Rep:Mod:Em2316
Molecular Modelling Lab 2: Ella McKay
NH3
Optimisations
N-H bond distance, Å= 1.01798
H-N-H bond angle, ° = 105.741
Calculation method= RB3LYP
Basis set= 6-31G(d,p)
Final energy, (a.u)= -56.55776873
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
Optimised Ammonia Molecule |
The NH3 log file is linked here
Vibrations
How many modes do you expect from the 3N-6 rule?
From the 3N-6 rule 6 vibrational modes could be expected, as N=4 for this molecule. 6 modes have been found thus optimisation is correct.
Which modes are degenerate (i.e have the same energy)?
With reference to the image above, there are 2 sets of degenerate modes: 2 and 3 (at 3589.82cm-1) are degenerate, and 5 and 6 (at 1693.95cm-1) are also degenerate.
Which modes are "bending" vibrations and which are "bond stretch" vibrations?
From the above image, modes 1, 2 and 3 are 'bending' vibrations and modes 4, 5 and 6 are 'bond stretch' vibrations.
Which mode is highly symmetric?
Mode 4 is highly symmetric, as it retains all symmetry functions during vibration.
One mode is known as the "umbrella" mode, which one is this?
Mode 1 is the umbrella mode.
How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 bands would be expected on the experimental spectrum for gaseous ammonia, conducted on an average quality spectrometer, corresponding to modes 1, 2 and 3, where 2 and 3 are degenerate. The remaining 3 modes would be concealed by the noise of the spectrometer, and would thus not be seen.
Charges
Charge on N atom= -1.125
Charge on each H atom= +0.375
We would expect the charge on the N atom to be negative, as it is more electronegative than the H atoms, and the charge on each H atom to be positive.
N2 Molecule
Optimisations
N-N bond distance, Å= 1.10550
N-N bond angle, ° = 180
Calculation method= RB3LYP
Basis set= 6-31G(d,p)
Final energy, (a.u)= -109.52412868
Point group= D∞H
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
Optimised Nitrogen Molecule |
The N2 log file is linked here
Vibrations
N2 shows no vibrational modes as it is not an infrared active species, thus the intensity of the mode shown above is 0.0000. The intensity of an infrared absorption band depends on the change in dipole moment, which depends on the charge difference between the atoms in the molecule. N2 is homonuclear and diatomic, so there is no charge difference and thus no change in dipole moment, making it unsuitable for IR spectroscopy. [1]
Charges
Charge on N atoms= 0.000
As this is a homonuclear diatomic species, the two atoms have equal electronegativity, therefore it could be expected that the charge on each atom would be zero.
H2 Molecule
Optimisations
H-H bond distance, Å= 0.74279
H-H bond angle, ° = 180
Calculation method= RB3LYP
Basis set= 6-31G(d,p)
Final energy, (a.u)= -1.17853936
Point group= D∞H
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
Optimised Hydrogen Molecule |
The H2 log file is linked here
Vibrations
Like N2, H2 shows no vibrational modes as it is not an infrared active species.
Charges
Charge on H atoms= 0.000
As this is a homonuclear diatomic species, the two atoms have equal electronegativity, therefore it could be expected that the charge on each atom would be zero.
The Haber-Bosch Process
N2 + 3H2 -> 2NH3
Energy Values
E(NH3)= -56.55776873
2*E(NH3)= -113.11553746
E(N2)= -109.52412868
E(H2)= -1.17853936
3*E(H2)= -3.53561808
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.0557907
ΔE(kJmol-1)= -146.47849401
The ammonia product is more stable than the gaseous reactants, as the Haber-Bosch process is exothermic thus energy is given out lowering the energy of the system.
A value for the enthalpy change of reaction for the Haber-Bosch process was found in literature to be -91.4 kJmol-1.[2] The difference between this value and the value calculated computationally is due to the literature value being calculated experimentally, thus the value given will vary with the conditions used. The experiment was likely conducted in the presence of air as opposed to in a vacuum, thus the enthalpy change value will be different to that given computationally which doesn't take into account these external conditions.
