File:HighHF.gs2016.png

2018-05-11T14:51:49Z

Gs2016:

MRD:gs2016

2018-05-11T14:43:20Z

Gs2016:

==Exercise 1==

'''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface?'''

The gradient for both the transition state and a minimum will be 0. The 2nd derivative for a transition state will be negative and that of a minimum will be positive.

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

At the transition state Rab=Rbc=0.90774255. The forces AB and BC at this point are 0 and no oscillation is seen. A single dot is seen at the minimum of the Potential energy graph on the Surface plot. The internuclear graph shows that the distances between all 3 atoms remain the same over time and there is no interaction and therefore reaction at this position.

[[File:IDG.gs2016.png|thumb|center|Internuclear Distance graph of when AB=BC=0.90774255]]

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

The reaction continues to the products when BC (r1) is increased by 0.01. The mep internuclear graph shows that the distance AB decreases while BC increases so the reaction moves towards the product. No vibrational energy is seen. However, with the dynamic calculations the distance AB decreases and then vibrational energy is seen after the reaction goes to products.

'''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! '''p1''' !! '''p2''' !! '''Total energy''' !! '''Reactive(R)/Unreactive(U)'''
|-
| -1.25 || -2.5 || -99.018 || R
|-
| -1.5 || -2.0 || 100.456 || U
|-
| -1.5 || -2.5 || -98.956 || R
|-
| -2.5 || -5.0 || -84.965 || U
|-
| -2.5 || -5.2 || -83.416 || R
|}

[[File:Traj1.gs2016.png|thumb|center|A combines with BC with the right momentum to react so AB is formed and vibrations in AB are seen.]]

[[File:Traj2.gs2016.png|thumb|center|A does not combine with BC at the transition state so only the reactants are seen with BC vibrations.]]

[[File:Traj3.gs2016.png|thumb|center|A combines with BC at the transition state and AB is formed.]]

[[File:Traj4.gs2016.png|thumb|center|Penultimate conditions where the product forms and then dissociates as BC is reformed and A moves away.]]

[[File:Traj5.gs2016.png|thumb|center|AB is formed, then dissociates as BC is reformed, and then AB is formed again as C moves away.]]

'''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?'''

Atomic nuclei behave according to classical mechanics.

If atoms do not collide with enough energy to form the transition state, the reaction will not occur.

With this said,

==Exercise 2==

'''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?'''

[[File:FinalSP.gs2016.png|thumb|center|Surface plot of H-H-F reaction]]

F + H2 is exothermic because the energy of the reactants is more than the energy of the products. H + HF is endothermic as it is the reverse of the first reaction and this time the products have more energy than the reactants. The H-F bond is stronger than H-H because more energy has to be put into it to break the bond.

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

AB=1.8115

BC=0.7445

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

Hammond's postulate states that in an exothermic reaction the energy of the transition state closely resembles the energy of the reactants as there is likely to be less rearrangement of the reactants. It also states that in an endothermic reaction the energy of the transition state resembles the energy of the products more closely as there has been more rearrangement of the reactants.

Activation energy of F + H2 = 0.15

Activation energy of H + HF = 17.002

Plotting the internuclear distance against time graph using mep helps show whether products have been formed, the reactants are present or the reaction is at the transition state.

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

Energy from breaking down of H2 to form HF is used as firstly transitional energy for formation of the HF bond as the H moves away and the energy left is used for vibration of the HF molecule. IR of the resulting product could be taken, peaks are evidence of vibration in the molecule. (NMR could be used to confirm that the HF molecule has been formed initially).

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

For exothermic reactions, translational energy is more useful for the formation of products than vibrational energy. This is seen from the little change when the momentum of vibration of H2 is increased but much larger effect (reaction going to products--HF formed) when the momentum with which F hits H2 is increased.

[[File:HighHHCP.gs2016.png|thumb|center|Reactants not moving to products when HH vibrational energy is increased.]]

[[File:HighHFCP.gs2016.png|thumb|center|Reactants moving to products when F momentum is increased, translational energy seen.]]

For endothermic reactions,

File:HighHHCP.gs2016.png

2018-05-11T14:36:34Z

Gs2016:

File:HighHFCP.gs2016.png

2018-05-11T14:36:07Z

Gs2016:

==Exercise 1==

What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface?

The gradient for both the transition state and a minimum will be 0. The 2nd derivative for a transition state will be negative and that of a minimum will be positive.

Rep:Mod:217GASE

2017-02-24T12:17:02Z

Gs2016: /* Molecular Orbitals */

==NH3 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -56.55776873

The RMS gradient was 0.00000485

Point group of the molecule = C3v

The N-H bond distance is 1.01798 angstroms

The H-N-H bong angle is 105.741 degrees

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986272D-10
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016_NH3_OPT.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[Media:GS2016_NH3_OPT.LOG| here]]

[[File:gs2016_Display_Vibrations.png]]

6 modes are expected from the 3N-6 rule.

Modes 2 and 3,as well as 5 and 6 have degenerate energies.

Modes 1,2 and 3 are bending. Modes 4,5 and 6 are stretching.

Modes 1,3 and 4 are symmetric.

Mode 1 is the umbrella one.

4 bands would be degenerate.

The charge on the Nitrogen atom is -1.125 and the charge on each hydrogen atom was 0.375.

I would have expected the Nitrogen charge to be -3 and each hydrogen having +1 charge.

===The Hydrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -1.17853936

The RMS gradient was 0.00000017

Point group of the molecule = D*H

The H-H bond distance is 0.74279 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164080D-13
</pre>

The only frequency is 4465.68, so there are no negative frequencies.

===The Nitrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -109.52412868

The RMS gradient was 0.00000060

Point group of the molecule = D*H

The N-N bond distance is 1.10550 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.400879D-13
</pre>

The only frequency is 2457.33, so there are no negative frequencies.

===Energy Changes===

The equation of ammonia formation is as follows:

3H2 + N2 → 2NH3

E(NH3)= -56.55776873
2*E(NH3)= -113.1155375
E(N2)= -109.52412868
E(H2)= -1.17853936
3*E(H2)= -3.53561808

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of ammonia as follows:

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579074

Energy change in kJ/mol = -146.49

The ammonia product is more stable than its gaseous reactants because the energy change calculated is negative, an exothermic reaction. The ammonia has a lower energy.
The calculated value is however much higher than experimentally calculated as shown in the following [[Enthalpies of formation|https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation]]. This is because the conditions used in the Haber process make it possible to reduce the energy used in the formation of ammonia.

==SiH4 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -291.88802760

The RMS gradient was 0.00000002

Point group of the molecule = TD

The Si-H bond distance is 1.48485 angstroms

The H-Si-H bong angle is 109.471 degrees, so the molecule has a terahedral shape.

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-2.452659D-14
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016SIH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[https://wiki.ch.ic.ac.uk/wiki/images/7/75/GS2016_SIH4_OPT.LOG | here]]

[[File:gs2016_SiH4_vibrations.png]]

There are 9 vibrational modes observed.

The modes 1,2,3 are degenerate. As well as 4,5 and 7,8,9, with 3 different values.

2 bands would be seen because there are only 2 infrared active modes (figures).

Modes 1 to 5 are all bending, and modes 6 to 9 are stretching.

Modes 4,5 and 6 are symmetric.

The Si atom has a charge of 0.629, while each hydrogen atom has a charge of -0.157. This is different from the expected values which were Si:-4 and H:+1.

===Energy Changes===

The equation of SiH4 formation is as follows:

Si + 2H2 → SiH4

E(H2)= -1.17853936
2*E(H2)= -2.35707872
E(Si)= 0
E(SiH4)= -291.88802760

Note that the energy of Si is 0 because it is a unimolecular atom with no bonds.

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of SiH4 as follows:

ΔE=E(SiH4)-[E(Si)+2*E(H2)]= -289.5309519

Energy change in kJ/mol = -760163.51

This show that SiH4 is more stable than its reactants as the reaction is exothermic.

===Molecular Orbitals===

[[File:gs2016_MO6.png]]

This is a 3s molecular orbital. It does not contribute to the bonding because its energy is lower than that of the 3p, i.e, there is no degeneracy so the orbital is not hybridised.

[[File:gs2016_MO2.png]]

[[File:gs2016_MO3.png]]

[[File:gs2016_MO4.png]]

These are all degenerate 2p orbitals. They have the same energies and are facing different directions (2px, 2py and 2pz). The 3p orbitals have the same shapes and directions but are much larger. The 2p orbitals have no effects on the MOs. The 3p orbitals however bond with the hydrogen electrons and are the LUMO.

[[File:gs2016_MO7.png]]

This is the orbital with the deepest energy. It is a filled 1s orbital and has no effect on bonding. It is not visible because it is very small and within the Si atom.

