MRD:YLS15

2017-05-12T12:42:08Z

== Exercise 1: H + H2 system ==

===What value does the total 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 total gradient of the potential energy surface is 0 both at the transition state and the minima. However, these can be distinguished using the value of the second derivatives.
The second derivative of the minima would be greater than zero as the minima is the lowest value on the potential energy surface.
The second derivative of the transition structure of the transition state would be less than zero. It would appear as a characteristic ‘saddle point’ on the potential energy surface.

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

Since the H+H2 surface is symmetric, the transition state must have r1 = r2. Figure 1 shows as the time goes the 0.378 time units, the internuclear distance rAB and rBC are the same (r1 = r2). The value of the rts estimated to be 0.9131 Å at this point of intersection.

[[File:yls151.PNG|thumb|450px|none|Figure 1. Intermolecular Distance vs Time]]

By testing different initial conditions where r1 = r2and p1 = p2 = 0, figure 2 shows the actual value of rts is determined to be 0.908 Å

[[File:yls152.PNG|thumb|450px|none|Figure 2. Intermolecular Distance vs Time with (r1 = r2)]]

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

The minimum energy path (mep) always resets its velocity to zero. The trajectory follows the minima of the ‘valley’ towards the transition state. This gives a straight line trajectory where as in the dynamics calculations, the vibrational energies are considered so the trajectory by dynamic calculations follow an oscillating path.

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable"
|-
! p1
! p2
! Does the reaction happen?
! Trajectory
|-
| -1.25
| -2.5
| Yes
|[[File:yls153.PNG|thumb|The energy is enough for the atom to reach and pass through the transition state forming a B-C bond.]]
|-
| -1.5
| -2.0
| No
|[[File:yls154.PNG|thumb|
The energy is only enough for the atoms to approach to each other. Insufficient energy for the reaction to happen.]]
|-
| -1.5
| -2.5
| Yes
|[[File:yls155.PNG|thumb|The atoms have sufficient energy to pass through the transition state.
A bond is formed between B-C, which oscillates due to vibrations.]]
|-
| -2.5
| -5.0
| No
|[[File:yls156.PNG|thumb|The transition state is reached however it rolls back down towards the reactants.]]
|-
| -2.5
| -5.2
| Yes
|[[File:yls157.PNG|thumb|
The atoms pass through the transition state twice and the bond between B-C gets formed.]]
|}

===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 the motion of the atoms follows classical mechanics as the mass of electrons are negligible compared to the atoms. However, this assumption ignores quantum effects such as quantum tunneling.

== Exercise 2: F - H - H system ==

===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?===

The reaction F + H2 is exothermic and H + HF is endothermic as the bond enthalpy of H-F is higher than H-H.

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

According to Hammond’s postulate, in an exothermic reaction, the transition state resembles the reactants.
Therefore, the transition state exists where rH-F = 1.813 Å and rH-H = 0.744 Å which is similar to the bond length of H-H (reactants).

[[File:yls158.PNG|thumb|450px|none|]]

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

Potential Energy for the Transition State = -103.75 kcal/mol
Potential Energy for the H-H / F = -104.0kcal/mol
Potential Energy for the H-F / H = -133.7 kcal/mol
Therefore;
The activation energy for H + HF is +30.05 kcal/mol
The activation energy for F + H-H is +0.25 kcal/mol

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

===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.===

MRD:YLS15

2017-05-12T12:41:03Z

== Exercise 1: H + H2 system ==

===What value does the total 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 total gradient of the potential energy surface is 0 both at the transition state and the minima. However, these can be distinguished using the value of the second derivatives.
The second derivative of the minima would be greater than zero as the minima is the lowest value on the potential energy surface.
The second derivative of the transition structure of the transition state would be less than zero. It would appear as a characteristic ‘saddle point’ on the potential energy surface.

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

Since the H+H2 surface is symmetric, the transition state must have r1 = r2. Figure 1 shows as the time goes the 0.378 time units, the internuclear distance rAB and rBC are the same (r1 = r2). The value of the rts estimated to be 0.9131 Å at this point of intersection.

[[File:yls151.PNG|thumb|450px|none|Figure 1. Intermolecular Distance vs Time]]

By testing different initial conditions where r1 = r2and p1 = p2 = 0, figure (X) shows the actual value of rts is determined to be 0.908 Å

[[File:yls152.PNG|thumb|450px|none|]]

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

The minimum energy path (mep) always resets its velocity to zero. The trajectory follows the minima of the ‘valley’ towards the transition state. This gives a straight line trajectory where as in the dynamics calculations, the vibrational energies are considered so the trajectory by dynamic calculations follow an oscillating path.

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable"
|-
! p1
! p2
! Does the reaction happen?
! Trajectory
|-
| -1.25
| -2.5
| Yes
|[[File:yls153.PNG|thumb|The energy is enough for the atom to reach and pass through the transition state forming a B-C bond.]]
|-
| -1.5
| -2.0
| No
|[[File:yls154.PNG|thumb|
The energy is only enough for the atoms to approach to each other. Insufficient energy for the reaction to happen.]]
|-
| -1.5
| -2.5
| Yes
|[[File:yls155.PNG|thumb|The atoms have sufficient energy to pass through the transition state.
A bond is formed between B-C, which oscillates due to vibrations.]]
|-
| -2.5
| -5.0
| No
|[[File:yls156.PNG|thumb|The transition state is reached however it rolls back down towards the reactants.]]
|-
| -2.5
| -5.2
| Yes
|[[File:yls157.PNG|thumb|
The atoms pass through the transition state twice and the bond between B-C gets formed.]]
|}

===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 the motion of the atoms follows classical mechanics as the mass of electrons are negligible compared to the atoms. However, this assumption ignores quantum effects such as quantum tunneling.

== Exercise 2: F - H - H system ==

===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?===

The reaction F + H2 is exothermic and H + HF is endothermic as the bond enthalpy of H-F is higher than H-H.

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

According to Hammond’s postulate, in an exothermic reaction, the transition state resembles the reactants.
Therefore, the transition state exists where rH-F = 1.813 Å and rH-H = 0.744 Å which is similar to the bond length of H-H (reactants).

[[File:yls158.PNG|thumb|450px|none|]]

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

Potential Energy for the Transition State = -103.75 kcal/mol
Potential Energy for the H-H / F = -104.0kcal/mol
Potential Energy for the H-F / H = -133.7 kcal/mol
Therefore;
The activation energy for H + HF is +30.05 kcal/mol
The activation energy for F + H-H is +0.25 kcal/mol

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

===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.===

MRD:YLS15

2017-05-12T12:40:20Z

== Exercise 1: H + H2 system ==

===What value does the total 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 total gradient of the potential energy surface is 0 both at the transition state and the minima. However, these can be distinguished using the value of the second derivatives.
The second derivative of the minima would be greater than zero as the minima is the lowest value on the potential energy surface.
The second derivative of the transition structure of the transition state would be less than zero. It would appear as a characteristic ‘saddle point’ on the potential energy surface.

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

Since the H+H2 surface is symmetric, the transition state must have r1 = r2. Figure (x) shows as the time goes the 0.378 time units, the internuclear distance rAB and rBC are the same (r1 = r2). The value of the rts estimated to be 0.9131 Å at this point of intersection.

[[File:yls151.PNG|thumb|450px|none|Figure 1. Intermolecular Distance vs Time]]

By testing different initial conditions where r1 = r2and p1 = p2 = 0, figure (X) shows the actual value of rts is determined to be 0.908 Å

[[File:yls152.PNG|thumb|450px|none|]]

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

The minimum energy path (mep) always resets its velocity to zero. The trajectory follows the minima of the ‘valley’ towards the transition state. This gives a straight line trajectory where as in the dynamics calculations, the vibrational energies are considered so the trajectory by dynamic calculations follow an oscillating path.

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable"
|-
! p1
! p2
! Does the reaction happen?
! Trajectory
|-
| -1.25
| -2.5
| Yes
|[[File:yls153.PNG|thumb|The energy is enough for the atom to reach and pass through the transition state forming a B-C bond.]]
|-
| -1.5
| -2.0
| No
|[[File:yls154.PNG|thumb|
The energy is only enough for the atoms to approach to each other. Insufficient energy for the reaction to happen.]]
|-
| -1.5
| -2.5
| Yes
|[[File:yls155.PNG|thumb|The atoms have sufficient energy to pass through the transition state.
A bond is formed between B-C, which oscillates due to vibrations.]]
|-
| -2.5
| -5.0
| No
|[[File:yls156.PNG|thumb|The transition state is reached however it rolls back down towards the reactants.]]
|-
| -2.5
| -5.2
| Yes
|[[File:yls157.PNG|thumb|
The atoms pass through the transition state twice and the bond between B-C gets formed.]]
|}

===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 the motion of the atoms follows classical mechanics as the mass of electrons are negligible compared to the atoms. However, this assumption ignores quantum effects such as quantum tunneling.

== Exercise 2: F - H - H system ==

===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?===

The reaction F + H2 is exothermic and H + HF is endothermic as the bond enthalpy of H-F is higher than H-H.

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

According to Hammond’s postulate, in an exothermic reaction, the transition state resembles the reactants.
Therefore, the transition state exists where rH-F = 1.813 Å and rH-H = 0.744 Å which is similar to the bond length of H-H (reactants).

[[File:yls158.PNG|thumb|450px|none|]]

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

Potential Energy for the Transition State = -103.75 kcal/mol
Potential Energy for the H-H / F = -104.0kcal/mol
Potential Energy for the H-F / H = -133.7 kcal/mol
Therefore;
The activation energy for H + HF is +30.05 kcal/mol
The activation energy for F + H-H is +0.25 kcal/mol

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

===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.===

MRD:YLS15

2017-05-12T12:38:44Z

Yls15: /* Locate the approximate position of the transition state. */

== Exercise 1: H + H2 system ==

===What value does the total 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 total gradient of the potential energy surface is 0 both at the transition state and the minima. However, these can be distinguished using the value of the second derivatives.
The second derivative of the minima would be greater than zero as the minima is the lowest value on the potential energy surface.
The second derivative of the transition structure of the transition state would be less than zero. It would appear as a characteristic ‘saddle point’ on the potential energy surface.

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

Since the H+H2 surface is symmetric, the transition state must have r1 = r2. Figure (x) shows as the time goes the 0.378 time units, the internuclear distance rAB and rBC are the same (r1 = r2). The value of the rts estimated to be 0.9131 Å at this point of intersection.

[[File:yls151.PNG|thumb|450px|none|]]

By testing different initial conditions where r1 = r2and p1 = p2 = 0, figure (X) shows the actual value of rts is determined to be 0.908 Å

[[File:yls152.PNG|thumb|450px|none|]]

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

The minimum energy path (mep) always resets its velocity to zero. The trajectory follows the minima of the ‘valley’ towards the transition state. This gives a straight line trajectory where as in the dynamics calculations, the vibrational energies are considered so the trajectory by dynamic calculations follow an oscillating path.

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable"
|-
! p1
! p2
! Does the reaction happen?
! Trajectory
|-
| -1.25
| -2.5
| Yes
|[[File:yls153.PNG|thumb|The energy is enough for the atom to reach and pass through the transition state forming a B-C bond.]]
|-
| -1.5
| -2.0
| No
|[[File:yls154.PNG|thumb|
The energy is only enough for the atoms to approach to each other. Insufficient energy for the reaction to happen.]]
|-
| -1.5
| -2.5
| Yes
|[[File:yls155.PNG|thumb|The atoms have sufficient energy to pass through the transition state.
A bond is formed between B-C, which oscillates due to vibrations.]]
|-
| -2.5
| -5.0
| No
|[[File:yls156.PNG|thumb|The transition state is reached however it rolls back down towards the reactants.]]
|-
| -2.5
| -5.2
| Yes
|[[File:yls157.PNG|thumb|
The atoms pass through the transition state twice and the bond between B-C gets formed.]]
|}

===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 the motion of the atoms follows classical mechanics as the mass of electrons are negligible compared to the atoms. However, this assumption ignores quantum effects such as quantum tunneling.

== Exercise 2: F - H - H system ==

===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?===

The reaction F + H2 is exothermic and H + HF is endothermic as the bond enthalpy of H-F is higher than H-H.

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

According to Hammond’s postulate, in an exothermic reaction, the transition state resembles the reactants.
Therefore, the transition state exists where rH-F = 1.813 Å and rH-H = 0.744 Å which is similar to the bond length of H-H (reactants).

[[File:yls158.PNG|thumb|450px|none|]]

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

Potential Energy for the Transition State = -103.75 kcal/mol
Potential Energy for the H-H / F = -104.0kcal/mol
Potential Energy for the H-F / H = -133.7 kcal/mol
Therefore;
The activation energy for H + HF is +30.05 kcal/mol
The activation energy for F + H-H is +0.25 kcal/mol

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

===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.===

2016-02-25T16:13:12Z

Yls15: /* Methane, CH4 */

== Ammonia, NH3 ==

==== Gaussian Calculation Summary ====
* Molecule Name: Ammonia, NH3
* Calculation Method: RB3LYP
* Basis set:6-31G(d,p)
* Final energy E(RB3LYP): -56.55776873 a.u.
* RMS Gradient: 0.00000485 a.u.
* The point group of your molecule: C3v
* H-N-H bond angle:105.7412°
* H-N bond length: 1.018Å

The optimisation file of NH3 is linked [[Media:YLS15_NH3_OPTIMISATION.LOG| here]]
<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=-5.986284D-10

Optimization completed.

-- Stationary point found.

----------------------------

! Optimized Parameters !

! (Angstroms and Degrees) !

-------------------------- --------------------------

! Name Definition Value Derivative Info. !

--------------------------------------------------------------------------------

! R1 R(1,2) 1.018 -DE/DX = 0.0 !

! R2 R(1,3) 1.018 -DE/DX = 0.0 !

! R3 R(1,4) 1.018 -DE/DX = 0.0 !

! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !

! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !

! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !

! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !

--------------------------------------------------------------------------------</pre>
<jmol><jmolApplet>
<title>NH3 Molecule</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>YLS15_NH3_OPTIMISATION.LOG</uploadedFileContents>
</jmolApplet></jmol>

====Vibration Modes====

[[File:YLS15_Vibration_Frequencies.PNG]]

'''The animations of the different vibrations modes of the NH3 molecule [[Media:YLS15_vibrations_1.gif| Mode 1]][[Media:YLS15_vibrations_2.gif| Mode 2]][[Media:YLS15_vibrations_3.gif| Mode 3]][[Media:YLS15_vibrations_4.gif| Mode 4]][[Media:YLS15_vibrations_5.gif| Mode 5]][[Media:YLS15_vibrations_6.gif| Mode 6]]'''

'''How many modes do you expect from the 3N-6 rule?'''

6 Vibration modes.

'''Which modes are degenerate (ie have the same energy)?'''

Modes 2 and 3 are degenerate. Also modes 5 and 6 are degenerate.

'''Which modes are "bending" vibrations and which are "bond stretch" vibrations?'''

Modes 2 and 3 are "bending" vibrations, mode 4 is symmetrical stretching, modes 5 and 6 are asymmetrical stretching.

'''Which mode is highly symmetric?'''

Modes 1 and 4 is highly symmetrical.

'''One mode is known as the "umbrella" mode, which one is this?'''

Mode 1

'''How many bands would you expect to see in an experimental spectrum of gaseous ammonia?'''

2 Bands at 1089.54 cm-1 and 1693.95 cm-1

====Charge Distribution====

The charge on the Nitrogen atom is expected to be negative and the charge on the Hydrogen atom is expected to be positive as Nitrogen is more electronegative than Hydrogen.

[[File:YLS15_Charge_Distribution.PNG|thumb|left|Charge Distribution of a NH3 molecule]]

Charge on the Hydrogen atom: +0.375

Charge on the Nitrogen atom: -1.125

== Nitrogen, N2 ==

==== Gaussian Calculation Summary ====
* Molecule Name: Nitrogen, N2
* Calculation Method: RB3LYP
* Basis set:6-31G(d,p)
* Final energy E(RB3LYP): -109.52412868 a.u.
* RMS Gradient: 0.00000217 a.u.
* The point group of your molecule: D∞h
* N≡N bond angle: 180°
* N≡N bond length: 1.1055Å

The optimisation file of N2 is linked [[Media:YLS15_N2_OPTIMISATION.LOG| here]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000001 0.001800 YES
RMS Displacement 0.000002 0.001200 YES
Predicted change in Energy=-4.428714D-12
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------</pre>

One vibration mode occurs at 2457.34 cm-1

== Hydrogen, H2 ==

==== Gaussian Calculation Summary ====
* Molecule Name: Hydrogen, H2
* Calculation Method: RB3LYP
* Basis set:6-31G(d,p)
* Final energy E(RB3LYP): -1.17853936 a.u.
* RMS Gradient: 0.00000017 a.u.
* The point group of your molecule: D∞h
* H-H bond angle: 180°
* H-H bond length: 0.7428Å

The optimisation file of H2 is linked [[Media:YLS15_H2_OPTIMISATION.LOG| here]]

<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
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------</pre>

One vibration mode occurs at 4465.68 cm-1

== Reaction Energy of the Haber-Bosch process ==

* E(NH3)= -56.55776873 a.u.
* 2*E(NH3)= -113.11553746 a.u.
* E(N2)= -109.52412868 a.u.
* E(H2)= -1.17853936 a.u.
* 3*E(H2)= -3.53561808 a.u.
* ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.0557907 a.u. = -146.4784828 kJ moɭ -1

ΔE is a negative value which indicated that the gaseous product is more stable.

== Methane, CH4 ==

==== Gaussian Calculation Summary ====
* Molecule Name: Methane, CH4
* Calculation Method: RB3LYP
* Basis set:6-31G(d,p)
* Final energy E(RB3LYP): -40.52401404 a.u.
* RMS Gradient: 0.00003263 a.u.
* The point group of your molecule: Td
* H-C-H bond angle:109.4712°
* C-H bond length: 1.092Å

The optimisation file of CH4 is linked [[Media:YLS15_CH4_OPTIMISED.LOG| here]]

<pre> Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.256106D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) 120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) -120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) 120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) -120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------</pre>

<jmol><jmolApplet>
<title>CH4 Molecule</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>YLS15_CH4_OPTIMISED.LOG</uploadedFileContents>
</jmolApplet></jmol>

====Vibration Modes====

[[File:CH4_Vibrations.PNG]]

Using the 3N-6 rule, the methane molecule is expected to have 9 vibrational modes.

'''The animations of the different vibrations modes of the CH4 molecule '''[[Media:YLS15_CH4_Vibration_1.gif| Mode 1]][[Media:YLS15_CH4_Vibration_2.gif| Mode 2]][[Media:YLS15_CH4_Vibration_3.gif| Mode 3]][[Media:YLS15_CH4_Vibration_4.gif| Mode 4]][[Media:YLS15_CH4_Vibration_5.gif| Mode 5]][[Media:YLS15_CH4_Vibration_6.gif| Mode 6]][[Media:YLS15_CH4_Vibration_7.gif| Mode 7]][[Media:YLS15_CH4_Vibration_8.gif| Mode 8]][[Media:YLS15_CH4_Vibration_9.gif| Mode 9]]'''

Modes 1, 2 and 3 are degenerate. Modes 4 and 5 are degenerate. Modes 7, 8 and 9 are also degenerate.

̈Modes 1, 2, 3, 4 and 5 are "bending vibrations. Mode 6 is symmetrical stretching. Modes 7 and 8 are asymmetrical stretching.

====Charge Distribution====

The charge on the Carbon atom is expected to be negative and the charge on the Hydrogen atom is expected to be positive as Carbon is more electronegative than Hydrogen.

[[File:YLS15_CH4_CHARGE_DISTRIBUTION.PNG|thumb|left|Charge Distribution of a CH4 molecule]]

Charge on the Hydrogen atom: +0.233

Charge on the Carbon atom: -0.930

==== Molecular Orbital ====

Yls_15_MO_6.PNG

1= 1s sigma bond
2= 2s sigma bond
3=s-p sigma bond
4=s-p sigma bond
5=s-p sigma bond
6= 2s sigma star bond

File:Yls 15 MO 6.PNG

2016-02-25T16:12:48Z

Yls15:

File:Yls 15 MO 5.PNG

2016-02-25T16:12:29Z

Yls15:

File:Yls 15 MO 4.PNG

2016-02-25T16:12:19Z

Yls15: