Rep:Mod:JZ14418
NH3 molecule
Optimisation
Optimised NH3 |
The optimisation file can be found here
Molecule name: NH3
Calculation method: RB3LYP
Basis set: 6-31G(d,p)
Final energy E(RB3LYP) in atomic units (au): -56.5577687
RMS gradient: 0.00000485 a.u.
Point group: C3V
Optimised N-H bond length: 1.02Å
Optimised H-N-H bond angle: 106°
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
Vibrations Summary
| wavenumber cm-1 | 1090 | 1694 | 1694 | 3461 | 3590 | 3590 |
| symmetry | A1 | E | E | A1 | E | E |
| intensity arbitrary units | 145 | 14 | 14 | 1 | 0 | 0 |
| images |  |
 |
 |
 |
 |
|
How many modes do you expect from the 3N-6 rule? Number of modes= 3(4)-6=6
Which modes are degenerate (ie have the same energy)?
The modes with the same frequencies are degenerate, 2 modes at 1694cm-1 and 2 modes at 3590cm-1.
Which modes are "bending" vibrations and which are "bond stretch" vibrations?
Bending vibrations: 1090cm-1 and 1694cm-1 Bond stretch vibrations: 3461cm-1 and 3590cm-1.
Which mode is highly symmetric? The mode with a frequency of 1090cm-1 and 3461cm-1.
One mode is known as the "umbrella" mode, which one is this? The mode with a frequency of 1090cm-1.
How many bands would you expect to see in an experimental spectrum of gaseous ammonia? 3 different bands should be observed. This is because 2 of the vibrations have 0 intensity and is therefore unlikely to be detected. Out of the remaining 4 vibrations, 2 are degenerate with a vibrational wavenumber of 1694cm-1 which will only appear as 1 band. Hence, a total of 3 bands are observed.
Charge Analysis
Charge on N: -1.125 Charge on H: 0.375
The charge on N is expected to be negative while that of the H is expected to be positive. This is because nitrogen is a much more electronegative atom compared to hydrogen and therefore draws the electron density towards it, making it more negative while hydrogen becomes more positive.
N2 molecule
Optimisation
Optimised N2 |
The optimisation file can be found here
Molecule name: N2
Calculation method: RB3LYP
Basis set: 6-31G(d,p)
Final energy E(RB3LYP) in atomic units (au): -109.5241287
RMS gradient: 0.00000060 a.u.
Point group: D∞h
Optimised N≡N bond length: 1.11Å
NBO charge: 0.000 (N2 is linear)
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
Vibrations Summary
| wavenumber cm-1 | 2457 |
| symmetry | SGG |
| intensity arbitrary units | 0 |
Within the crystal structure DAMSUG, the length of the N≡N bond is 1.128Å which is longer than the N≡N bond in a nitrogen molecule. In the formation of the crystal structure, the N2 acts as a ligand which is a σ-donor, π-acceptor. Electron density is donated from the HOMO of N2 into an empty d orbital of the transition metal Mo while a π-interaction occurs from the overlap of a occupied metal-based orbital and the π* orbital of N2 that can accept electron density. In this crystal structure DAMSUG, the effect of backbonding with the donation of electrons from the metal to the π* orbital of N2 dominates. The addition of electrons into the π* orbital weakens the bond in N2, thus causing the bond length to increase.[1]
More information on DAMSUG can be found at: https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=DAMSUG&DatabaseToSearch=Published
H2 molecule
Optimisation
optimised H2 |
The optimisation file can be found here
Molecule name: H2
Calculation method: RB3LYP
Basis set: 6-31G(d,p)
Final energy E(RB3LYP) in atomic units (au): -1.1785394
RMS gradient: 0.00000017 a.u.
Point group: D∞h
Optimised H-H bond length: 0.74Å
NBO charge: 0.000 (H2 is linear)
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
Vibrations Summary
| wavenumber cm-1 | 4466 |
| symmetry | SGG |
| intensity arbitrary units | 0 |
Haber-Bosch process
N2 + 3H2 -> 2NH3
E(NH3)= -56.5577687au
2*E(NH3)= -113.1155375au
E(N2)= -109.5241287au
E(H2)= -1.1785394au
3*E(H2)= -3.5356181au
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]=-0.05579074au = -146.5kJ/mol
The ammonia product is more stable as the reaction of the formation of ammonia is exothermic. Energy is released and the product ammonia is at a lower energy level compared to the gaseous reactants, thus making it more stable.