Rep:Mod:217GASE

2017-02-24T12:14:22Z

Gs2016:

==NH3 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -56.55776873

The RMS gradient was 0.00000485

Point group of the molecule = C3v

The N-H bond distance is 1.01798 angstroms

The H-N-H bong angle is 105.741 degrees

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986272D-10
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016_NH3_OPT.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[Media:GS2016_NH3_OPT.LOG| here]]

[[File:gs2016_Display_Vibrations.png]]

6 modes are expected from the 3N-6 rule.

Modes 2 and 3,as well as 5 and 6 have degenerate energies.

Modes 1,2 and 3 are bending. Modes 4,5 and 6 are stretching.

Modes 1,3 and 4 are symmetric.

Mode 1 is the umbrella one.

4 bands would be degenerate.

The charge on the Nitrogen atom is -1.125 and the charge on each hydrogen atom was 0.375.

I would have expected the Nitrogen charge to be -3 and each hydrogen having +1 charge.

===The Hydrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -1.17853936

The RMS gradient was 0.00000017

Point group of the molecule = D*H

The H-H bond distance is 0.74279 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164080D-13
</pre>

The only frequency is 4465.68, so there are no negative frequencies.

===The Nitrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -109.52412868

The RMS gradient was 0.00000060

Point group of the molecule = D*H

The N-N bond distance is 1.10550 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.400879D-13
</pre>

The only frequency is 2457.33, so there are no negative frequencies.

===Energy Changes===

The equation of ammonia formation is as follows:

3H2 + N2 → 2NH3

E(NH3)= -56.55776873
2*E(NH3)= -113.1155375
E(N2)= -109.52412868
E(H2)= -1.17853936
3*E(H2)= -3.53561808

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of ammonia as follows:

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579074

Energy change in kJ/mol = -146.49

The ammonia product is more stable than its gaseous reactants because the energy change calculated is negative, an exothermic reaction. The ammonia has a lower energy.
The calculated value is however much higher than experimentally calculated as shown in the following [[Enthalpies of formation|https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation]]. This is because the conditions used in the Haber process make it possible to reduce the energy used in the formation of ammonia.

==SiH4 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -291.88802760

The RMS gradient was 0.00000002

Point group of the molecule = TD

The Si-H bond distance is 1.48485 angstroms

The H-Si-H bong angle is 109.471 degrees, so the molecule has a terahedral shape.

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-2.452659D-14
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016SIH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[https://wiki.ch.ic.ac.uk/wiki/images/7/75/GS2016_SIH4_OPT.LOG | here]]

[[File:gs2016_SiH4_vibrations.png]]

There are 9 vibrational modes observed.

The modes 1,2,3 are degenerate. As well as 4,5 and 7,8,9, with 3 different values.

2 bands would be seen because there are only 2 infrared active modes (figures).

Modes 1 to 5 are all bending, and modes 6 to 9 are stretching.

Modes 4,5 and 6 are symmetric.

The Si atom has a charge of 0.629, while each hydrogen atom has a charge of -0.157. This is different from the expected values which were Si:-4 and H:+1.

===Energy Changes===

The equation of SiH4 formation is as follows:

Si + 2H2 → SiH4

E(H2)= -1.17853936
2*E(H2)= -2.35707872
E(Si)= 0
E(SiH4)= -291.88802760

Note that the energy of Si is 0 because it is a unimolecular atom with no bonds.

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of SiH4 as follows:

ΔE=E(SiH4)-[E(Si)+2*E(H2)]= -289.5309519

Energy change in kJ/mol = -760163.51

This show that SiH4 is more stable than its reactants as the reaction is exothermic.

===Molecular Orbitals===

[[File:gs2016_MO6.png]]

This is a 3s molecular orbital.

[[File:gs2016_MO2.png]]

[[File:gs2016_MO3.png]]

[[File:gs2016_MO4.png]]

These are all degenerate 2p orbitals. They have the same energies and are facing different directions (2px, 2py and 2pz). The 3p orbitals have the same shapes and directions but are much larger. The 2p orbitals have no effects on the MOs. The 3p orbitals however bond with the hydrogen electrons and are the LUMO.

[[File:gs2016_MO7.png]]

This is the orbital with the deepest energy. It is a filled 1s orbital and has no effect on bonding. It is not visible because it is very small and within the Si atom.

