ChemWiki - User contributions [en]

MRD:kt3817

2019-05-24T16:58:45Z

Kt3817:

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products. Second derivative negative for local minimum, positive for transition state which is a maximum.

2. Estimated position of a transition state is r(ts)=0.9078.
[[File:ktProblem1disttime.png]]

'''3. How do the mep and dynamics trajectory differ?

The minimum energy path is a straight line whereas the trajectory calculated with the dynamics method oscillates (vibrations are visible).

'''4. Table'''

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration
|-
| -1.25 || -2.5 || -99.018 || Yes || The reaction is taking place. The reaction path is a relatively straight line going through the lowest energy pathway. || [[File:ktP1table1.png|250px]]
|-
| -1.5 || -2.0 || -100.456 || No || AB distance goes from 2.0 to about 1.1, but then starts to increase again and the reaction does not happen. || [[File:ktP1table2.png|250px]]
|-
| -1.5 || -2.5 || -98.956 || Yes || Reaction is happening. The trajectory goes through the low energy region and is a relatively straight line with a few oscillations. || [[File:ktP1table3.png|250px]]
|-
| -2.5 || -5.0 || -84.956 || No || The AB distance gets to about 0.6 but then increases again, oscillating quite a lot. The reaction is not happening even though A and B do come close to one another. || [[File:ktP1table4.png|250px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reaction is happening. However, the trajectory has quite large oscillations in it and does not just stay in the lowest energy region. In addition, when the A-B distance first gets as small as to have a bond between the two molecules, it increases again for a short period of time and then starts oscillating between 0.6 and 1.0. || [[File:ktP1table5.png|250px]]
|}

'''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 assumes that particles move according to classical mechanics and that if there is not enough kinetic energy for particles to react, they will not react. Transition state is a saddle point on a potential energy surface and a maximum of the minimum energy path between reactants and products. In a transition state, reactants and products are in an equilibrium state between the two states and can be moved to either reactants or products with a small change in geometry.

Transition State Theory assumes that when particles collide with an enough energy to react, they will react regardless of which kind of kinetic energy it is. Results showed, however, that when pathway is reactive with certain amount of vibrational and translational kinetic energy, when the amount of vibrational energy, for example, is decreased and the amount of translational energy is increased so that the overall energy remains the same, it does not mean that the second reaction path is also reactive.

== Problem 2 ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic, H + HF reaction is endothermic. H-F bond is stronger than H-H bond so when F + HH reaction takes place, the weaker bond is broken and stronger bond is formed which makes reaction exothermic, with H + HF reaction, stronger bond is broken and reaction is endothermic.

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

When atom A is F and atoms B and C are H atoms:

AB distance is 1.81

BC distance is 0.745

[[File:ktP2figure.png]]

'''3. Activation energies'''

Activation energy can be calculated by taking the energy of the transition state and substracting the energy of the respective species, either H2 + F or HF + H. The energies of the molecules can be found by giving the distance of the other atom to be infinetely large.

Transition state energy is -103.752

For F + H2: 103.752- 33.154 = 70.598

For H + HF: 104.190- 103.752 = 0.348

'''4.'''

Conditions:

r1 = 1.0

r2 = 2.0

p1 = -2.5

p2 = -5.0

First, H2 molecule approaches the F atom and H-H atoms are vibrating. As the H atom closest to the F atom reaches the F atom, H-H bond will get weaker and simultaneously, H-F bond is formed. The vibration starts in H-F molecule and the other H atom will distance from HF.
From momenta vs time plot, similar thing can be seen as first, there is vibration between the two H atoms as they approach F atom and when the bond breaks, momentum remains constant. For H-F bond, momentum is constant while H2 molecule approaches it, then drops and increases again as the bond forms and the vibration pattern starts.
The overall momentum has increased throughout the reaction therefore there is more kinetic energy and less potential energy in the system. The reaction has an early transition state.

'''5.'''

Both translational and vibrational energy affect the efficiency of the reaction, however, according to examples done above about H-H-H system, translational energy seems to have a bit larger role in whether or not the reaction is efficient. For example, when going from first to second row of the table, translational energy is decreased and vibrational energy is increased by the same amount, however, the reaction only happens during the first but not during the second case.

Transition state is always higher in energy than the reactant or product state. If the reaction is exothermic, it has an early transition state and when reaction is endothermic, it has a late transition state. This means that in case of an exothermic reaction, translational energy is converted into vibrational earlier than in the case of an endothermic reaction which has a late transition state.

MRD:kt3817

2019-05-24T16:58:28Z

Kt3817: /* Problem 1 */

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products. Second derivative negative for local minimum, positive for transition state which is a maximum.

2. Estimated position of a transition state is r(ts)=0.9078.
[[File:ktProblem1disttime.png]]

'''3. How do the mep and dynamics trajectory differ?

The minimum energy path is a straight line whereas the trajectory calculated with the dynamics method oscillates (vibrations are visible).

'''4. Table'''

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration
|-
| -1.25 || -2.5 || -99.018 || Yes || The reaction is taking place. The reaction path is a relatively straight line going through the lowest energy pathway. || [[File:ktP1table1.png|250px]]
|-
| -1.5 || -2.0 || -100.456 || No || AB distance goes from 2.0 to about 1.1, but then starts to increase again and the reaction does not happen. || [[File:ktP1table2.png|250px]]
|-
| -1.5 || 2.5 || -98.956 || Yes || Reaction is happening. The trajectory goes through the low energy region and is a relatively straight line with a few oscillations. || [[File:ktP1table3.png|250px]]
|-
| -2.5 || -5.0 || -84.956 || No || The AB distance gets to about 0.6 but then increases again, oscillating quite a lot. The reaction is not happening even though A and B do come close to one another. || [[File:ktP1table4.png|250px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reaction is happening. However, the trajectory has quite large oscillations in it and does not just stay in the lowest energy region. In addition, when the A-B distance first gets as small as to have a bond between the two molecules, it increases again for a short period of time and then starts oscillating between 0.6 and 1.0. || [[File:ktP1table5.png|250px]]
|}

'''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 assumes that particles move according to classical mechanics and that if there is not enough kinetic energy for particles to react, they will not react. Transition state is a saddle point on a potential energy surface and a maximum of the minimum energy path between reactants and products. In a transition state, reactants and products are in an equilibrium state between the two states and can be moved to either reactants or products with a small change in geometry.

Transition State Theory assumes that when particles collide with an enough energy to react, they will react regardless of which kind of kinetic energy it is. Results showed, however, that when pathway is reactive with certain amount of vibrational and translational kinetic energy, when the amount of vibrational energy, for example, is decreased and the amount of translational energy is increased so that the overall energy remains the same, it does not mean that the second reaction path is also reactive.

== Problem 2 ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic, H + HF reaction is endothermic. H-F bond is stronger than H-H bond so when F + HH reaction takes place, the weaker bond is broken and stronger bond is formed which makes reaction exothermic, with H + HF reaction, stronger bond is broken and reaction is endothermic.

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

When atom A is F and atoms B and C are H atoms:

AB distance is 1.81

BC distance is 0.745

[[File:ktP2figure.png]]

'''3. Activation energies'''

Activation energy can be calculated by taking the energy of the transition state and substracting the energy of the respective species, either H2 + F or HF + H. The energies of the molecules can be found by giving the distance of the other atom to be infinetely large.

Transition state energy is -103.752

For F + H2: 103.752- 33.154 = 70.598

For H + HF: 104.190- 103.752 = 0.348

'''4.'''

Conditions:

r1 = 1.0

r2 = 2.0

p1 = -2.5

p2 = -5.0

First, H2 molecule approaches the F atom and H-H atoms are vibrating. As the H atom closest to the F atom reaches the F atom, H-H bond will get weaker and simultaneously, H-F bond is formed. The vibration starts in H-F molecule and the other H atom will distance from HF.
From momenta vs time plot, similar thing can be seen as first, there is vibration between the two H atoms as they approach F atom and when the bond breaks, momentum remains constant. For H-F bond, momentum is constant while H2 molecule approaches it, then drops and increases again as the bond forms and the vibration pattern starts.
The overall momentum has increased throughout the reaction therefore there is more kinetic energy and less potential energy in the system. The reaction has an early transition state.

'''5.'''

Both translational and vibrational energy affect the efficiency of the reaction, however, according to examples done above about H-H-H system, translational energy seems to have a bit larger role in whether or not the reaction is efficient. For example, when going from first to second row of the table, translational energy is decreased and vibrational energy is increased by the same amount, however, the reaction only happens during the first but not during the second case.

Transition state is always higher in energy than the reactant or product state. If the reaction is exothermic, it has an early transition state and when reaction is endothermic, it has a late transition state. This means that in case of an exothermic reaction, translational energy is converted into vibrational earlier than in the case of an endothermic reaction which has a late transition state.

MRD:kt3817

2019-05-24T16:57:15Z

Kt3817: /* Problem 2 */

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products. Second derivative negative for local minimum, positive for transition state which is a maximum.

2. Estimated position of a transition state is r(ts)=0.9078.
[[File:ktProblem1disttime.png]]

'''3. How do the mep and dynamics trajectory differ?

The minimum energy path is a straight line whereas the trajectory calculated with the dynamics method oscillates (vibrations are visible).

'''4. Table'''

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration
|-
| -1.25 || -2.5 || -99.018 || Yes || The reaction path is a relatively straight line going through the lowest energy pathway. || [[File:ktP1table1.png|250px]]
|-
| -1.5 || -2.0 || -100.456 || No || AB distance goes from 2.0 to about 1.1, but then starts to increase again and the reaction does not happen. || [[File:ktP1table2.png|250px]]
|-
| -1.5 || 2.5 || -98.956 || Yes || Reaction is happening. The trajectory goes through the low energy region and is a relatively straight line with a few oscillations. || [[File:ktP1table3.png|250px]]
|-
| -2.5 || -5.0 || -84.956 || No || The AB distance gets to about 0.6 but then increases again, oscillating quite a lot. The reaction is not happening even though A and B do come close to one another. || [[File:ktP1table4.png|250px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reaction is happening. However, the trajectory has quite large oscillations in it and does not just stay in the lowest energy region. In addition, when the A-B distance first gets as small as to have a bond between the two molecules, it increases again for a short period of time and then starts oscillating between 0.6 and 1.0. || [[File:ktP1table5.png|250px]]
|}

'''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 assumes that particles move according to classical mechanics and that if there is not enough kinetic energy for particles to react, they will not react. Transition state is a saddle point on a potential energy surface and a maximum of the minimum energy path between reactants and products. In a transition state, reactants and products are in an equilibrium state between the two states and can be moved to either reactants or products with a small change in geometry.

Transition State Theory assumes that when particles collide with an enough energy to react, they will react regardless of which kind of kinetic energy it is. Results showed, however, that when pathway is reactive with certain amount of vibrational and translational kinetic energy, when the amount of vibrational energy, for example, is decreased and the amount of translational energy is increased so that the overall energy remains the same, it does not mean that the second reaction path is also reactive.

== Problem 2 ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic, H + HF reaction is endothermic. H-F bond is stronger than H-H bond so when F + HH reaction takes place, the weaker bond is broken and stronger bond is formed which makes reaction exothermic, with H + HF reaction, stronger bond is broken and reaction is endothermic.

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

When atom A is F and atoms B and C are H atoms:

AB distance is 1.81

BC distance is 0.745

[[File:ktP2figure.png]]

'''3. Activation energies'''

Activation energy can be calculated by taking the energy of the transition state and substracting the energy of the respective species, either H2 + F or HF + H. The energies of the molecules can be found by giving the distance of the other atom to be infinetely large.

Transition state energy is -103.752

For F + H2: 103.752- 33.154 = 70.598

For H + HF: 104.190- 103.752 = 0.348

'''4.'''

Conditions:

r1 = 1.0

r2 = 2.0

p1 = -2.5

p2 = -5.0

First, H2 molecule approaches the F atom and H-H atoms are vibrating. As the H atom closest to the F atom reaches the F atom, H-H bond will get weaker and simultaneously, H-F bond is formed. The vibration starts in H-F molecule and the other H atom will distance from HF.
From momenta vs time plot, similar thing can be seen as first, there is vibration between the two H atoms as they approach F atom and when the bond breaks, momentum remains constant. For H-F bond, momentum is constant while H2 molecule approaches it, then drops and increases again as the bond forms and the vibration pattern starts.
The overall momentum has increased throughout the reaction therefore there is more kinetic energy and less potential energy in the system. The reaction has an early transition state.

'''5.'''

Both translational and vibrational energy affect the efficiency of the reaction, however, according to examples done above about H-H-H system, translational energy seems to have a bit larger role in whether or not the reaction is efficient. For example, when going from first to second row of the table, translational energy is decreased and vibrational energy is increased by the same amount, however, the reaction only happens during the first but not during the second case.

Transition state is always higher in energy than the reactant or product state. If the reaction is exothermic, it has an early transition state and when reaction is endothermic, it has a late transition state. This means that in case of an exothermic reaction, translational energy is converted into vibrational earlier than in the case of an endothermic reaction which has a late transition state.

MRD:kt3817

2019-05-24T16:50:56Z

Kt3817: /* Problem 1 */

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products. Second derivative negative for local minimum, positive for transition state which is a maximum.

2. Estimated position of a transition state is r(ts)=0.9078.
[[File:ktProblem1disttime.png]]

'''3. How do the mep and dynamics trajectory differ?

The minimum energy path is a straight line whereas the trajectory calculated with the dynamics method oscillates (vibrations are visible).

'''4. Table'''

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration
|-
| -1.25 || -2.5 || -99.018 || Yes || The reaction path is a relatively straight line going through the lowest energy pathway. || [[File:ktP1table1.png|250px]]
|-
| -1.5 || -2.0 || -100.456 || No || AB distance goes from 2.0 to about 1.1, but then starts to increase again and the reaction does not happen. || [[File:ktP1table2.png|250px]]
|-
| -1.5 || 2.5 || -98.956 || Yes || Reaction is happening. The trajectory goes through the low energy region and is a relatively straight line with a few oscillations. || [[File:ktP1table3.png|250px]]
|-
| -2.5 || -5.0 || -84.956 || No || The AB distance gets to about 0.6 but then increases again, oscillating quite a lot. The reaction is not happening even though A and B do come close to one another. || [[File:ktP1table4.png|250px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reaction is happening. However, the trajectory has quite large oscillations in it and does not just stay in the lowest energy region. In addition, when the A-B distance first gets as small as to have a bond between the two molecules, it increases again for a short period of time and then starts oscillating between 0.6 and 1.0. || [[File:ktP1table5.png|250px]]
|}

'''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 assumes that particles move according to classical mechanics and that if there is not enough kinetic energy for particles to react, they will not react. Transition state is a saddle point on a potential energy surface and a maximum of the minimum energy path between reactants and products. In a transition state, reactants and products are in an equilibrium state between the two states and can be moved to either reactants or products with a small change in geometry.

Transition State Theory assumes that when particles collide with an enough energy to react, they will react regardless of which kind of kinetic energy it is. Results showed, however, that when pathway is reactive with certain amount of vibrational and translational kinetic energy, when the amount of vibrational energy, for example, is decreased and the amount of translational energy is increased so that the overall energy remains the same, it does not mean that the second reaction path is also reactive.

== Problem 2 ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic, H + HF reaction is endothermic. H-F bond is stronger than H-H bond so when F + HH reaction takes place, the weaker bond is broken and stronger bond is formed which makes reaction exothermic, with H + HF reaction, stronger bond is broken and reaction is endothermic.

Microsoft Windows [Version 10.0.16299.1087]
(c) 2017 Microsoft Corporation. All rights reserved.

H:\>cd Desktop

H:\Desktop>cd mrd-demo-master

H:\Desktop\mrd-demo-master>python lepsgui.py

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

When atom A is F and atoms B and C are H atoms:

AB distance is 1.81

BC distance is 0.745

[[File:ktP2figure.png]]

'''3. Activation energies'''

Activation energy can be calculated by taking the energy of the transition state and substracting the energy of the respective species, either H2 + F or HF + H. The energies of the molecules can be found by giving the

Transition state energy is -103.752

For F + H2:

For H + HF:

'''4.'''

Conditions:

r1 = 1.0

r2 = 2.0

p1 = -2.5

p2 = -5.0

First, H2 molecule approaches the F atom and H-H atoms are vibrating. As the H atom closest to the F atom reaches the F atom, H-H bond will get weaker and simultaneously, H-F bond is formed. The vibration starts in H-F molecule and the other H atom will distance from HF.
From momenta vs time plot, similar thing can be seen as first, there is vibration between the two H atoms as they approach F atom and when the bond breaks, momentum remains constant. For H-F bond, momentum is constant while H2 molecule approaches it, then drops and increases again as the bond forms and the vibration pattern starts.
The overall momentum has increased throughout the reaction therefore there is more kinetic energy and less potential energy in the system.

'''5.'''

Both translational and vibrational energy affect the efficiency of the reaction, however, according to examples done above about H-H-H system, translational energy seems to have a bit larger role in whether or not the reaction is efficient. For example, when going from first to second row of the table, translational energy is decreased and vibrational energy is increased by the same amount, however, the reaction only happens during the first but not during the second case.

Transition state is always higher in energy than the reactant or product state. If the reaction is exothermic, it has an early transition state and when reaction is endothermic, it has a late transition state. This means that in case of an exothermic reaction, translational energy is converted into vibrational earlier than in the case of an endothermic reaction which has a late transition state.

MRD:kt3817

2019-05-24T16:20:20Z

Kt3817: /* Problem 2 */

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products. Second derivative negative for local minimum, positive for transition state which is a maximum.

2. Estimated position of a transition state is r(ts)=0.9078.
[[File:ktProblem1disttime.png]]

'''3. How do the mep and dynamics trajectory differ?

The minimum energy path is a straight line whereas the trajectory calculated with the dynamics method oscillates.

'''4. Table'''

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration
|-
| -1.25 || -2.5 || -99.018 || Yes || The reaction path is a relatively straight line going through the lowest energy pathway. || [[File:ktP1table1.png|250px]]
|-
| -1.5 || -2.0 || -100.456 || No || AB distance goes from 2.0 to about 1.1, but then starts to increase again and the reaction does not happen. || [[File:ktP1table2.png|250px]]
|-
| -1.5 || 2.5 || -98.956 || Yes || Reaction is happening. The trajectory goes through the low energy region and is a relatively straight line with a few oscillations. || [[File:ktP1table3.png|250px]]
|-
| -2.5 || -5.0 || -84.956 || No || The AB distance gets to about 0.6 but then increases again, oscillating quite a lot. The reaction is not happening even though A and B do come close to one another. || [[File:ktP1table4.png|250px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reaction is happening. However, the trajectory has quite large oscillations in it and does not just stay in the lowest energy region. In addition, when the A-B distance first gets as small as to have a bond between the two molecules, it increases again for a short period of time and then starts oscillating between 0.6 and 1.0. || [[File:ktP1table5.png|250px]]
|}

'''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?'''

== Problem 2 ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic, H + HF reaction is endothermic. H-F bond is stronger than H-H bond so when F + HH reaction takes place, the weaker bond is broken and stronger bond is formed which makes reaction exothermic, with H + HF reaction, stronger bond is broken and reaction is endothermic.

Microsoft Windows [Version 10.0.16299.1087]
(c) 2017 Microsoft Corporation. All rights reserved.

H:\>cd Desktop

H:\Desktop>cd mrd-demo-master

H:\Desktop\mrd-demo-master>python lepsgui.py

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

When atom A is F and atoms B and C are H atoms:

AB distance is 1.81

BC distance is 0.745

[[File:ktP2figure.png]]

'''3. Activation energies'''

Activation energy can be calculated by taking the energy of the transition state and substracting the energy of the respective species, either H2 + F or HF + H. The energies of the molecules can be found by giving the

Transition state energy is -103.752

For F + H2:

For H + HF:

'''4.'''

Conditions:

r1 = 1.0

r2 = 2.0

p1 = -2.5

p2 = -5.0

First, H2 molecule approaches the F atom and H-H atoms are vibrating. As the H atom closest to the F atom reaches the F atom, H-H bond will get weaker and simultaneously, H-F bond is formed. The vibration starts in H-F molecule and the other H atom will distance from HF.
From momenta vs time plot, similar thing can be seen as first, there is vibration between the two H atoms as they approach F atom and when the bond breaks, momentum remains constant. For H-F bond, momentum is constant while H2 molecule approaches it, then drops and increases again as the bond forms and the vibration pattern starts.
The overall momentum has increased throughout the reaction therefore there is more kinetic energy and less potential energy in the system.

'''5.'''

Both translational and vibrational energy affect the efficiency of the reaction, however, according to examples done above about H-H-H system, translational energy seems to have a bit larger role in whether or not the reaction is efficient. For example, when going from first to second row of the table, translational energy is decreased and vibrational energy is increased by the same amount, however, the reaction only happens during the first but not during the second case.

Transition state is always higher in energy than the reactant or product state. If the reaction is exothermic, it has an early transition state and when reaction is endothermic, it has a late transition state. This means that in case of an exothermic reaction, translational energy is converted into vibrational earlier than in the case of an endothermic reaction which has a late transition state.

MRD:kt3817

2019-05-24T16:18:42Z

Kt3817: /* Problem 2 */

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products. Second derivative negative for local minimum, positive for transition state which is a maximum.

2. Estimated position of a transition state is r(ts)=0.9078.
[[File:ktProblem1disttime.png]]

'''3. How do the mep and dynamics trajectory differ?

The minimum energy path is a straight line whereas the trajectory calculated with the dynamics method oscillates.

'''4. Table'''

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration
|-
| -1.25 || -2.5 || -99.018 || Yes || The reaction path is a relatively straight line going through the lowest energy pathway. || [[File:ktP1table1.png|250px]]
|-
| -1.5 || -2.0 || -100.456 || No || AB distance goes from 2.0 to about 1.1, but then starts to increase again and the reaction does not happen. || [[File:ktP1table2.png|250px]]
|-
| -1.5 || 2.5 || -98.956 || Yes || Reaction is happening. The trajectory goes through the low energy region and is a relatively straight line with a few oscillations. || [[File:ktP1table3.png|250px]]
|-
| -2.5 || -5.0 || -84.956 || No || The AB distance gets to about 0.6 but then increases again, oscillating quite a lot. The reaction is not happening even though A and B do come close to one another. || [[File:ktP1table4.png|250px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reaction is happening. However, the trajectory has quite large oscillations in it and does not just stay in the lowest energy region. In addition, when the A-B distance first gets as small as to have a bond between the two molecules, it increases again for a short period of time and then starts oscillating between 0.6 and 1.0. || [[File:ktP1table5.png|250px]]
|}

'''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?'''

== Problem 2 ==
== Problem 2 ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic, H + HF reaction is endothermic. H-F bond is stronger than H-H bond so when F + HH reaction takes place, the weaker bond is broken and stronger bond is formed which makes reaction exothermic, with H + HF reaction, stronger bond is broken and reaction is endothermic.

Microsoft Windows [Version 10.0.16299.1087]
(c) 2017 Microsoft Corporation. All rights reserved.

H:\>cd Desktop

H:\Desktop>cd mrd-demo-master

H:\Desktop\mrd-demo-master>python lepsgui.py

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

When atom A is F and atoms B and C are H atoms:

AB distance is 1.81

BC distance is 0.745

[[File:ktP2figure.png]]

'''3. Activation energies'''

Activation energy can be calculated by taking the energy of the transition state and substracting the energy of the respective species, either H2 + F or HF + H. The energies of the molecules can be found by giving the

Transition state energy is -103.752

For F + H2:

For H + HF:

'''4.'''

Conditions:

r1 = 1.0

r2 = 2.0

p1 = -2.5

p2 = -5.0

First, H2 molecule approaches the F atom and H-H atoms are vibrating. As the H atom closest to the F atom reaches the F atom, H-H bond will get weaker and simultaneously, H-F bond is formed. The vibration starts in H-F molecule and the other H atom will distance from HF.
From momenta vs time plot, similar thing can be seen as first, there is vibration between the two H atoms as they approach F atom and when the bond breaks, momentum remains constant. For H-F bond, momentum is constant while H2 molecule approaches it, then drops and increases again as the bond forms and the vibration pattern starts.
The overall momentum has increased throughout the reaction therefore there is more kinetic energy and less potential energy in the system.

'''5.'''

Both translational and vibrational energy affect the efficiency of the reaction, however, according to examples done above about H-H-H system, translational energy seems to have a bit larger role in whether or not the reaction is efficient. For example, when going from first to second row of the table, translational energy is decreased and vibrational energy is increased by the same amount, however, the reaction only happens during the first but not during the second case.

Transition state is always higher in energy than the reactant or product state. If the reaction is exothermic, it has an early transition state and when reaction is endothermic, it has a late transition state. This means that in case of an exothermic reaction, translational energy is converted into vibrational earlier than in the case of an endothermic reaction which has a late transition state.

MRD:kt3817

2019-05-24T16:05:25Z

Kt3817: /* Problem 2 */

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products. Second derivative negative for local minimum, positive for transition state which is a maximum.

2. Estimated position of a transition state is r(ts)=0.9078.
[[File:ktProblem1disttime.png]]

'''3. How do the mep and dynamics trajectory differ?

The minimum energy path is a straight line whereas the trajectory calculated with the dynamics method oscillates.

'''4. Table'''

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration
|-
| -1.25 || -2.5 || -99.018 || Yes || The reaction path is a relatively straight line going through the lowest energy pathway. || [[File:ktP1table1.png|250px]]
|-
| -1.5 || -2.0 || -100.456 || No || AB distance goes from 2.0 to about 1.1, but then starts to increase again and the reaction does not happen. || [[File:ktP1table2.png|250px]]
|-
| -1.5 || 2.5 || -98.956 || Yes || Reaction is happening. The trajectory goes through the low energy region and is a relatively straight line with a few oscillations. || [[File:ktP1table3.png|250px]]
|-
| -2.5 || -5.0 || -84.956 || No || The AB distance gets to about 0.6 but then increases again, oscillating quite a lot. The reaction is not happening even though A and B do come close to one another. || [[File:ktP1table4.png|250px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reaction is happening. However, the trajectory has quite large oscillations in it and does not just stay in the lowest energy region. In addition, when the A-B distance first gets as small as to have a bond between the two molecules, it increases again for a short period of time and then starts oscillating between 0.6 and 1.0. || [[File:ktP1table5.png|250px]]
|}

'''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?'''

== Problem 2 ==
== Problem 2 ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic, H + HF reaction is endothermic. H-F bond is stronger than H-H bond so when F + HH reaction takes place, the weaker bond is broken and stronger bond is formed which makes reaction exothermic, with H + HF reaction, stronger bond is broken and reaction is endothermic.

Microsoft Windows [Version 10.0.16299.1087]
(c) 2017 Microsoft Corporation. All rights reserved.

H:\>cd Desktop

H:\Desktop>cd mrd-demo-master

H:\Desktop\mrd-demo-master>python lepsgui.py

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

When atom A is F and atoms B and C are H atoms:

AB distance is 1.81

BC distance is 0.745

[[File:ktP2figure.png]]

'''3. Activation energies'''

Activation energy can be calculated by taking the energy of the transition state and substracting the energy of the respective species, either H2 + F or HF + H. The energies of the molecules can be found by giving the

Transition state energy is -103.752

For F + H2:

For H + HF:

'''4.'''

Conditions:

r1 = 1.0

r2 = 2.0

p1 = -2.5

p2 = -5.0

First, H2 molecule approaches the F atom and H-H atoms are vibrating. As the H atom closest to the F atom reaches the F atom, H-H bond will get weaker and simultaneously, H-F bond is formed. The vibration starts in H-F molecule and the other H atom will distance from HF.
From momenta vs time plot, similar thing can be seen as first, there is vibration between the two H atoms as they approach F atom and when the bond breaks, momentum remains constant. For H-F bond, momentum is constant while H2 molecule approaches it, then drops and increases again as the bond forms and the vibration pattern starts.
The overall momentum has increased throughout the reaction therefore there is more kinetic energy and less potential energy in the system.

'''5.'''

Both translational and vibrational energy affect the efficiency of the reaction, however, according to examples done above,

MRD:kt3817

2019-05-24T15:10:11Z

Kt3817: /* Problem 2 */

MRD:kt3817

2019-05-24T15:09:44Z

Kt3817: /* Problem 2 */

File:KtP2figure.png

2019-05-24T15:08:42Z

Kt3817:

MRD:kt3817

2019-05-24T15:06:48Z

Kt3817: /* Problem 2 */

MRD:kt3817

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Kt3817: /* Problem 2 */

MRD:kt3817

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Kt3817: /* Problem 2 */

MRD:kt3817

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Kt3817: /* Problem 2 */

MRD:kt3817

2019-05-24T12:52:25Z

Kt3817: /* Problem 1 */

MRD:kt3817

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Kt3817: /* Problem 1 */

MRD:kt3817

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Kt3817: /* Problem 1 */

MRD:kt3817

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Kt3817: /* Problem 2 */

MRD:kt3817

2019-05-24T12:49:01Z

Kt3817: /* Problem 2 */

MRD:kt3817

2019-05-24T11:53:01Z

Kt3817: /* Problem 2 */

MRD:kt3817

2019-05-21T16:05:52Z

Kt3817:

MRD:kt3817

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Kt3817:

MRD:kt3817

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Kt3817:

MRD:kt3817

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Kt3817:

File:KtP1table5.png

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Kt3817:

MRD:kt3817

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Kt3817:

File:KtP1table4.png

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Kt3817:

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Kt3817:

MRD:kt3817

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Kt3817:

File:KtP1table3.png

2019-05-21T15:22:24Z

Kt3817:

MRD:kt3817

2019-05-21T15:21:02Z

Kt3817:

MRD:kt3817

2019-05-21T15:16:12Z

Kt3817:

MRD:kt3817

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Kt3817:

File:KtP1table2.png

2019-05-21T15:15:01Z

Kt3817:

MRD:kt3817

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Kt3817:

MRD:kt3817

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Kt3817:

File:KtP1table1.png

2019-05-21T15:09:48Z

Kt3817:

MRD:kt3817

2019-05-21T15:09:25Z

Kt3817: /* Problem 1 */

MRD:kt3817

2019-05-21T14:27:04Z

Kt3817:

File:KtProblem1disttime.png

2019-05-21T14:26:26Z

Kt3817:

MRD:kt3817

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Kt3817:

MRD:kt3817

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Kt3817:

MRD:kt3817

2019-05-21T14:03:50Z

Kt3817: Created page with "== Problem 1 == '''1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it..."

== Problem 1 ==

'''1. 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?'''

Transition state is mathematically defined as aV(ri)/ari=0 or that it is the point where the gradient of the potential is zero.

Transition state can be identified by starting trajectories close to the transition state and see if they go towards reactants or the products. Transition state is the maximum of the minimum energy path so it is not a local minimum. If the point is a local minimum and not a transition state then the trajectory, when started close to the minimum, will always go to that minimum, whereas trajectory started on either side of the transition state will go to either reactants or products.

Rep:Mod:kta3817

2019-05-20T16:58:51Z

Kt3817: /* NH3BH3 */

== BH3 molecule ==

B3LYP/6-31G(d,p)

[[File:ktBH3table2.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000011 0.000450 YES
RMS Force 0.000007 0.000300 YES
Maximum Displacement 0.000043 0.001800 YES
RMS Displacement 0.000028 0.001200 YES
Predicted change in Energy=-6.856714D-10
Optimization completed.
-- Stationary point found.</pre>

BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/1/18/KERLITALI_BH3_FREQ.log | BH3.log file]]

<pre>Low frequencies --- -7.5936 -1.5614 -0.0055 0.6514 6.9319 7.1055
Low frequencies --- 1162.9677 1213.1634 1213.1661 </pre>

<jmol><jmolApplet>
<title>Optimized BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>KERLITALI_BH3_FREQ.log</uploadedFileContents>
</jmolApplet></jmol>

=== Vibrations ===

[[File:Kt3817bh3ir1.png]]

{| class="wikitable" border="1"
|+ IR vibrations of BH3
! No !! Frequency !! Intensity !! Bend/stretch !! Symmetry !! IR active?
|-
| 1 || 1163 || 93 || Bend || A2 || Yes
|-
| 2 || 1213 || 14 || Bend || E || Yes
|-
| 3 || 1213 || 14 || Bend || E || Yes
|-
| 4 || 2582 || 0 ||Symmetric stretch || A1 || No
|-
| 5 || 2716 || 126 || Asymmetric stretch || E || Yes
|-
| 6 || 2716 || 1126 || Asymmetric stretch || E || Yes
|}

Only 3 peaks are visible in the IR spectrum. Vibration 4 in the table above is not visible because in order for a bend/stretch to be IR active, the dipole moment of the molecule must change but during symmetric stretch, dipole moment remains the same. Vibrations 2 and 3 have the same frequency and peaks are therefore overlapping, the same happens with vibrations 5 and 6.

=== MOs ===

[[File:ktBH3MOdiagram1.png]]

2a1' LCAO MO is exactly the same as the computed version, however for 1e', the two computed MOs are the same, just in different directions, whereas the LCAO MOs indicate that the MOs are different and that one of them has a B-H bond that is outside both MO phases when this is not actually the case. 1a2'' and 3a1' LCAO MOs are exactly the same as their computed versions. 2e' has the same difference that 1e' had: computed MOs are simply different directions but LCAO theory predicts that they have bigger differences and that one of them has a B-H bond that is away from both phases when this is not the case.

== NH3 ==

B3LYP/6-31G(d,p)

[[File:kt3817nh3table.jpg]]

NH3 frequency .log file: [[Media:KT3817_NH3OPT_FREQ.LOG]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000012 0.001800 YES
RMS Displacement 0.000008 0.001200 YES
Predicted change in Energy=-9.844611D-11
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -8.5646 -8.5588 -0.0041 0.0455 0.1784 26.4183
Low frequencies --- 1089.7603 1694.1865 1694.1865 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_nh3opt.log</uploadedFileContents>
</jmolApplet></jmol>

== NH3BH3 ==

B3LYP/6-31G(d,p)

[[File:ktNH3BH3table3.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000114 0.000450 YES
RMS Force 0.000064 0.000300 YES
Maximum Displacement 0.000617 0.001800 YES
RMS Displacement 0.000362 0.001200 YES
Predicted change in Energy=-1.787894D-07
Optimization completed.
-- Stationary point found.</pre>

<pre>Low frequencies --- -827.7026 -33.9159 -19.3033 -3.1260 0.0002 0.0010
Low frequencies --- 0.0011 716.1684 743.0238 </pre>

NH3BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/b/be/REALNH3BH3OPT_FREQ1.LOG | NH3BH3 .log file]]

<jmol><jmolApplet>
<title>Optimized NH3BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>REALNH3BH3OPT_FREQ1.LOG</uploadedFileContents>
</jmolApplet></jmol>

E(BH3)= -26.61532 hartree

E(NH3)= -56.55777 hartree

E(NH3BH3)= -83.22469 hartree

ΔE = -83.22469 + 56.5577 + 26.61532 = -0.05160 hartree = -136 kJ/mol

<pre>Is the B-N bond weak, medium or strong?</pre>
Based on the calculation, I would expect B-N bond to be medium as it is the only forming bond in the reaction and the reaction is exothermic but not as exothermic as the reactions where very strong bonds are formed. For example, formation energy for C=O is -393.5 kJ/mol.

== NI3 ==

B3LYP/GEN

[[File:ktni3table.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000139 0.000450 YES
RMS Force 0.000090 0.000300 YES
Maximum Displacement 0.001129 0.001800 YES
RMS Displacement 0.000796 0.001200 YES
Predicted change in Energy=-1.675756D-07
Optimization completed.
-- Stationary point found.</pre>

NI3 frequency .log file: [[Media:ktni3_freq.log]]

<pre>Low frequencies --- -12.7179 -12.7118 -6.4125 -0.0039 0.0189 0.0621
Low frequencies --- 101.0754 101.0761 147.4556</pre>

<jmol><jmolApplet>
<title>Optimized NI3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>ktni3_freq.log</uploadedFileContents>
</jmolApplet></jmol>

Optimized N-I distance is 2.184 Å.

== Ionic liquids ==

=== N(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktNCtable.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000023 0.000450 YES
RMS Force 0.000008 0.000300 YES
Maximum Displacement 0.001012 0.001800 YES
RMS Displacement 0.000229 0.001200 YES
Predicted change in Energy=-1.320388D-08
Optimization completed.
-- Stationary point found. </pre>

N(CH4)+ frequency .log file: [[Media:kt3817_n(ch3)4+opt1_freq.log]]

<pre>Low frequencies --- -0.0014 -0.0012 -0.0012 34.9548 34.9548 34.9548
Low frequencies --- 217.5087 316.5877 316.5877 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_n(ch3)4+opt1_freq.log</uploadedFileContents>
</jmolApplet></jmol>

=== P(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktPCtable.png]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000138 0.000450 YES
RMS Force 0.000032 0.000300 YES
Maximum Displacement 0.000792 0.001800 YES
RMS Displacement 0.000286 0.001200 YES
Predicted change in Energy=-1.621425D-07
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -0.0034 -0.0032 -0.0023 51.2606 51.2606 51.2606
Low frequencies --- 186.5808 211.3947 211.3947 </pre>

P(CH4)+ frequency .log file: [[Media:Kt3817_P(CH3)4OPT_SYM1.LOG]]

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>Kt3817_P(CH3)4OPT_SYM1.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution===

{| class="wikitable" border="1"
|+
! N(CH3)4+ !! P(CH3)4+
|-
| [[File:kt3817_charge_n1.jpg|250px]] || [[File:kt3817_charge_p.jpg|250px]]
|}

In the N(CH4)+ image above, N has a charge of -0.295, C has a charge of -0.483 and H has a charge of 0.269.
In the P(CH4)+ image above, P has a charge of 1.666, C has a charge of -1.050 and H has a charge of 0.298.

In N(CH3)4+, central atom N has a negative charge, however in P(CH3)4+, central atom P has a positive charge.
In both molecules, C has the largest negative charge of all the atoms and H atoms have positive charge. In P(CH3)4+, P has the largest positive charge located on it, whereas H atoms are the only ones in N(CH3)4+ to have positive charge. In N(CH3)4+, atoms with a negative charge are bonded (N and C) whereas in P(CH3)4+, the charges are alternating as P has positive charge, C that is bonded with it has negative charge and H that is bonded with C has positive charge.

In the classical N(R)4+ charge distribution, N is depicted to have a positive charge since N in ammonium ion has 1 less free electron pair than the neutral N atom. Computed charge distribution, however, shows that N has negative charge and the only atoms carrying positive charge are H atoms.

=== MOs ===

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 9 || [[File:ktNMO9.png]] || [[File:ktNLCAOMO9.png]]
|}

MO 9 pictured above is bonding and occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 10 || [[File:ktNMO10.png]] || [[File:ktNMOLCAO101.png]]
|}

MO 10 pictured above is antibonding if the molecule is pictured as NL4+ but is bonding for methyl group fragment orbitals. Methyl group FOs are in the opposite phase in the molecule compared to the figure which shows how they are formed. MO is occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 19 || [[File:ktNMO19.png]] || [[File:ktNLCAOMO19.png]]
|}

MO 19 pictured above is bonding overall and occupied.

Rep:Mod:kta3817

2019-05-20T16:55:35Z

Kt3817: /* MOs */

== BH3 molecule ==

B3LYP/6-31G(d,p)

[[File:ktBH3table2.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000011 0.000450 YES
RMS Force 0.000007 0.000300 YES
Maximum Displacement 0.000043 0.001800 YES
RMS Displacement 0.000028 0.001200 YES
Predicted change in Energy=-6.856714D-10
Optimization completed.
-- Stationary point found.</pre>

BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/1/18/KERLITALI_BH3_FREQ.log | BH3.log file]]

<pre>Low frequencies --- -7.5936 -1.5614 -0.0055 0.6514 6.9319 7.1055
Low frequencies --- 1162.9677 1213.1634 1213.1661 </pre>

<jmol><jmolApplet>
<title>Optimized BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>KERLITALI_BH3_FREQ.log</uploadedFileContents>
</jmolApplet></jmol>

=== Vibrations ===

[[File:Kt3817bh3ir1.png]]

{| class="wikitable" border="1"
|+ IR vibrations of BH3
! No !! Frequency !! Intensity !! Bend/stretch !! Symmetry !! IR active?
|-
| 1 || 1163 || 93 || Bend || A2 || Yes
|-
| 2 || 1213 || 14 || Bend || E || Yes
|-
| 3 || 1213 || 14 || Bend || E || Yes
|-
| 4 || 2582 || 0 ||Symmetric stretch || A1 || No
|-
| 5 || 2716 || 126 || Asymmetric stretch || E || Yes
|-
| 6 || 2716 || 1126 || Asymmetric stretch || E || Yes
|}

Only 3 peaks are visible in the IR spectrum. Vibration 4 in the table above is not visible because in order for a bend/stretch to be IR active, the dipole moment of the molecule must change but during symmetric stretch, dipole moment remains the same. Vibrations 2 and 3 have the same frequency and peaks are therefore overlapping, the same happens with vibrations 5 and 6.

=== MOs ===

[[File:ktBH3MOdiagram1.png]]

2a1' LCAO MO is exactly the same as the computed version, however for 1e', the two computed MOs are the same, just in different directions, whereas the LCAO MOs indicate that the MOs are different and that one of them has a B-H bond that is outside both MO phases when this is not actually the case. 1a2'' and 3a1' LCAO MOs are exactly the same as their computed versions. 2e' has the same difference that 1e' had: computed MOs are simply different directions but LCAO theory predicts that they have bigger differences and that one of them has a B-H bond that is away from both phases when this is not the case.

== NH3 ==

B3LYP/6-31G(d,p)

[[File:kt3817nh3table.jpg]]

NH3 frequency .log file: [[Media:KT3817_NH3OPT_FREQ.LOG]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000012 0.001800 YES
RMS Displacement 0.000008 0.001200 YES
Predicted change in Energy=-9.844611D-11
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -8.5646 -8.5588 -0.0041 0.0455 0.1784 26.4183
Low frequencies --- 1089.7603 1694.1865 1694.1865 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_nh3opt.log</uploadedFileContents>
</jmolApplet></jmol>

== NH3BH3 ==

B3LYP/6-31G(d,p)

[[File:ktNH3BH3table3.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000114 0.000450 YES
RMS Force 0.000064 0.000300 YES
Maximum Displacement 0.000617 0.001800 YES
RMS Displacement 0.000362 0.001200 YES
Predicted change in Energy=-1.787894D-07
Optimization completed.
-- Stationary point found.</pre>

<pre>Low frequencies --- -827.7026 -33.9159 -19.3033 -3.1260 0.0002 0.0010
Low frequencies --- 0.0011 716.1684 743.0238 </pre>

NH3BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/b/be/REALNH3BH3OPT_FREQ1.LOG | NH3BH3 .log file]]

<jmol><jmolApplet>
<title>Optimized NH3BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>REALNH3BH3OPT_FREQ1.LOG</uploadedFileContents>
</jmolApplet></jmol>

E(BH3)= -26.61532 hartree

E(NH3)= -56.55777 hartree

E(NH3BH3)= -83.22469 hartree

ΔE = -83.22469 + 56.5577 + 26.61532 = -0.05160 hartree = -136 kJ/mol

<pre>Is the B-N bond weak, medium or strong?</pre>
Based on the calculation, I would expect B-N bond to be medium as it is the only forming bond in the reaction and the reaction is exothermic but not as exothermic as the reactions where very strong bonds are formed.

== NI3 ==

B3LYP/GEN

[[File:ktni3table.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000139 0.000450 YES
RMS Force 0.000090 0.000300 YES
Maximum Displacement 0.001129 0.001800 YES
RMS Displacement 0.000796 0.001200 YES
Predicted change in Energy=-1.675756D-07
Optimization completed.
-- Stationary point found.</pre>

NI3 frequency .log file: [[Media:ktni3_freq.log]]

<pre>Low frequencies --- -12.7179 -12.7118 -6.4125 -0.0039 0.0189 0.0621
Low frequencies --- 101.0754 101.0761 147.4556</pre>

<jmol><jmolApplet>
<title>Optimized NI3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>ktni3_freq.log</uploadedFileContents>
</jmolApplet></jmol>

Optimized N-I distance is 2.184 Å.

== Ionic liquids ==

=== N(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktNCtable.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000023 0.000450 YES
RMS Force 0.000008 0.000300 YES
Maximum Displacement 0.001012 0.001800 YES
RMS Displacement 0.000229 0.001200 YES
Predicted change in Energy=-1.320388D-08
Optimization completed.
-- Stationary point found. </pre>

N(CH4)+ frequency .log file: [[Media:kt3817_n(ch3)4+opt1_freq.log]]

<pre>Low frequencies --- -0.0014 -0.0012 -0.0012 34.9548 34.9548 34.9548
Low frequencies --- 217.5087 316.5877 316.5877 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_n(ch3)4+opt1_freq.log</uploadedFileContents>
</jmolApplet></jmol>

=== P(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktPCtable.png]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000138 0.000450 YES
RMS Force 0.000032 0.000300 YES
Maximum Displacement 0.000792 0.001800 YES
RMS Displacement 0.000286 0.001200 YES
Predicted change in Energy=-1.621425D-07
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -0.0034 -0.0032 -0.0023 51.2606 51.2606 51.2606
Low frequencies --- 186.5808 211.3947 211.3947 </pre>

P(CH4)+ frequency .log file: [[Media:Kt3817_P(CH3)4OPT_SYM1.LOG]]

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>Kt3817_P(CH3)4OPT_SYM1.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution===

{| class="wikitable" border="1"
|+
! N(CH3)4+ !! P(CH3)4+
|-
| [[File:kt3817_charge_n1.jpg|250px]] || [[File:kt3817_charge_p.jpg|250px]]
|}

In the N(CH4)+ image above, N has a charge of -0.295, C has a charge of -0.483 and H has a charge of 0.269.
In the P(CH4)+ image above, P has a charge of 1.666, C has a charge of -1.050 and H has a charge of 0.298.

In N(CH3)4+, central atom N has a negative charge, however in P(CH3)4+, central atom P has a positive charge.
In both molecules, C has the largest negative charge of all the atoms and H atoms have positive charge. In P(CH3)4+, P has the largest positive charge located on it, whereas H atoms are the only ones in N(CH3)4+ to have positive charge. In N(CH3)4+, atoms with a negative charge are bonded (N and C) whereas in P(CH3)4+, the charges are alternating as P has positive charge, C that is bonded with it has negative charge and H that is bonded with C has positive charge.

In the classical N(R)4+ charge distribution, N is depicted to have a positive charge since N in ammonium ion has 1 less free electron pair than the neutral N atom. Computed charge distribution, however, shows that N has negative charge and the only atoms carrying positive charge are H atoms.

=== MOs ===

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 9 || [[File:ktNMO9.png]] || [[File:ktNLCAOMO9.png]]
|}

MO 9 pictured above is bonding and occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 10 || [[File:ktNMO10.png]] || [[File:ktNMOLCAO101.png]]
|}

MO 10 pictured above is antibonding if the molecule is pictured as NL4+ but is bonding for methyl group fragment orbitals. Methyl group FOs are in the opposite phase in the molecule compared to the figure which shows how they are formed. MO is occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 19 || [[File:ktNMO19.png]] || [[File:ktNLCAOMO19.png]]
|}

MO 19 pictured above is bonding overall and occupied.

Rep:Mod:kta3817

2019-05-20T16:54:25Z

Kt3817: /* NH3BH3 */

== BH3 molecule ==

B3LYP/6-31G(d,p)

[[File:ktBH3table2.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000011 0.000450 YES
RMS Force 0.000007 0.000300 YES
Maximum Displacement 0.000043 0.001800 YES
RMS Displacement 0.000028 0.001200 YES
Predicted change in Energy=-6.856714D-10
Optimization completed.
-- Stationary point found.</pre>

BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/1/18/KERLITALI_BH3_FREQ.log | BH3.log file]]

<pre>Low frequencies --- -7.5936 -1.5614 -0.0055 0.6514 6.9319 7.1055
Low frequencies --- 1162.9677 1213.1634 1213.1661 </pre>

<jmol><jmolApplet>
<title>Optimized BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>KERLITALI_BH3_FREQ.log</uploadedFileContents>
</jmolApplet></jmol>

=== Vibrations ===

[[File:Kt3817bh3ir1.png]]

{| class="wikitable" border="1"
|+ IR vibrations of BH3
! No !! Frequency !! Intensity !! Bend/stretch !! Symmetry !! IR active?
|-
| 1 || 1163 || 93 || Bend || A2 || Yes
|-
| 2 || 1213 || 14 || Bend || E || Yes
|-
| 3 || 1213 || 14 || Bend || E || Yes
|-
| 4 || 2582 || 0 ||Symmetric stretch || A1 || No
|-
| 5 || 2716 || 126 || Asymmetric stretch || E || Yes
|-
| 6 || 2716 || 1126 || Asymmetric stretch || E || Yes
|}

Only 3 peaks are visible in the IR spectrum. Vibration 4 in the table above is not visible because in order for a bend/stretch to be IR active, the dipole moment of the molecule must change but during symmetric stretch, dipole moment remains the same. Vibrations 2 and 3 have the same frequency and peaks are therefore overlapping, the same happens with vibrations 5 and 6.

=== MOs ===

[[File:ktBH3MOdiagram1.png]]

2a1' LCAO MO is exactly the same as the computed version, however for 1e', the two computed MOs are the same, just in different directions, whereas the LCAO MOs indicate that the MOs are different and that one of them has a B-H bond that is outside both MO phases when this is not actually the case. 1a2'' and 3a1' LCAO MOs are exactly the same as their computed versions. 2e' has the same difference that 1e' had: computed MOs are simply different directions but LCAO theory predicts that they have bigger differences and that one of them has a B-H bond that is away from both phases when this is not the case.

== NH3 ==

B3LYP/6-31G(d,p)

[[File:kt3817nh3table.jpg]]

NH3 frequency .log file: [[Media:KT3817_NH3OPT_FREQ.LOG]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000012 0.001800 YES
RMS Displacement 0.000008 0.001200 YES
Predicted change in Energy=-9.844611D-11
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -8.5646 -8.5588 -0.0041 0.0455 0.1784 26.4183
Low frequencies --- 1089.7603 1694.1865 1694.1865 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_nh3opt.log</uploadedFileContents>
</jmolApplet></jmol>

== NH3BH3 ==

B3LYP/6-31G(d,p)

[[File:ktNH3BH3table3.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000114 0.000450 YES
RMS Force 0.000064 0.000300 YES
Maximum Displacement 0.000617 0.001800 YES
RMS Displacement 0.000362 0.001200 YES
Predicted change in Energy=-1.787894D-07
Optimization completed.
-- Stationary point found.</pre>

<pre>Low frequencies --- -827.7026 -33.9159 -19.3033 -3.1260 0.0002 0.0010
Low frequencies --- 0.0011 716.1684 743.0238 </pre>

NH3BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/b/be/REALNH3BH3OPT_FREQ1.LOG | NH3BH3 .log file]]

<jmol><jmolApplet>
<title>Optimized NH3BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>REALNH3BH3OPT_FREQ1.LOG</uploadedFileContents>
</jmolApplet></jmol>

E(BH3)= -26.61532 hartree

E(NH3)= -56.55777 hartree

E(NH3BH3)= -83.22469 hartree

ΔE = -83.22469 + 56.5577 + 26.61532 = -0.05160 hartree = -136 kJ/mol

<pre>Is the B-N bond weak, medium or strong?</pre>
Based on the calculation, I would expect B-N bond to be medium as it is the only forming bond in the reaction and the reaction is exothermic but not as exothermic as the reactions where very strong bonds are formed.

== NI3 ==

B3LYP/GEN

[[File:ktni3table.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000139 0.000450 YES
RMS Force 0.000090 0.000300 YES
Maximum Displacement 0.001129 0.001800 YES
RMS Displacement 0.000796 0.001200 YES
Predicted change in Energy=-1.675756D-07
Optimization completed.
-- Stationary point found.</pre>

NI3 frequency .log file: [[Media:ktni3_freq.log]]

<pre>Low frequencies --- -12.7179 -12.7118 -6.4125 -0.0039 0.0189 0.0621
Low frequencies --- 101.0754 101.0761 147.4556</pre>

<jmol><jmolApplet>
<title>Optimized NI3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>ktni3_freq.log</uploadedFileContents>
</jmolApplet></jmol>

Optimized N-I distance is 2.184 Å.

== Ionic liquids ==

=== N(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktNCtable.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000023 0.000450 YES
RMS Force 0.000008 0.000300 YES
Maximum Displacement 0.001012 0.001800 YES
RMS Displacement 0.000229 0.001200 YES
Predicted change in Energy=-1.320388D-08
Optimization completed.
-- Stationary point found. </pre>

N(CH4)+ frequency .log file: [[Media:kt3817_n(ch3)4+opt1_freq.log]]

<pre>Low frequencies --- -0.0014 -0.0012 -0.0012 34.9548 34.9548 34.9548
Low frequencies --- 217.5087 316.5877 316.5877 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_n(ch3)4+opt1_freq.log</uploadedFileContents>
</jmolApplet></jmol>

=== P(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktPCtable.png]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000138 0.000450 YES
RMS Force 0.000032 0.000300 YES
Maximum Displacement 0.000792 0.001800 YES
RMS Displacement 0.000286 0.001200 YES
Predicted change in Energy=-1.621425D-07
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -0.0034 -0.0032 -0.0023 51.2606 51.2606 51.2606
Low frequencies --- 186.5808 211.3947 211.3947 </pre>

P(CH4)+ frequency .log file: [[Media:Kt3817_P(CH3)4OPT_SYM1.LOG]]

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>Kt3817_P(CH3)4OPT_SYM1.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution===

{| class="wikitable" border="1"
|+
! N(CH3)4+ !! P(CH3)4+
|-
| [[File:kt3817_charge_n1.jpg|250px]] || [[File:kt3817_charge_p.jpg|250px]]
|}

In the N(CH4)+ image above, N has a charge of -0.295, C has a charge of -0.483 and H has a charge of 0.269.
In the P(CH4)+ image above, P has a charge of 1.666, C has a charge of -1.050 and H has a charge of 0.298.

In N(CH3)4+, central atom N has a negative charge, however in P(CH3)4+, central atom P has a positive charge.
In both molecules, C has the largest negative charge of all the atoms and H atoms have positive charge. In P(CH3)4+, P has the largest positive charge located on it, whereas H atoms are the only ones in N(CH3)4+ to have positive charge. In N(CH3)4+, atoms with a negative charge are bonded (N and C) whereas in P(CH3)4+, the charges are alternating as P has positive charge, C that is bonded with it has negative charge and H that is bonded with C has positive charge.

In the classical N(R)4+ charge distribution, N is depicted to have a positive charge since N in ammonium ion has 1 less free electron pair than the neutral N atom. Computed charge distribution, however, shows that N has negative charge and the only atoms carrying positive charge are H atoms.

=== MOs ===

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 9 || [[File:ktNMO9.png]] || [[File:ktNLCAOMO9.png]]
|}

MO 9 pictured above is bonding and occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 10 || [[File:ktNMO10.png]] || [[File:ktNMOLCAO101.png]]
|}

MO 10 pictured above is antibonding if the molecule is pictured as NL4+ but is bonding for methyl group fragment orbitals. Methyl group FOs are in the opposite phase in the molecule compared to the figure which shows how they are formed.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 19 || [[File:ktNMO19.png]] || [[File:ktNLCAOMO19.png]]
|}

MO 19 pictured above is bonding overall and occupied.

Rep:Mod:kta3817

2019-05-20T16:49:53Z

Kt3817: /* MOs */

== BH3 molecule ==

B3LYP/6-31G(d,p)

[[File:ktBH3table2.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000011 0.000450 YES
RMS Force 0.000007 0.000300 YES
Maximum Displacement 0.000043 0.001800 YES
RMS Displacement 0.000028 0.001200 YES
Predicted change in Energy=-6.856714D-10
Optimization completed.
-- Stationary point found.</pre>

BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/1/18/KERLITALI_BH3_FREQ.log | BH3.log file]]

<pre>Low frequencies --- -7.5936 -1.5614 -0.0055 0.6514 6.9319 7.1055
Low frequencies --- 1162.9677 1213.1634 1213.1661 </pre>

<jmol><jmolApplet>
<title>Optimized BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>KERLITALI_BH3_FREQ.log</uploadedFileContents>
</jmolApplet></jmol>

=== Vibrations ===

[[File:Kt3817bh3ir1.png]]

{| class="wikitable" border="1"
|+ IR vibrations of BH3
! No !! Frequency !! Intensity !! Bend/stretch !! Symmetry !! IR active?
|-
| 1 || 1163 || 93 || Bend || A2 || Yes
|-
| 2 || 1213 || 14 || Bend || E || Yes
|-
| 3 || 1213 || 14 || Bend || E || Yes
|-
| 4 || 2582 || 0 ||Symmetric stretch || A1 || No
|-
| 5 || 2716 || 126 || Asymmetric stretch || E || Yes
|-
| 6 || 2716 || 1126 || Asymmetric stretch || E || Yes
|}

Only 3 peaks are visible in the IR spectrum. Vibration 4 in the table above is not visible because in order for a bend/stretch to be IR active, the dipole moment of the molecule must change but during symmetric stretch, dipole moment remains the same. Vibrations 2 and 3 have the same frequency and peaks are therefore overlapping, the same happens with vibrations 5 and 6.

=== MOs ===

[[File:ktBH3MOdiagram1.png]]

2a1' LCAO MO is exactly the same as the computed version, however for 1e', the two computed MOs are the same, just in different directions, whereas the LCAO MOs indicate that the MOs are different and that one of them has a B-H bond that is outside both MO phases when this is not actually the case. 1a2'' and 3a1' LCAO MOs are exactly the same as their computed versions. 2e' has the same difference that 1e' had: computed MOs are simply different directions but LCAO theory predicts that they have bigger differences and that one of them has a B-H bond that is away from both phases when this is not the case.

== NH3 ==

B3LYP/6-31G(d,p)

[[File:kt3817nh3table.jpg]]

NH3 frequency .log file: [[Media:KT3817_NH3OPT_FREQ.LOG]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000012 0.001800 YES
RMS Displacement 0.000008 0.001200 YES
Predicted change in Energy=-9.844611D-11
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -8.5646 -8.5588 -0.0041 0.0455 0.1784 26.4183
Low frequencies --- 1089.7603 1694.1865 1694.1865 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_nh3opt.log</uploadedFileContents>
</jmolApplet></jmol>

== NH3BH3 ==

B3LYP/6-31G(d,p)

[[File:ktNH3BH3table3.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000114 0.000450 YES
RMS Force 0.000064 0.000300 YES
Maximum Displacement 0.000617 0.001800 YES
RMS Displacement 0.000362 0.001200 YES
Predicted change in Energy=-1.787894D-07
Optimization completed.
-- Stationary point found.</pre>

<pre>Low frequencies --- -827.7026 -33.9159 -19.3033 -3.1260 0.0002 0.0010
Low frequencies --- 0.0011 716.1684 743.0238 </pre>

NH3BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/b/be/REALNH3BH3OPT_FREQ1.LOG | NH3BH3 .log file]]

<jmol><jmolApplet>
<title>Optimized NH3BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>REALNH3BH3OPT_FREQ1.LOG</uploadedFileContents>
</jmolApplet></jmol>

E(BH3)= -26.61532 hartree

E(NH3)= -56.55777 hartree

E(NH3BH3)= -83.22469 hartree

ΔE = -83.22469 + 56.5577 + 26.61532 = -0.05160 hartree = -136 kJ/mol

<pre>Is the B-N bond weak, medium or strong?</pre>
Based on the calculation, I would expect B-N bond to be weak as it is the only forming bond in the reaction and the reaction is

== NI3 ==

B3LYP/GEN

[[File:ktni3table.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000139 0.000450 YES
RMS Force 0.000090 0.000300 YES
Maximum Displacement 0.001129 0.001800 YES
RMS Displacement 0.000796 0.001200 YES
Predicted change in Energy=-1.675756D-07
Optimization completed.
-- Stationary point found.</pre>

NI3 frequency .log file: [[Media:ktni3_freq.log]]

<pre>Low frequencies --- -12.7179 -12.7118 -6.4125 -0.0039 0.0189 0.0621
Low frequencies --- 101.0754 101.0761 147.4556</pre>

<jmol><jmolApplet>
<title>Optimized NI3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>ktni3_freq.log</uploadedFileContents>
</jmolApplet></jmol>

Optimized N-I distance is 2.184 Å.

== Ionic liquids ==

=== N(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktNCtable.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000023 0.000450 YES
RMS Force 0.000008 0.000300 YES
Maximum Displacement 0.001012 0.001800 YES
RMS Displacement 0.000229 0.001200 YES
Predicted change in Energy=-1.320388D-08
Optimization completed.
-- Stationary point found. </pre>

N(CH4)+ frequency .log file: [[Media:kt3817_n(ch3)4+opt1_freq.log]]

<pre>Low frequencies --- -0.0014 -0.0012 -0.0012 34.9548 34.9548 34.9548
Low frequencies --- 217.5087 316.5877 316.5877 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_n(ch3)4+opt1_freq.log</uploadedFileContents>
</jmolApplet></jmol>

=== P(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktPCtable.png]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000138 0.000450 YES
RMS Force 0.000032 0.000300 YES
Maximum Displacement 0.000792 0.001800 YES
RMS Displacement 0.000286 0.001200 YES
Predicted change in Energy=-1.621425D-07
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -0.0034 -0.0032 -0.0023 51.2606 51.2606 51.2606
Low frequencies --- 186.5808 211.3947 211.3947 </pre>

P(CH4)+ frequency .log file: [[Media:Kt3817_P(CH3)4OPT_SYM1.LOG]]

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>Kt3817_P(CH3)4OPT_SYM1.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution===

{| class="wikitable" border="1"
|+
! N(CH3)4+ !! P(CH3)4+
|-
| [[File:kt3817_charge_n1.jpg|250px]] || [[File:kt3817_charge_p.jpg|250px]]
|}

In the N(CH4)+ image above, N has a charge of -0.295, C has a charge of -0.483 and H has a charge of 0.269.
In the P(CH4)+ image above, P has a charge of 1.666, C has a charge of -1.050 and H has a charge of 0.298.

In N(CH3)4+, central atom N has a negative charge, however in P(CH3)4+, central atom P has a positive charge.
In both molecules, C has the largest negative charge of all the atoms and H atoms have positive charge. In P(CH3)4+, P has the largest positive charge located on it, whereas H atoms are the only ones in N(CH3)4+ to have positive charge. In N(CH3)4+, atoms with a negative charge are bonded (N and C) whereas in P(CH3)4+, the charges are alternating as P has positive charge, C that is bonded with it has negative charge and H that is bonded with C has positive charge.

In the classical N(R)4+ charge distribution, N is depicted to have a positive charge since N in ammonium ion has 1 less free electron pair than the neutral N atom. Computed charge distribution, however, shows that N has negative charge and the only atoms carrying positive charge are H atoms.

=== MOs ===

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 9 || [[File:ktNMO9.png]] || [[File:ktNLCAOMO9.png]]
|}

MO 9 pictured above is bonding and occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 10 || [[File:ktNMO10.png]] || [[File:ktNMOLCAO101.png]]
|}

MO 10 pictured above is antibonding if the molecule is pictured as NL4+ but is bonding for methyl group fragment orbitals. Methyl group FOs are in the opposite phase in the molecule compared to the figure which shows how they are formed.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 19 || [[File:ktNMO19.png]] || [[File:ktNLCAOMO19.png]]
|}

MO 19 pictured above is bonding overall and occupied.

File:KtBH3MOdiagram1.png

2019-05-20T16:49:30Z

Kt3817:

Rep:Mod:kta3817

2019-05-20T16:49:07Z

Kt3817: /* MOs */

== BH3 molecule ==

B3LYP/6-31G(d,p)

[[File:ktBH3table2.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000011 0.000450 YES
RMS Force 0.000007 0.000300 YES
Maximum Displacement 0.000043 0.001800 YES
RMS Displacement 0.000028 0.001200 YES
Predicted change in Energy=-6.856714D-10
Optimization completed.
-- Stationary point found.</pre>

BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/1/18/KERLITALI_BH3_FREQ.log | BH3.log file]]

<pre>Low frequencies --- -7.5936 -1.5614 -0.0055 0.6514 6.9319 7.1055
Low frequencies --- 1162.9677 1213.1634 1213.1661 </pre>

<jmol><jmolApplet>
<title>Optimized BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>KERLITALI_BH3_FREQ.log</uploadedFileContents>
</jmolApplet></jmol>

=== Vibrations ===

[[File:Kt3817bh3ir1.png]]

{| class="wikitable" border="1"
|+ IR vibrations of BH3
! No !! Frequency !! Intensity !! Bend/stretch !! Symmetry !! IR active?
|-
| 1 || 1163 || 93 || Bend || A2 || Yes
|-
| 2 || 1213 || 14 || Bend || E || Yes
|-
| 3 || 1213 || 14 || Bend || E || Yes
|-
| 4 || 2582 || 0 ||Symmetric stretch || A1 || No
|-
| 5 || 2716 || 126 || Asymmetric stretch || E || Yes
|-
| 6 || 2716 || 1126 || Asymmetric stretch || E || Yes
|}

Only 3 peaks are visible in the IR spectrum. Vibration 4 in the table above is not visible because in order for a bend/stretch to be IR active, the dipole moment of the molecule must change but during symmetric stretch, dipole moment remains the same. Vibrations 2 and 3 have the same frequency and peaks are therefore overlapping, the same happens with vibrations 5 and 6.

=== MOs ===

[[File:ktBH3MOdiagram.png]]

2a1' LCAO MO is exactly the same as the computed version, however for 1e', the two computed MOs are the same, just in different directions, whereas the LCAO MOs indicate that the MOs are different and that one of them has a B-H bond that is outside both MO phases when this is not actually the case. 1a2'' and 3a1' LCAO MOs are exactly the same as their computed versions. 2e' has the same difference that 1e' had: computed MOs are simply different directions but LCAO theory predicts that they have bigger differences and that one of them has a B-H bond that is away from both phases when this is not the case.

== NH3 ==

B3LYP/6-31G(d,p)

[[File:kt3817nh3table.jpg]]

NH3 frequency .log file: [[Media:KT3817_NH3OPT_FREQ.LOG]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000012 0.001800 YES
RMS Displacement 0.000008 0.001200 YES
Predicted change in Energy=-9.844611D-11
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -8.5646 -8.5588 -0.0041 0.0455 0.1784 26.4183
Low frequencies --- 1089.7603 1694.1865 1694.1865 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_nh3opt.log</uploadedFileContents>
</jmolApplet></jmol>

== NH3BH3 ==

B3LYP/6-31G(d,p)

[[File:ktNH3BH3table3.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000114 0.000450 YES
RMS Force 0.000064 0.000300 YES
Maximum Displacement 0.000617 0.001800 YES
RMS Displacement 0.000362 0.001200 YES
Predicted change in Energy=-1.787894D-07
Optimization completed.
-- Stationary point found.</pre>

<pre>Low frequencies --- -827.7026 -33.9159 -19.3033 -3.1260 0.0002 0.0010
Low frequencies --- 0.0011 716.1684 743.0238 </pre>

NH3BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/b/be/REALNH3BH3OPT_FREQ1.LOG | NH3BH3 .log file]]

<jmol><jmolApplet>
<title>Optimized NH3BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>REALNH3BH3OPT_FREQ1.LOG</uploadedFileContents>
</jmolApplet></jmol>

E(BH3)= -26.61532 hartree

E(NH3)= -56.55777 hartree

E(NH3BH3)= -83.22469 hartree

ΔE = -83.22469 + 56.5577 + 26.61532 = -0.05160 hartree = -136 kJ/mol

<pre>Is the B-N bond weak, medium or strong?</pre>
Based on the calculation, I would expect B-N bond to be weak as it is the only forming bond in the reaction and the reaction is

== NI3 ==

B3LYP/GEN

[[File:ktni3table.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000139 0.000450 YES
RMS Force 0.000090 0.000300 YES
Maximum Displacement 0.001129 0.001800 YES
RMS Displacement 0.000796 0.001200 YES
Predicted change in Energy=-1.675756D-07
Optimization completed.
-- Stationary point found.</pre>

NI3 frequency .log file: [[Media:ktni3_freq.log]]

<pre>Low frequencies --- -12.7179 -12.7118 -6.4125 -0.0039 0.0189 0.0621
Low frequencies --- 101.0754 101.0761 147.4556</pre>

<jmol><jmolApplet>
<title>Optimized NI3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>ktni3_freq.log</uploadedFileContents>
</jmolApplet></jmol>

Optimized N-I distance is 2.184 Å.

== Ionic liquids ==

=== N(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktNCtable.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000023 0.000450 YES
RMS Force 0.000008 0.000300 YES
Maximum Displacement 0.001012 0.001800 YES
RMS Displacement 0.000229 0.001200 YES
Predicted change in Energy=-1.320388D-08
Optimization completed.
-- Stationary point found. </pre>

N(CH4)+ frequency .log file: [[Media:kt3817_n(ch3)4+opt1_freq.log]]

<pre>Low frequencies --- -0.0014 -0.0012 -0.0012 34.9548 34.9548 34.9548
Low frequencies --- 217.5087 316.5877 316.5877 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_n(ch3)4+opt1_freq.log</uploadedFileContents>
</jmolApplet></jmol>

=== P(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktPCtable.png]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000138 0.000450 YES
RMS Force 0.000032 0.000300 YES
Maximum Displacement 0.000792 0.001800 YES
RMS Displacement 0.000286 0.001200 YES
Predicted change in Energy=-1.621425D-07
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -0.0034 -0.0032 -0.0023 51.2606 51.2606 51.2606
Low frequencies --- 186.5808 211.3947 211.3947 </pre>

P(CH4)+ frequency .log file: [[Media:Kt3817_P(CH3)4OPT_SYM1.LOG]]

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>Kt3817_P(CH3)4OPT_SYM1.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution===

{| class="wikitable" border="1"
|+
! N(CH3)4+ !! P(CH3)4+
|-
| [[File:kt3817_charge_n1.jpg|250px]] || [[File:kt3817_charge_p.jpg|250px]]
|}

In the N(CH4)+ image above, N has a charge of -0.295, C has a charge of -0.483 and H has a charge of 0.269.
In the P(CH4)+ image above, P has a charge of 1.666, C has a charge of -1.050 and H has a charge of 0.298.

