Rep:Mod:MAlberts
Molecular Modeling 2
NH3
Optimisation
Summary Information
Molecule: NH3
Calculation Method: B3LYP
Basic Set: 6-31G (d,p)
Final Energy: -56.55776873 a. u.
RMS Gradient: 0.00000485 a.u.
Point Group: C3V
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=-5.986333D-10 Optimization completed.
All the items in the above table signify the slope of either the force in case of the Maximum Force or the displacement in case of the Maximum Displacement. The optimisation tries to minimise the force and the displacement in the molecule. A minimum is found when the slope of both curves is equal to zero. As Gaussian can never attain a perfect zero due to rounding errors a threshold is set.
In the table above all items are under the threshold so it is said that a minimum has been reached.
Structural Information
Bond length: 1.02 Å
Bond Angle: 106°
Structure: Trigonal pyrimidal
NH3 Optimised |
Vibrations
Number of vibrations according to 3N-6: 6
Degenerates: 2&3 and 5&6 are degenerate
Bending: 1, 2, 3
Stretching: 4, 5, 6
Highly Symmetric: 4
Umbrella: 1
IR Bands: 4 as two of the 6 are degenerate
| Symmetry | Wavenumber cm-1 | IR Intensity |
|---|---|---|
| A1 | 1089 | 145 |
| E | 1694 | 14 |
| E | 1694 | 14 |
| A1 | 3461 | 1 |
| E | 3589 | 0 |
| E | 3589 | 0 |
The IR intensities varies with the amount of dipole change in the molecule. In this case the first vibrations has the strongest intensity because in this vibration the dipole of the molecule changes the most.
Atomic Charge
I would expect the nitrogen to have a partial negative charge as it is more electronegative than hydrogen.
N2
Summary Informations
Molecule: N2
Calculation Method: RB3LYP
Basis Set: 6-31G(d,p)
Final Energy: -109.52412868 a.u.
RMS Gradient Norm: 0.00000060 a.u.
Point Group: Dinf H
File:MARVIN ALBERTS N2 OPT+FREQ.LOG
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=-3.401002D-13 Optimization completed.
Refer to section NH3 for explanation. A minimum has been reached in this optimisation.
Structural Informations
Bond length: 1.11 Å
Bond Angle: 0°
Structure: Linear
==
N2 |
Vibrations
| Mode | Frequency | IR Intensity |
|---|---|---|
| SGG | 2457 | 0 |
Number of Vibrations according to 3N-5: 1
Degenerates: None
Bending: None
Stretching: 1
IR Bands: 0 because no change in dipol moment
Charges
The partial charges on both atoms are the same as they have the same electronegativity.
N2 Optimised |
H2
Summary Informations
Molecule: H2
Calculation Method: RB3LYP
Basis Set: 6-31G(d,p)
Final Energy: -1.17853936 a.u.
RMS Gradient Norm: 0.00000017 a.u.
Point Group: Dinf H
File:MARVIN ALBERTS H2 OPT+FREQ.LOG
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.
Refer to section NH3 for explanation. A minimum has been reached in this optimisation.
Structural Informations
Bond length: 0.74Å
Bond Angle: 0°
Structure: Linear
Vibrations
| Mode | Frequency | IR Intensity |
|---|---|---|
| SGG | 4465 | 0 |
Number of Vibrations according to 3N-5: 1
Degenerates: None
Bending: None
Stretching: 1
IR Bands: 0 because no change in dipol moment
Charges
The partial charges on both atoms are the same as they have the same electronegativity.
H2 Optimised |
Transition Metal Complex
Identifier: VEJSEL [1]
Computational bond distance: 1.10550 Å
Bond distance in complex: 1.131(8)Å
The bond distance is the same different in the computational model and the crystallography data. Normally if a metal bonds to a ligand a sharing of electrons occurs. Either electrons from the complex or electrons from the ligand are shared. Due to this sharing of electrons the force constant is changes and analogously the bond distance changes. A change in bond distance is expected.
In this example there is a change in bond distance. This is due to the bonding of the nitrogen molecule to the complex. Another factor for the difference might be inaccuracies in the computational model and the crystallography data but as a change in bond distance is expected it doesn't have a large impact on the results.
Haber-Bosch
E(NH3)= -56.5577687 a.u.
2*E(NH3)= -113.1155374 a.u.
E(N2)= -109.5241286 a.u.
E(H2)= -1.1785393 a.u.
3*E(H2)= -3.5356179 a.u.
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.0557909 a.u.
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -146.5 kJ/mol
The ammonia is the more stable molecule as this is a exothermic reaction.
Small Molecule
SH2
Summary Informations
Molecule: SH2
Calculation Method: RB3LYP
Basis Set: 6-31G(d,p)
Final Energy: -399.3916241 a.u.
RMS Gradient Norm: 0.00012068a.u.
Point Group: C2V
Item Value Threshold Converged? Maximum Force 0.000175 0.000450 YES RMS Force 0.000145 0.000300 YES Maximum Displacement 0.000386 0.001800 YES RMS Displacement 0.000386 0.001200 YES Predicted change in Energy=-1.208488D-07 Optimization completed.
Refer to section NH3 for explanation. A minimum has been reached in this optimisation.
Structural Informations
Bond length: 1.3Å
Bond Angle: 93°
Structure: Bent
Vibrations
| Mode | Frequency | IR Intensity |
|---|---|---|
| A1 | 1224 | 5 |
| A1 | 2692 | 7 |
| B2 | 2712 | 9 |
The intensities vary according to the change in dipole moment. In this case B2 has the strongest IR intensity as there is the highest change in dipole moment during this vibration
Number of Vibrations according to 3N-6: 3
Degenerates: None
Bending: 1
Stretching: 2 and 3
IR Bands: 3 as all are IR active
Charges
As sulfur is more electronegative than hydrogen it has a partial negative charge while the hydrogen atoms have a partial positive charge.
SH2 Optimised |
MOs of SH2
Orbital 1
Atomic Orbitals: 2s of Sulfur (Core)
Energy: -7.95115 a.u.
Occupied: 2e
Type: Core
This molecular orbital doesn't have an effect on bonding as it is a core orbital and doesn't interact with the orbitals of hydrogen. The orbital depicted consists mostly of the 2s orbital of the sulfur.
As this orbital is a core orbital it is low in energy.
Orbital 2
Atomic Orbitals: 2p of Sulfur (Core)
Energy: -5.91588 a.u.
Occupied: 2e
Type: Core
This molecular orbital doesn't have an effect on bonding as it is a core orbital and doesn't interact with the orbitals of hydrogen. The orbital depicted consists mostly of the 2p orbital of the sulfur. The two lobes of the p-orbital are visible in green and red.
This is also a core orbital but it is higher in energy than the previous 2s core orbital as it is a 2p orbital.
Orbital 3
Atomic Orbitals: 3s of Sulfur and 1s of Hydrogen
Energy: -0.74654 a.u.
Occupied: 2e
Type: Bonding
This molecular orbital is mostly formed from the mixing of the 3s orbital of the sulfur and the 1s orbital of the hydrogen. As it is bonding it shortens the bond length and increases the force constant of the bond. Due to the sperical nature and lack of lobes of both orbitals the molecular obital takes the shape of a large solid encompassing the molecule.
Orbital 4
Atomic Orbitals: 3p of Sulfur + 1s of Hydrogen
Energy: -0.44963 a.u.
Occupied: 2e
Type: Bonding
This molecular orbital is formed from the mixing of the 3p orbital of the sulfur and the 1s orbital of the hydrogen. As it is bonding it shortens the bond length and increases the force constant of the bond. Due to the worse overlap of the 3p and the 1s this orbital is lower in energy than the previous one.
As the two 1s orbitals of the hydrogen overlap out of phase but with matching spin with a 3p orbital of the sulfur the molecular orbital takes the shape of two lobes. These lobes originate from the central sulfur atom and connect to one of the hydrogen atoms.
Orbital 5
Atomic Orbitals: 3p of Sulfur + 1s of Hydrogen
Energy: 0.02126 a.u.
Occupied: No
Type: Antibonding
LUMO
The orbital depicted here is the LUMO, meaning the lowest unoccupied orbital. It is mostly formed from the overlap of the out of phase 1s orbitals of the hydrogens and a 3p orbital of the sulfur. In comparison to orbital 4 here the spins of the two orbitals don't overlap. This results in four distinct lobes of different spin forming.
Independent Work
SbF5
Summary Informations
Molecule: SbF5
Calculation Method: RB3LYP
Basis Set: LANL2DZ
Final Energy: -504.72072922 a.u.
RMS Gradient Norm: 0.00000316 a.u.
Point Group: D3H
Item Value Threshold Converged? Maximum Force 0.000009 0.000450 YES RMS Force 0.000003 0.000300 YES Maximum Displacement 0.000038 0.001800 YES RMS Displacement 0.000012 0.001200 YES Predicted change in Energy=-3.803784D-10 Optimization completed.
Refer to section NH3 for explanation. A minimum has been reached in this optimisation.
Structural Informations
Bond length (axial): 1.908Å
Bond length (equatorial): 1.896Å
Bond Angle (equatorial fluorines): 120°
Bond Angle (axial fluorines): 180°
Bond Angle (axial to equatorial fluorines): 180°
Structure: Trigonal Bipyrimidal
Vibrations
| Mode | Frequency | IR Intensity |
|---|---|---|
| E' | 94 | 0 |
| E' | 94 | 0 |
| E' | 237 | 64 |
| E' | 237 | 64 |
| E | 247 | 0 |
| E | 247 | 0 |
| A2 | 259 | 66 |
| A1' | 572 | 0 |
| A1' | 572 | 0 |
| A1' | 587 | 0 |
| A2 | 641 | 70 |
| E' | 646 | 61 |
| E' | 646 | 61 |
The intensities vary according to the change in dipole moment. In this case some have an IR intensity of zero because there is no change in dipole moment while other exhibit a large IR intensity due to a large change in dipole moment.
Number of Vibrations according to 3N-6: 12
Degenerates: 1&2, 3&4, 5&6 and 11&12
Bending: 1, 2, 3, 4, 5, 6, 7
Stretching: 8, 9, 10, 11, 12
IR Bands: 4 as all other have either an IR activity of zero or are degenerate
Charges
As shown in the picture the partial charge on the antimony atom is 3.04. The relative charges on the fluorine atoms vary depending on wether they are in the axial or equatorial position. In the equatorial position the charge is -0.605 while the charge for the axial ones is -0.612. This difference can be explained by the different bond lengths of the two types. As the fluorines in the axial position are further removed from the core antimony atom their charge is higher than that of their counterparts in the equatorial.
SbF5 Optimised |
Marking
Note: All grades and comments are provisional and subjecct to change until your grades are officially returned via blackboard. Please do not contact anyone about anything to do with the marking of this lab until you have recieved your grade from blackboard.
Wiki structure and presentation 1/1
Is your wiki page clear and easy to follow, with consistent formatting?
YES
Do you effectively use tables, figures and subheadings to communicate your work?
YES
NH3 1/1
Have you completed the calculation and given a link to the file?
YES
Have you included summary and item tables in your wiki?
YES
Have you included a 3d jmol file or an image of the finished structure?
YES
Have you included the bond lengths and angles asked for?
YES
Have you included the “display vibrations” table?
YES
Have you added a table to your wiki listing the wavenumber and intensity of each vibration?
YES
Did you do the optional extra of adding images of the vibrations?
YES
Have you included answers to the questions about vibrations and charges in the lab script?
YES, the explanation about the dipole change was particularly good. However due to the low intensity of bands 4, 5, and 6 we only expect to see 2 peaks in the spectrum.
N2 and H2 0/0.5
Have you completed the calculations and included all relevant information? (summary, item table, structural information, jmol image, vibrations and charges)
YES well done on your explanation of the electronegativites.
However you have a bond angle for both diatomics - this is incorrect you need at least 3 atoms for a bond angle!
Crystal structure comparison 0.5/0.5
Have you included a link to a structure from the CCDC that includes a coordinated N2 or H2 molecule?
YES
Have you compared your optimised bond distance to the crystal structure bond distance?
YES
Your first sentence in this section doesn't make sense. You could have explained that the bond distance is longer in the crystal structure as the bonding interaction between the N2 and the metal centre reduces the bonding between the N atoms.
Haber-Bosch reaction energy calculation 1/1
Have you correctly calculated the energies asked for? ΔE=2*E(NH3)-[E(N2)+3*E(H2)]
YES
Have you reported your answers to the correct number of decimal places?
YES
Do your energies have the correct +/- sign?
YES
Have you answered the question, Identify which is more stable the gaseous reactants or the ammonia product?
YES
Your choice of small molecule 4/5
Have you completed the calculation and included all relevant information?
YES - however you have written hydrogen cyanide instead of hydrogen sulfide.
Have you added information about MOs and charges on atoms?
YES: You have done a good job of presenting this information, well done! You have explained the charges well using an electronegativity argument.
However you have referred to the different coloured parts of the MOs as having different spins - this is incorrect. MOs don't have spin, electrons do. MOs have diferent phases. So MO 3 is strongly bonding (as you explained) due the interactions between AOs of the same phase.
Independence 1/1
If you have finished everything else and have spare time in the lab you could: Check one of your results against the literature, or Do an extra calculation on another small molecule, or
YES - well done, very good explanations!
Do some deeper analysis on your results so far