Rep:Mod:217GASE

2017-02-24T12:07:11Z

Gs2016:

==NH3 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -56.55776873

The RMS gradient was 0.00000485

Point group of the molecule = C3v

The NH bond distance is 1.01798 angstroms

The H-N-H bong angle is 105.741 degrees

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986272D-10
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016_NH3_OPT.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[Media:GS2016_NH3_OPT.LOG| here]]

[[File:gs2016_Display_Vibrations.png]]

6 modes are expected from the 3N-6 rule.

Modes 2 and 3,as well as 5 and 6 have degenerate energies.

Modes 1,2 and 3 are bending. Modes 4,5 and 6 are stretching.

Modes 1,3 and 4 are symmetric.

Mode 1 is the umbrella one.

4 bands would be degenerate.

The charge on the Nitrogen atom is -1.125 and the charge on each hydrogen atom was 0.375.

I would have expected the Nitrogen charge to be -3 and each hydrogen having +1 charge.

===The Hydrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -1.17853936

The RMS gradient was 0.00000017

Point group of the molecule = D*H

The H-H bond distance is 0.74279 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164080D-13
</pre>

The only frequency is 4465.68, so there are no negative frequencies.

===The Nitrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -109.52412868

The RMS gradient was 0.00000060

Point group of the molecule = D*H

The N-N bond distance is 1.10550 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.400879D-13
</pre>

The only frequency is 2457.33, so there are no negative frequencies.

===Energy Changes===

The equation of ammonia formation is as follows:

3H2 + N2 → 2NH3

E(NH3)= -56.55776873
2*E(NH3)= -113.1155375
E(N2)= -109.52412868
E(H2)= -1.17853936
3*E(H2)= -3.53561808

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of ammonia as follows:

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579074

Energy change in kJ/mol = -146.49

The ammonia product is more stable than its gaseous reactants because the energy change calculated is negative, an exothermic reaction. The ammonia has a lower energy.
The calculated value is however much higher than experimentally calculated as shown in the following [[Enthalpies of formation|https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation]]. This is because the conditions used in the Haber process make it possible to reduce the energy used in the formation of ammonia.

==SiH4 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -291.88802760

The RMS gradient was 0.00000002

Point group of the molecule = TD

The Si-H bond distance is 1.48485 angstroms

The H-N-H bong angle is 109.471 degrees, so the molecule has a terahedral shape.

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-2.452659D-14
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016SIH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[https://wiki.ch.ic.ac.uk/wiki/images/7/75/GS2016_SIH4_OPT.LOG | here]]

[[File:gs2016_SiH4_vibrations.png]]

There are 9 vibrational modes observed.

The modes 1,2,3 are degenerate. As well as 4,5 and 7,8,9, with 3 different values.

2 bands would be seen because there are only 2 infrared active modes (figures).

Modes 1 to 5 are all bending, and modes 6 to 9 are stretching.

Modes ... are symmetric.

The Si atom has a charge of 0.629, while each hydrogen atom has a charge of -0.157. This is different from the expected values which were Si:-4 and H:+1.

===Energy Changes===

The equation of SiH4 formation is as follows:

Si + 2H2 → SiH4

E(H2)= -1.17853936
2*E(H2)= -2.35707872
E(Si)= 0
E(SiH4)= -291.88802760

Note that the energy of Si is 0 because it is a unimolecular atom with no bonds.

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of SiH4 as follows:

ΔE=E(SiH4)-[E(Si)+2*E(H2)]= -289.5309519

Energy change in kJ/mol = -760163.51

This show that SiH4 is more stable than its reactants as the reaction is exothermic.

===Molecular Orbitals===

[[File:gs2016_MO6.png]]

This is a 3s molecular orbital.

[[File:gs2016_MO2.png]]

[[File:gs2016_MO3.png]]

[[File:gs2016_MO4.png]]

These are all degenerate 2p orbitals. They have the same energies and are facing different directions (2px, 2py and 2pz). The 3p orbitals have the same shapes and directions but are much larger. The 2p orbitals have no effects on the MOs. The 3p orbitals however bond with the hydrogen electrons and are the LUMO.

[[File:gs2016_MO7.png]]

This is the orbital with the deepest energy. It is a filled 1s orbital and has no effect on bonding. It is not visible because it is very small and within the Si atom.

File:Gs2016 MO7.png

2017-02-24T11:57:32Z

Gs2016:

Rep:Mod:217GASE

2017-02-24T11:53:04Z

Gs2016:

==NH3 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -56.55776873

The RMS gradient was 0.00000485

Point group of the molecule = C3v

The NH bond distance is 1.01798 angstroms

The H-N-H bong angle is 105.741 degrees

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986272D-10
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016_NH3_OPT.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[Media:GS2016_NH3_OPT.LOG| here]]

[[File:gs2016_Display_Vibrations.png]]

6 modes are expected from the 3N-6 rule.

Modes 2 and 3,as well as 5 and 6 have degenerate energies.

Modes 1,2 and 3 are bending. Modes 4,5 and 6 are stretching.

Modes 1,3 and 4 are symmetric.

Mode 1 is the umbrella one.

4 bands would be degenerate.

The charge on the Nitrogen atom is -1.125 and the charge on each hydrogen atom was 0.375.

I would have expected the Nitrogen charge to be -3 and each hydrogen having +1 charge.

===The Hydrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -1.17853936

The RMS gradient was 0.00000017

Point group of the molecule = D*H

The H-H bond distance is 0.74279 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164080D-13
</pre>

The only frequency is 4465.68, so there are no negative frequencies.

===The Nitrogen Molecule===

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -109.52412868

The RMS gradient was 0.00000060

Point group of the molecule = D*H

The N-N bond distance is 1.10550 angstroms

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.400879D-13
</pre>

The only frequency is 2457.33, so there are no negative frequencies.

===Energy Changes===

The equation of ammonia formation is as follows:

3H2 + N2 → 2NH3

E(NH3)= -56.55776873
2*E(NH3)= -113.1155375
E(N2)= -109.52412868
E(H2)= -1.17853936
3*E(H2)= -3.53561808

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of ammonia as follows:

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579074

Energy change in kJ/mol = -146.49

The ammonia product is more stable than its gaseous reactants because the energy change calculated is negative, an exothermic reaction. The ammonia has a lower energy.
The calculated value is however much higher than experimentally calculated as shown in the following link. This is because ....

==SiH4 Molecule==

RB3LYP was the method used for this molecule

6-31G(d.p) was the basis set used

The final energy in au was -291.88802760

The RMS gradient was 0.00000002

Point group of the molecule = TD

The Si-H bond distance is 1.48485 angstroms

The H-N-H bong angle is 109.471 degrees, so the molecule has a terahedral shape.

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-2.452659D-14
</pre>

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>GS2016SIH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

The optimisation file is linked to [[https://wiki.ch.ic.ac.uk/wiki/images/7/75/GS2016_SIH4_OPT.LOG | here]]

[[File:gs2016_SiH4_vibrations.png]]

There are 9 vibrational modes observed.

The modes 1,2,3 are degenerate. As well as 4,5 and 7,8,9, with 3 different values.

2 bands would be seen because there are only 2 infrared active modes...

Modes 1 to 5 are all bending, and modes 6 to 9 are stretching.

The Si atom has a charge of 0.629, while each hydrogen atom has a charge of -0.157. This is different from the expected values which were Si:-4 and H:+1.

===Energy Changes===

The equation of SiH4 formation is as follows:

Si + 2H2 → SiH4

E(H2)= -1.17853936
2*E(H2)= -2.35707872
E(Si)= 0
E(SiH4)= -291.88802760

Note that the energy of Si is 0 because it is a unimolecular atom with no bonds.

From these energy values, we can then use the stoichiometry of the equation to find the enthalpy of formation of SiH4 as follows:

ΔE=E(SiH4)-[E(Si)+2*E(H2)]= -289.5309519

Energy change in kJ/mol = -760163.51

===Molecular Orbitals===

[[File:gs2016_MO6.png]]

This is a 3s molecular orbital.

[[File:gs2016_MO2.png]]

[[File:gs2016_MO3.png]]

[[File:gs2016_MO4.png]]

These are all degenerate 2p orbitals. They have the same energies and are facing different directions (2px, 2py and 2pz). The 3p orbitals have the same shapes and directions but are much larger. The 2p orbitals have no effects on the MOs. The 3p orbitals however bond with the hydrogen electrons and are the LUMO.

[[File:gs2016_MO1.png]]

This is the orbital with the deepest energy. It is a filled 1s orbital and has no effect on bonding.

File:Gs2016 MO6.png

2017-02-24T11:28:58Z

Gs2016:

File:Gs2016 MO5.png

2017-02-24T11:28:27Z

Gs2016: