Rep:Mod:ROBFINN
NH3 Molecule
Summary
| Calculation Method | RB3LYP |
| Basic set | 6-31G(d,p) |
| Final Energy / au | -56.55776873 |
| RSM Gradient | 0.00000485 |
| Point Group | C3V |
| N-H bond distance / Å | 1.01798 |
| H-N-H bond angle / ° | 105.741 |
Item Table
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=-5.986260D-10
Optimization completed.
-- Stationary point found.
Jmol Image
NH3 Molecule |
The optimisation file is liked to here
Frequency Analysis
| wavenumber / cm-1 | intensity / arbitrary units | image |
|---|---|---|
| 1090 | 145 | 
|
| 1694 | 14 | 
|
| 1694 | 14 | 
|
| 3461 | 1 | 
|
| 3590 | 0 | 
|
| 3590 | 0 | 
|
From the 3N-6 rule, six modes would be expected. Mode 2 and 3, and 5 and 6 are degenerate. Modes 1, 2 and 3 are bending vibrations. Modes 4, 5 and 6 are stretching vibrations. Mode 4 is highly symmetrical. Mode 1 is the umbrella mode. You would expect to see two bands in an experimental spectrum of gaseous ammonia. Mode four is symmetrical, so won't produce a dipole moment so won't be seen. Modes 2&3 and 5&6 are degenerate so will appear as two peaks. Mode 5&6 have an intensity of zero so won't be seen.
Charge Analysis
The charge on the nitrogen is -1.125 and the charge on the hydrogen is +0.375. This is due to nitrogen being more electronegative than hydrogen, so nitrogen has a higher tendency to attract the electron density from the covalent bond, making the nitrogen more negative.
H2 Molecule
Summary
| Calculation Method | RB3LYP |
| Basic set | 6-31G(d,p) |
| Final Energy / au | -1.17853936 |
| RSM Gradient | 0.00000017 |
| Point Group | Dinfh |
| H-H bond distance / Å | 0.74279 |
Item Table
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.
Jmol Image
H2 Molecule |
The optimisation file is liked to here
Frequency Analysis
| wavenumber / cm-1 | intensity / arbitrary units |
|---|---|
| 4466 | 0 |
Linear molecules use the rule 3N-5. Therefore, we can see there should only be one vibrational mode.
Charge Analysis
The charge on both hydrogens is zero, as they have the same electronegativity.
N2 Molecule
Summary
| Calculation Method | RB3LYP |
| Basic set | 6-31G(d,p) |
| Final Energy / au | -109.52412868 |
| RSM Gradient | 0.00000060 |
| Point Group | Dinfh |
| N-N bond distance / Å | 1.10550 |
Item Table
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=-3.400991D-13
Optimization completed.
-- Stationary point found.
Jmol Image
N2 Molecule |
The optimisation file is liked to here
Frequency Analysis
| wavenumber / cm-1 | intensity / arbitrary units |
|---|---|
| 2457 | 0 |
Linear molecules use the rule 3N-5. Therefore, we can see there should only be one vibrational mode.
Charge Analysis
The charge on both nitrogens is zero, as they have the same electronegativity.
trans-bis(1-(diethylphosphino)-N-((diethylphosphino)methyl)-N-(2,6-difluorobenzyl)methanamine)-bis(dinitrogen)-chromium [AQEZIH]
N-N bond distance is equal to 1.127 Å. The crystal structure has a longer N-N bond length when compared to the computational distance. This implies that the bond is weaker. This could be due to the central chromium being electron withdrawing, which makes the bond weaker and consequently longer. The crystal structure is also a solid, so more electron withdrawing groups could interact with the N-N bond, spreading electron density over more atoms, making the bond weaker and longer.
Haber-Bosch Process
| Molecule | Energy / au |
|---|---|
| NH3 | -56.55777 |
| N2 | -109.52413 |
| H2 | -1.17854 |
N2 + 3H2 → 2NH3
ΔE = [2xE(NH3)] - [E(N2) + 3xE(H2)]
= [-113.11554] - [-109.52413 - 3.53562] = -0.05579 au = -146.5 kJ/mol
The reaction is exothermic, therefore the product (ammonia) is more stable as energy is released during the reaction.
CH4 Molecule
Summary
| Calculation Method | RB3LYP |
| Basic set | 6-31G(d,p) |
| Final Energy / au | -40.52401404 |
| RSM Gradient | 0.00003263 |
| Point Group | Td |
| C-H bond distance / Å | 1.09197 |
| H-C-H bond angle / ° | 109.471 |
Item Table
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=-2.256043D-08
Optimization completed.
-- Stationary point found.
Jmol Image
CH4 Molecule |
The optimisation file is liked to here
Frequency Analysis
| wavenumber / cm-1 | intensity / arbitrary units |
|---|---|
| 1356 | 14 |
| 1356 | 14 |
| 1356 | 14 |
| 1579 | 0 |
| 1579 | 0 |
| 3046 | 0 |
| 3162 | 25 |
| 3162 | 25 |
| 3162 | 25 |
From the 3N-6 rule, nine modes would be expected. Mode 1&2&3, and 4&5, and 7&8&9 are degenerate. Modes 1, 2, 3, 4 and 5 are bending vibrations. Modes 6, 7, 8 and 9 are stretching vibrations. Mode 6 is highly symmetrical. You would see two peaks for methane. Modes 1&2&3 and 7&8&9 are degenerate so will produce two peaks. The rest of the modes have zero intensity so won't be seen.
Charge Analysis
The charge on the carbon is -0.930 and the charge on the hydrogen is +0.233. This is due to carbon being more electronegative than hydrogen, so carbon has a higher tendency to attract the electron density from the covalent bond, making the carbon more negative.
Molecular Orbitals
HOMO
These three molecular orbitals are degenerate with an energy of -0.38831 au. These are the highest occupied molecular orbital in the molecule, and are each occupied by two electrons with opposite spins. These are bonding orbitals and form σ bonds. Each MO has contribution from 1s atomic orbital from hydrogen and a 2p atomic orbital from carbon, which also forms an antibonding σ MO. This does contribute to the bond order of the molecule.
LUMO
This is the lowest unoccupied molecular orbital in the molecule, with an energy of 0.11824 au. It is unoccupied. The 2s AO of carbon and the 1s AO of hydrogen combind to form this MO and a σ MO. It is an antibonding orbital. This doesn't contribute to the bond order of the molecule.
Carbon 1s
This MO is contributed to by the 1s AO in carbon, with an energy of -10.16707 au. Due to it's highly negative energy, it doesn't contribute to the bonding.
1σ
This id the first σ molecular orbital with an energy of -0.69041 au. This is occupied with two electrons with opposite spins and is a bonding orbital. The 2s AO from carbon and the 1s AO from hydrogen contribute to this. This does contribute to the bond order of the molecule.
7σ* 8σ* 9σ*
These are three degenerate σ* molecular orbitals. They are unoccupied and each have an energy of 0.17677 au. A 2p AO from carbon and a 1s AO from hydrogen combind to form each MO. This doesn't contribute to the bond order of the molecule.
Independent Cl2 Molecule
Summary
| Calculation Method | RB3LYP |
| Basic set | 6-31G(d,p) |
| Final Energy / au | -920.34987886 |
| RSM Gradient | 0.00002511 |
| Point Group | Dinfh |
| Cl-Cl bond distance / Å | 2.04174 |
Item Table
Item Value Threshold Converged?
Maximum Force 0.000043 0.000450 YES
RMS Force 0.000043 0.000300 YES
Maximum Displacement 0.000121 0.001800 YES
RMS Displacement 0.000172 0.001200 YES
Predicted change in Energy=-5.277255D-09
Optimization completed.
-- Stationary point found.
Jmol Image
Cl2 Molecule |
The optimisation file is liked to here
Frequency Analysis
| wavenumber / cm-1 | intensity / arbitrary units |
|---|---|
| 520 | 0 |
Linear molecules use the rule 3N-5. Therefore, we can see there should only be one vibrational mode.
Charge Analysis
The charge on both chlorines is zero, as they have the same electronegativity.
Marking
Note: All grades and comments are provisional and subject 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 received your grade from blackboard.
Wiki structure and presentation 1/1
Is your wiki page clear and easy to follow, with consistent formatting?
YES, overall a very good wiki well done!
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
N2 and H2 0.5/0.5
Have you completed the calculations and included all relevant information? (summary, item table, structural information, jmol image, vibrations and charges)
YES
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
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
Have you added information about MOs and charges on atoms?
YES, and overall your explanations are correct, clear and well presented, well done! The aspect you could have improved upon would be to explain that the LUMO and the 1 sigma MOs are bonding/antibonding counterparts. And the same for the HOMOs and the 7,8,9 sigma star MOs.
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
Do some deeper analysis on your results so far
You calculated Cl2 independently well done!