In N(CH3)4+, central atom N has a negative charge, however in P(CH3)4+, central atom P has a positive charge.
In both molecules, C has the largest negative charge of all the atoms and H atoms have positive charge. In P(CH3)4+, P has the largest positive charge located on it, whereas H atoms are the only ones in N(CH3)4+ to have positive charge. In N(CH3)4+, atoms with a negative charge are bonded (N and C) whereas in P(CH3)4+, the charges are alternating as P has positive charge, C that is bonded with it has negative charge and H that is bonded with C has positive charge.

In the classical N(R)4+ charge distribution, N is depicted to have a positive charge since N in ammonium ion has 1 less free electron pair than the neutral N atom. Computed charge distribution, however, shows that N has negative charge and the only atoms carrying positive charge are H atoms.

=== MOs ===

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 9 || [[File:ktNMO9.png]] || [[File:ktNLCAOMO9.png]]
|}

MO 9 pictured above is bonding and occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 10 || [[File:ktNMO10.png]] || [[File:ktNMOLCAO101.png]]
|}

MO 10 pictured above is antibonding if the molecule is pictured as NL4+ but is bonding for methyl group fragment orbitals. Methyl group FOs are in the opposite phase in the molecule compared to the figure which shows how they are formed.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 19 || [[File:ktNMO19.png]] || [[File:ktNLCAOMO19.png]]
|}

MO 19 pictured above is bonding overall and occupied.

Rep:Mod:kta3817

2019-05-20T16:36:34Z

Kt3817: /* MOs */

== BH3 molecule ==

B3LYP/6-31G(d,p)

[[File:ktBH3table2.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000011 0.000450 YES
RMS Force 0.000007 0.000300 YES
Maximum Displacement 0.000043 0.001800 YES
RMS Displacement 0.000028 0.001200 YES
Predicted change in Energy=-6.856714D-10
Optimization completed.
-- Stationary point found.</pre>

BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/1/18/KERLITALI_BH3_FREQ.log | BH3.log file]]

<pre>Low frequencies --- -7.5936 -1.5614 -0.0055 0.6514 6.9319 7.1055
Low frequencies --- 1162.9677 1213.1634 1213.1661 </pre>

<jmol><jmolApplet>
<title>Optimized BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>KERLITALI_BH3_FREQ.log</uploadedFileContents>
</jmolApplet></jmol>

=== Vibrations ===

[[File:Kt3817bh3ir1.png]]

{| class="wikitable" border="1"
|+ IR vibrations of BH3
! No !! Frequency !! Intensity !! Bend/stretch !! Symmetry !! IR active?
|-
| 1 || 1163 || 93 || Bend || A2 || Yes
|-
| 2 || 1213 || 14 || Bend || E || Yes
|-
| 3 || 1213 || 14 || Bend || E || Yes
|-
| 4 || 2582 || 0 ||Symmetric stretch || A1 || No
|-
| 5 || 2716 || 126 || Asymmetric stretch || E || Yes
|-
| 6 || 2716 || 1126 || Asymmetric stretch || E || Yes
|}

Only 3 peaks are visible in the IR spectrum. Vibration 4 in the table above is not visible because in order for a bend/stretch to be IR active, the dipole moment of the molecule must change but during symmetric stretch, dipole moment remains the same. Vibrations 2 and 3 have the same frequency and peaks are therefore overlapping, the same happens with vibrations 5 and 6.

=== MOs ===

[[File:ktBH3MOdiagram.png]]

== NH3 ==

B3LYP/6-31G(d,p)

[[File:kt3817nh3table.jpg]]

NH3 frequency .log file: [[Media:KT3817_NH3OPT_FREQ.LOG]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000012 0.001800 YES
RMS Displacement 0.000008 0.001200 YES
Predicted change in Energy=-9.844611D-11
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -8.5646 -8.5588 -0.0041 0.0455 0.1784 26.4183
Low frequencies --- 1089.7603 1694.1865 1694.1865 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_nh3opt.log</uploadedFileContents>
</jmolApplet></jmol>

== NH3BH3 ==

B3LYP/6-31G(d,p)

[[File:ktNH3BH3table3.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000114 0.000450 YES
RMS Force 0.000064 0.000300 YES
Maximum Displacement 0.000617 0.001800 YES
RMS Displacement 0.000362 0.001200 YES
Predicted change in Energy=-1.787894D-07
Optimization completed.
-- Stationary point found.</pre>

<pre>Low frequencies --- -827.7026 -33.9159 -19.3033 -3.1260 0.0002 0.0010
Low frequencies --- 0.0011 716.1684 743.0238 </pre>

NH3BH3 frequency .log file: [[https://wiki.ch.ic.ac.uk/wiki/images/b/be/REALNH3BH3OPT_FREQ1.LOG | NH3BH3 .log file]]

<jmol><jmolApplet>
<title>Optimized NH3BH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>REALNH3BH3OPT_FREQ1.LOG</uploadedFileContents>
</jmolApplet></jmol>

E(BH3)= -26.61532 hartree

E(NH3)= -56.55777 hartree

E(NH3BH3)= -83.22469 hartree

ΔE = -83.22469 + 56.5577 + 26.61532 = -0.05160 hartree = -136 kJ/mol

<pre>Is the B-N bond weak, medium or strong?</pre>
Based on the calculation, I would expect B-N bond to be weak as it is the only forming bond in the reaction and the reaction is

== NI3 ==

B3LYP/GEN

[[File:ktni3table.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000139 0.000450 YES
RMS Force 0.000090 0.000300 YES
Maximum Displacement 0.001129 0.001800 YES
RMS Displacement 0.000796 0.001200 YES
Predicted change in Energy=-1.675756D-07
Optimization completed.
-- Stationary point found.</pre>

NI3 frequency .log file: [[Media:ktni3_freq.log]]

<pre>Low frequencies --- -12.7179 -12.7118 -6.4125 -0.0039 0.0189 0.0621
Low frequencies --- 101.0754 101.0761 147.4556</pre>

<jmol><jmolApplet>
<title>Optimized NI3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>ktni3_freq.log</uploadedFileContents>
</jmolApplet></jmol>

Optimized N-I distance is 2.184 Å.

== Ionic liquids ==

=== N(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktNCtable.png]]

<pre>Item Value Threshold Converged?
Maximum Force 0.000023 0.000450 YES
RMS Force 0.000008 0.000300 YES
Maximum Displacement 0.001012 0.001800 YES
RMS Displacement 0.000229 0.001200 YES
Predicted change in Energy=-1.320388D-08
Optimization completed.
-- Stationary point found. </pre>

N(CH4)+ frequency .log file: [[Media:kt3817_n(ch3)4+opt1_freq.log]]

<pre>Low frequencies --- -0.0014 -0.0012 -0.0012 34.9548 34.9548 34.9548
Low frequencies --- 217.5087 316.5877 316.5877 </pre>

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>kt3817_n(ch3)4+opt1_freq.log</uploadedFileContents>
</jmolApplet></jmol>

=== P(CH3)4+ ===
B3LYP/6-31G(d,p)

[[File:ktPCtable.png]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000138 0.000450 YES
RMS Force 0.000032 0.000300 YES
Maximum Displacement 0.000792 0.001800 YES
RMS Displacement 0.000286 0.001200 YES
Predicted change in Energy=-1.621425D-07
Optimization completed.
-- Stationary point found. </pre>

<pre>Low frequencies --- -0.0034 -0.0032 -0.0023 51.2606 51.2606 51.2606
Low frequencies --- 186.5808 211.3947 211.3947 </pre>

P(CH4)+ frequency .log file: [[Media:Kt3817_P(CH3)4OPT_SYM1.LOG]]

<jmol><jmolApplet>
<title>Optimized NH3 molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>Kt3817_P(CH3)4OPT_SYM1.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution===

{| class="wikitable" border="1"
|+
! N(CH3)4+ !! P(CH3)4+
|-
| [[File:kt3817_charge_n1.jpg|250px]] || [[File:kt3817_charge_p.jpg|250px]]
|}

In the N(CH4)+ image above, N has a charge of -0.295, C has a charge of -0.483 and H has a charge of 0.269.
In the P(CH4)+ image above, P has a charge of 1.666, C has a charge of -1.050 and H has a charge of 0.298.

In N(CH3)4+, central atom N has a negative charge, however in P(CH3)4+, central atom P has a positive charge.
In both molecules, C has the largest negative charge of all the atoms and H atoms have positive charge. In P(CH3)4+, P has the largest positive charge located on it, whereas H atoms are the only ones in N(CH3)4+ to have positive charge. In N(CH3)4+, atoms with a negative charge are bonded (N and C) whereas in P(CH3)4+, the charges are alternating as P has positive charge, C that is bonded with it has negative charge and H that is bonded with C has positive charge.

In the classical N(R)4+ charge distribution, N is depicted to have a positive charge since N in ammonium ion has 1 less free electron pair than the neutral N atom. Computed charge distribution, however, shows that N has negative charge and the only atoms carrying positive charge are H atoms.

=== MOs ===

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 9 || [[File:ktNMO9.png]] || [[File:ktNLCAOMO9.png]]
|}

MO 9 pictured above is bonding and occupied.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 10 || [[File:ktNMO10.png]] || [[File:ktNMOLCAO101.png]]
|}

MO 10 pictured above is antibonding if the molecule is pictured as NL4+ but is bonding for methyl group fragment orbitals. Methyl group FOs are in the opposite phase in the molecule compared to the figure which shows how they are formed.

{| class="wikitable" border="1"
|+
! No !! MO !! LCAO MO
|-
| 19 || [[File:ktNMO19.png]] || [[File:ktNLCAOMO19.png]]
|}

MO 19 pictured above is bonding overall and occupied.