Project molecule: CO2
Optimisation
optimised CO2 |
The optimisation file can be found here
Molecule name: CO2
Calculation method: RB3LYP
Basis set: 6-31G(d,p)
Final energy E(RB3LYP) in atomic units (au): -188.5809395
RMS gradient: 0.00001154 a.u.
Point group: D∞h
Optimised C=O bond length: 1.17Å
Item Value Threshold Converged? Maximum Force 0.000024 0.000450 YES RMS Force 0.000017 0.000300 YES Maximum Displacement 0.000021 0.001800 YES RMS Displacement 0.000015 0.001200 YES
Vibrations Summary
| wavenumber cm-1 | 640 | 640 | 1372 | 2436 |
| symmetry | PIU | PIU | SGG | SGU |
| intensity arbitrary units | 31 | 31 | 0 | 546 |
| type of vibration | bending | bending | symmetric stretch | asymmetric stretch |
Charge Analysis
Charge on C: 1.022 Charge on O: -0.511
The charge on oxygen is negative while that of the carbon is positive as oxygen is more electronegative and draws electron density away from the carbon.
Molecular Orbitals
| MO | Information.[2] |
 |
This is a π*g MO. It is an anti-bonding molecular orbital and also the HOMO of the molecule. It is formed from the destructive combination of oxygen 2p AOs. There are 2 degenerate molecular orbitals at this energy level, one formed by two px orbitals and another from two py orbitals. |
 |
This is a π*u MO. It is also the LUMO of the molecule. It is an anti-bonding molecular orbital formed from the destructive combination of the oxygen and carbon p AOs. There are 2 degenerate MOs at this energy level due to possible combinations between the px orbitals and the py orbitals respectively. The addition of electrons into this π* anti-bonding MO will destabilise the π bonding MO, weakening the π bond in the molecule, thus leading to an increase in energy. |
 |
This is a σg MO. It is fully-occupied and is the molecular orbital with the deepest energy. It is a non-bonding orbital formed from the linear combination of the oxygen 1s AOs. As the 1s orbital of oxygen is deep in energy, it does not overlap with the orbitals of carbon. |
 |
This is a σu MO. It is a fully occupied bonding molecular orbital formed from the constructive linear combination between 2s orbitals of the oxygen atoms and the p orbital of carbon. The MO consists of 1 node. |
 |
This is a πu MO. There are 2 degenerate orbitals at this energy level. It is a fully occupied bonding orbital formed from the constructive linear combination of three px orbitals and between three py orbitals of oxygen and carbon respectively. |
Additional: O2 molecule
Optimisation
Optimised O2 |
The optimisation file can be found here
Molecule name: O2
Calculation method: RB3LYP
Basis set: 6-31G(d,p)
Final energy E(RB3LYP) in atomic units (au): -150.2574243
RMS gradient: 0.00007502 a.u.
Point group: D∞h
Optimised O=O bond length: 1.22Å
NBO charge: 0.000 (O2 is linear)
Item Value Threshold Converged? Maximum Force 0.000130 0.000450 YES RMS Force 0.000130 0.000300 YES Maximum Displacement 0.000080 0.001800 YES RMS Displacement 0.000113 0.001200 YES
Vibrations Summary
| wavenumber cm-1 | 1643 |
| symmetry | SGG |
| intensity arbitrary units | 0 |
References
- ↑ Chemistry LibreTexts, https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book%3A_Inorganic_Chemistry_(Wikibook)/Chapter_05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.04%3A_%5C(%5Cpi%5C)-bonding_between_metals_and_ligands, (accessed March 2019).
- ↑ C.Housecroft, A.G.Sharpe in Inorganic Chemistry, 4th edn, 2012, ch.5, pp. 139-171.
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 - your answers are correct overall/ Only the number of expected bands in the IR spectrum should be 2 rather than3 because the intensity of the symmetric stretching mode is too low to be detected.
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, you could have explained that the charges are 0 as the electronegativities are equal.
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 3.5/5
Have you completed the calculation and included all relevant information?
YES - however, you should have analysed the calculated vibrations as you did for NH3.
Have you added information about MOs and charges on atoms?
You have done a good job of presenting this information, well done! The HOMO is not anti-bonding. It is a non-bonding orbital and therefore does not contribute to the bonding. Besides this you descriptions of the MOs are correct as well as the contributing AOs! However for some MOs you missed to include information on their relative energies and if they are occupied/non-occupied.
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!
Do some deeper analysis on your results so far