Project Molecule: F2
Optimisations
F-F bond distance, Å =1.01798
F-F bond angle, ° =180
Calculation method= RB3LYP
Basis set= 6-31G(d,p)
Final energy, (a.u)= -199.49826218
Point group= D∞H
Item Value Threshold Converged? Maximum Force 0.000128 0.000450 YES RMS Force 0.000128 0.000300 YES Maximum Displacement 0.000156 0.001800 YES RMS Displacement 0.000221 0.001200 YES
Optimised Fluorine Molecule |
The F2 log file is linked here
Vibrations
The intensity of the vibrational mode shown above is 0.0000, thus F2 has no vibrational modes and no bands would be seen on its experimental spectrum. This is because like H2 and N2, F2 is not an infrared active species.
Charges
Charge on F atoms= 0.000
As this is a homonuclear diatomic species, the two atoms have equal electronegativity, therefore it could be expected that the charge on each atom would be zero.
Molecular Orbitals
This image shows the energies of all the MOs of F2, and can be used as a reference to find the corresponding energies for the images of the MOs shown below.
This image shows MO 2, which represents the occupied 1sσ* molecular orbital. It has sigma ungerade symmetry, σu, and is the anti-bonding combination of the core 1s atomic orbitals of each fluorine atom. This MO is only slightly higher in energy, 0.00007 a.u, than MO 1 which represents the bonding version of this orbital. The atomic orbitals that contribute to MO 2 barely overlap at all and are held very tightly to the surface of each nuclei. This is because the 1s electrons are the most penetrating, therefore experience a greater effective nuclear charge and are attracted strongly to the nucleus, and thus the atomic orbitals overlap less . Both MO 1 and MO 2 are significantly deeper in energy (23.46064 a.u) than the next lowest energy molecular orbital as they are generated from core atomic orbitals, whereas the other molecular orbitals are made from valence atomic orbitals, where electrons are shielded and experience a lower effective nuclear charge, thus are lower energy. [3]
This image shows MO 4, which represents the occupied 2sσ* molecular orbital. It has sigma ungerade symmetry, σu, and is the anti-bonding combination of the 2s atomic orbitals of each fluorine atom. It is deep in energy, and as it is an anti-bonding MO it will weaken the bonding between the fluorine atoms. This is because the areas of high electron density surrounding each nucleus in this MO are directed away from the F-F bond, and from the other positively charged nucleus in the molecule, thus the nuclei are attracted towards this electron density and pulled away from each other, weakening the bond.
This image shows MO 5, which represents the occupied 2pσ molecular orbital.
It has sigma gerade symmetry, σg, and is the σ-bonding combination of 2 of the 2p AOs of each fluorine atom. It is deep in energy, and as it is a bonding orbital will contribute constructively to chemical bonding between the F atoms. This is because the molecular orbitals containing the negatively charged electrons are held in between the positively charged nuclei, so that they are attracted towards this area of high electron density and are drawn towards each other, strengthening the bond.
This image shows MO 7, which is degenerate with MO 6, and they represent the occupied 2pπ molecular orbitals. They have pi ungerade symmetry, πu, and are the π-bonding combinations of the 2p atomic orbitals of the fluorine atoms. They are indistinguishable from each other and only differ in orientation due to the orientations of the 2px, 2py or 2pz atomic orbitals that combine to make them. They are still quite deep in energy but relatively close to the HOMO-LUMO region, and as they are bonding orbitals will contribute constructively to chemical bonding between the F atoms.
This image shows MO 9, which is degenerate with MO 8, and these represent the occupied 2pπ* molecular orbitals. They have pi ungerade symmetry, πu, and are the anti-bonding versions of MO 6 and MO 7. MO 8 and MO 9 are the π-anti-bonding combinations of the 2p atomic orbitals of the fluorine atoms. These orbitals are the HOMO for F2, thus are in the HOMO-LUMO region in energy. They are anti-bonding orbitals, so will contribute destructively to bonding between the F atoms, weakening the bond.
This image shows MO 10, which represents the unoccupied 2pσ* molecular orbital. It has sigma ungerade symmetry, σu, and is the anti-bonding combination of the 2px atomic orbitals of each fluorine atom. It is the antibonding version of MO 5, and is the LUMO of F2, and its energy is in the HOMO-LUMO region. As it is unoccupied it doesn't contribute to chemical bonding between the fluorine atoms.
References
- ↑ Morsch, L., New, O., User, P., Study, C., & Course, O. (2016). Some Vibrations Are Infrared Inactive. In UC Davis ChemWiki (pp. 1–10). File:Ir inactive.pdf
- ↑ Modak, J. M. (2002). Haber Process for Ammonia Synthesis Jayant M Modak. Resonance, 7(8), 69–77. File:Enthalpy of haber bosch.pdf
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