intensity || Symmetry || IR Active? || Type of Vibration
|-
|1164
|92
|A2' '
|Yes
|Bend
|-
|1214
|14
|E'
|Yes
|Bend
|-
|1214
|14
|E'
|Yes
|Bend
|-
|2580
|0
|A1'
|No
|Symmetric Stretch
|-
|2713
|126
|E'
|Yes
|Asymmetric Stretch
|-
|2713
|126
|E'
|Yes
|Asymmetric Stretch
|}

[[File:Ny517_BH3_Spectrum.jpeg|500px]]

There are 6 vibrations but only 3 peaks in the spectrum. There are two cases of degenerate vibrations where vibrations with the same energy superimpose their intensities. This gives rise to the peaks at 1214 cm-1 and 2713 cm-1.
The A1 vibration is a symmetric stretch so has no change in dipole moment and is therefore not IR active.

=== MO Diagram ===

[[File:NY_MO_final.JPG|500px|MO diagram for BH3]]

This diagram compares a MO diagram of BH3<ref>[http://www.huntresearchgroup.org.uk/teaching/teaching_MOs_year2/P1_BH3_MO_diagram.pdf" Dr. P. Hunt, presented in part at Molecular Orbitals in Inorganic Chemistry- Problem Class 1, 11/18"]</ref> and the LCAO with the computed MOs. The LCAO and real MOs are very similar since the LCAO MOs correctly predict the regions of electron density and nodes as well as correctly predict their relative energy levels. However, their shapes do not completely represent the overlapping of orbitals compared to heir associated computed MOs. This indicates that LCAO MOs for BH3 are very useful for predicting the general shapes and relative energies. They are accurate enough for most MO analysis and is a process that requires fewer resources than computing MOs.

== '''NH3''' ==

[[File:Ny517_NH3_Summary.jpeg|250px]]

<pre>
Item Value Threshold Converged?
Maximum Force 0.000013 0.000450 YES
RMS Force 0.000006 0.000300 YES
Maximum Displacement 0.000040 0.001800 YES
RMS Displacement 0.000013 0.001200 YES
</pre>

Frequency analysis log file: [[NY_NH3_631G_DP_FREQ2.LOG]]

<pre>
Low frequencies --- -8.5223 -8.4750 -0.0029 0.0335 0.1919 26.4067
Low frequencies --- 1089.7616 1694.1862 1694.1866
</pre>

<jmol><jmolApplet>
<title>Optimised ammonia molecule</title>
<color>black</color>
<size>250</size>
<uploadedFileContents>NY_NH3_631G_DP_FREQ2.LOG</uploadedFileContents>
</jmolApplet></jmol>

== '''NH3BH3''' ==

Method/Basis Set: '''B3LYP/6-31G(d,p)'''

[[File:Ny517 NH3BH3 Summary.JPG|250px]]

<pre>
Item Value Threshold Converged?
Maximum Force 0.000115 0.000450 YES
RMS Force 0.000060 0.000300 YES
Maximum Displacement 0.000569 0.001800 YES
RMS Displacement 0.000345 0.001200 YES
</pre>

Frequency analysis log file: [[NY_NH3BH3_631G_DP_FREQ.LOG]]

<pre>
Low frequencies --- -0.0249 -0.0031 0.0003 17.1475 17.1498 37.2156
Low frequencies --- 265.8048 632.2133 639.3570
</pre>

<jmol><jmolApplet>
<title>Optimised ammonia molecule</title>
<color>black</color>
<size>250</size>
<uploadedFileContents>NY_NH3BH3_631G_DP_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Energies ===

E(NH3)= -56.55776863 au

E(BH3)= -26.61532342 au

E(NH3BH3)= -83.22468893 au

ΔE= -0.05160 au x 2625.5 = -135 kJ mol-1

The B-N dative bond is weak since it’s bond enthalpy is lower than that for a C-N bond at 305 kJ mol-1

== '''NI3''' ==

Method/Basis Set: '''B3LYP/6-31G(d,p) for N and LanL2DZ for I'''

[[File:Ny517 NI3 Summary.jpeg|250px]]

<pre>
Item Value Threshold Converged?
Maximum Force 0.000116 0.000450 YES
RMS Force 0.000069 0.000300 YES
Maximum Displacement 0.001026 0.001800 YES
RMS Displacement 0.000645 0.001200 YES
</pre>

Frequency analysis log file: [[NY NH3 NI3 GEN OPT FREQ2.LOG]]

<pre>
Low frequencies --- -12.4683 -12.4620 -5.7337 -0.0040 0.0193 0.0695
Low frequencies --- 100.9420 100.9427 147.2614
</pre>

<jmol><jmolApplet>
<title>Optimised NI3 molecule</title>
<color>black</color>
<size>250</size>
<uploadedFileContents>NY NH3 NI3 GEN OPT FREQ2.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimised N-I bond length is 2.184 Å.

== '''Project: Metal Carbonyls''' ==

This project looks to investigate the trend in '''M-C bond lengths''' and '''vibrational frequencies''' of the CO ligand for the following metal complexes: '''[V(CO)6]-, [Cr(CO)6], [Mn(CO)6]+'''. These three complexes were chosen because they have metal centres that are successive along a period in the periodic table and a positive, neutral and negative complex are represented in this sample also. An initial hypothesis predicts that the bond lengths would decrease from V to Mn because their oxidation states increase and so will result in a larger Δoct and greater M-C bond strength. This increase in M-C bond strength will reduce the strength of the C-O bond (due to the electron density centring on the M-C bond) and therefore cause a reduction in their vibrational frequencies.

=== '''[V(CO)6]-''' ===

Method/Basis Set: '''B3LYP/6-31G(d,p) for C & O and LanL2DZ for V'''

[[File:Ny517_V_Summary.JPG|250px]]

<pre>
Item Value Threshold Converged?
Maximum Force 0.000198 0.000450 YES
RMS Force 0.000070 0.000300 YES
Maximum Displacement 0.001382 0.001800 YES
RMS Displacement 0.000660 0.001200 YES
</pre>

Frequency analysis log file: [[NY517_V2_OPTNFREQ.LOG]]

<pre>
Low frequencies --- 0.0010 0.0013 0.0014 14.1311 14.1311 14.1311
Low frequencies --- 52.8916 52.8916 52.8916
</pre>

<jmol><jmolApplet>
<title>Optimised [V(CO)6]- Complex</title>
<color>black</color>
<size>250</size>
<uploadedFileContents>NY517_V2_OPTNFREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== '''[Cr(CO)6]''' ===

Method/Basis Set: '''B3LYP/6-31G(d,p) for C & O and LanL2DZ for Cr'''

[[File:Ny517_Cr_Summary.JPG|250px]]

<pre>
Item Value Threshold Converged?
Maximum Force 0.000160 0.000450 YES
RMS Force 0.000057 0.000300 YES
Maximum Displacement 0.000225 0.001800 YES
RMS Displacement 0.000084 0.001200 YES
</pre>

Frequency analysis log file: [[NY_CR_OPTNFREQ.LOG]]

<pre>
Low frequencies --- 0.0012 0.0013 0.0014 10.8502 10.8502 10.8502
Low frequencies --- 66.4359 66.4359 66.4359
</pre>

<jmol><jmolApplet>
<title>Optimised [Cr(CO)6] Complex</title>
<color>black</color>
<size>250</size>
<uploadedFileContents>NY_CR_OPTNFREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== '''[Mn(CO)6]+''' ===

Method/Basis Set: '''B3LYP/6-31G(d,p) for C & O and LanL2DZ for Mn'''

[[File:Ny517_Mn_Summary.JPG|250px]]

<pre>
Item Value Threshold Converged?
Maximum Force 0.000054 0.000450 YES
RMS Force 0.000024 0.000300 YES
Maximum Displacement 0.000430 0.001800 YES
RMS Displacement 0.000204 0.001200 YES
</pre>

Frequency analysis log file: [[NY_MN_OPTNFREQ.LOG]]

<pre>
Low frequencies --- -0.0009 -0.0009 -0.0004 4.7607 4.7607 4.7607
Low frequencies --- 76.3202 76.3202 76.3202
</pre>

<jmol><jmolApplet>
<title>Optimised [Mn(CO)6]+ Complex</title>
<color>black</color>
<size>250</size>
<uploadedFileContents>NY_MN_OPTNFREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Evaluating Hypothesis ===

{| class="wikitable" border="1"
|+ M-C Bond Length and CO Frequency Vibrations
|-
|Metal Complex|| Frequency/ cm-1 || Intensity || Type of Vibration || Bond Length/ Å
|-
|[V(CO)6]-
|1969
|2383
|Symmetric Stretch
|1.954
|-
|[Cr(CO)6]
|2087
|1637
|Symmetric Stretch
|1.915
|-
|[Mn(CO)6]+
|2199
|879
|Symmetric Stretch
|1.908
|}

This table displays the ''bond lengths'' and frequency vibrations of the three metal complexes. The frequency values represent the ''stretching vibrations'' only of the CO ligands. There are no significant bending vibrations and will not be seen in an experimental spectrum due to their low intensities therefore they were not referred to in the table. Each stretching frequency has two additional degenerate vibrations which result in their large intensities seen in spectra (such as in the spectrum of [V(CO)6]- below).

[[File:Ny517_V_Sprectrum.JPG|500px]]

I incorrectly predicted that ''the vibrational frequencies would decrease'' from the Vanadium to Manganese complexes because I failed to account for the change in charges for the complexes. The CO bond frequency (and bond strength) in fact increased despite the oxidation state on the metal centre increasing which should result in a larger Δoct and a stronger M-C bond and hence a weaker C-O bond. This did not occur which due to the complexes becoming more positive (or less negative). Increasing negative charge leads to the expansion of d-orbitals on the metal and therefore results in a greater overlap of '''M(dπ)''' orbital with the '''CO π*''' orbital. This process, which strengthens the M-C bond and increases its bond order, is called π-backonding (and is illustrated in the diagram below). A decrease in M-C bond order from the Vanadium to Manganese complexes resulted in an increase in bond order of the C-O bond. Since the vibrational frequency for the CO ligand (ν(CO)) is proportional to the force constant (k), an increase in strength of the CO bond will result in an increase in its vibrational frequency.

[[File:Ny517_backbonding.JPG|thumb|500px|π backbonding between a metal centre and CO ligand<ref>[https://bb.imperial.ac.uk/bbcswebdav/pid-1537351-dt-content-rid-5135888_1/courses/DSS-CH2_IC-18_19/TM%20Lecture%205%202019%20Student%20Handouts.pdf" Prof. Nick Long, presented in part at Transition Metal, Coordination and Organometallic Chemistry- Lecture 5, 05/19"]</ref>]]

I correctly predicted that ''the M-C bond length would decrease'' but incorrectly justified this trend. The decrease in bond length directly contradicts the increase in vibrational frequency of the CO ligand and the theory that supports this. A decrease in M-C bond length implies an increase in the bond strength from the V to Mn complexes but the vibrational frequencies already confirmed that the M-C bond length decreases. This result could be a consequence of the theory used to justify this trend and requires further research to explain this.

It is ''not possible to analyse the totally symmetric C-O vibrations for these complexes'' because they do not appear on spectra as they are IR inactive. This is because a symmetric vibration does not cause a change in dipole moment in the molecule which is a fundamental selection rule for spectroscopy.

=== MO Analysis of [Cr(CO)6] ===

The three following molecular orbitals represent a mixture of bonding and non-bonding orbitals characterised by the different bonding and antibonding orbital interactions.

[[File:MO_43_Final_2.JPG|1000px]]

[[File:MO_49_Final.JPG|1000px]]

For MO 49, the metal d-orbital has a greater contribution to the MO than the ligand FO. This is because the contributing ligand orbital is the CO π* orbital which is higher in energy than the dyz orbital from the metal. Since MO 49 is a bonding orbital it lies closer in energy to the lower lying metal orbitals and is a metal-based MO.

[[File:MO_53_Final.JPG|1000px]]

== '''References''' ==

File:MO 43 Final 2.JPG

2019-05-20T12:53:58Z

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