NH3 Molecule

test molecule (NH3)

General

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=-5.986293D-10

Calculation Method	RB3LYP
Basis Set	6-31G(d,p)
E(RB3LYP)	-56.55776873	 a.u.
Dipole Moment      1.5008 Debye
Point Group	C3V

N-H bond distance = 1.02 a.u
H-N-H bond angle = 106

Vibration analysis


wavenumber cm^-1	1090	1694	1694	3461	3590	3590
symmetry	A1	E	E	A1	E	E
intensity(arbitary units)	145	14	14	1	0	0
Image

Number of modes= 6
Degenerate modes= modes at 1694, and 3590 cm^-1
bending vibrations= 1090,1694
stretching vibrations= 3461,3590
highly symmetric = 3461, and also 1090 (not as symmetric as 3461)
umbrella mode= 1090
bands in gaseous ammonia= 3
The number of bands is 3, as there are 2 that don't result in a change in dipole moment, and so do not show an IR peak. There are then 2 degenerate peaks; they will show the same wavenumber, and so there are only 3 visibile bands in gaseous ammonia. For condensed it might be different, considering the effect of disorder.

Charge output

Charge on N = -1.125
Charge on H = 0.375
This seems like a likely charge distribution, considering the electronegativity of Nitrogen is much greater than Hydrogen, resulting in a greater electron density around the nitrogen, and an overall much greater charge (x3).

N2 molecule

General

test molecule (N2)

Item               Value     Threshold  Converged?
 Maximum Force            0.000145     0.000450     YES
 RMS     Force            0.000145     0.000300     YES
 Maximum Displacement     0.000045     0.001800     YES
 RMS     Displacement     0.000064     0.001200     YES
 Predicted change in Energy=-6.585141D-09

Calculation Method	RB3LYP
Basis Set	6-31G(d,p)
E(RB3LYP)	-109.52412867	 a.u.
Dipole Moment      0.0000 Debye
Point Group	D*H

Vibration analysis

N-N bond length= 1.11 a.u
N-N bond angle= n/a. There are only 2 atoms, so no valid bond angle.


wavenumber cm^-1	2457
symmetry	SGG
intensity(arbitary units)	0
Image

Vibrational modes expected from 3N-5 = 1
Degenerate modes = n/a
bending vibrations = n/a
bond stretch vibrations = 2457
highly symmetric stretch = 2457
umbrella mode = n/a
how many bands= zero; no change of dipole in the stretch

TM co-ordination

N2 is co-ordinated to an iridium complex, in ''bis((adamantan-1-ylmethyl)(diisopropyl)phosphine)-(dinitrogen)-hydrido-iridium''. The unique identifier for this is ANAZEV, and the article for the molecule can be found here: https://pubs.acs.org/doi/10.1021/ja1037808. 

The NN bond length in this molecule is 1.108 Å. This is almost identical to the bond length for the calculated computational length of 1.11. This implies that the computational calculation works and has found the correct stationary point for the optimised bond length. However, there is still a chance of some error; there is a possibility that the bond length could be shorter in reality than the computed distance, as the computed distance could be the result of a local minimum potential energy rather than the global minimum. If the bond length is shorter in reality, it would imply that the bond length of the co-ordinated NN bond is greater than computed. This would make sense as the electron density of the NN bond could be reduced by coordination, resulting in an overall longer bond. 
The link to the CCDC structure is this:https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=ANAZEV&DatabaseToSearch=Published

Charge output

There is no charge on the atoms, as the electronegativities are equal.

Bis-((adamantan-1-ylmethyl)(diisopropyl)phosphine)-(dinitrogen)-hydrido-iridium

H2 molecule

General

test molecule (H2)

Item               Value     Threshold  Converged?
 Maximum Force            0.000066     0.000450     YES
 RMS     Force            0.000066     0.000300     YES
 Maximum Displacement     0.000087     0.001800     YES
 RMS     Displacement     0.000123     0.001200     YES
 Predicted change in Energy=-5.726834D-09

H-H bond length= 0.74 a.u
H-H bond angle= n/a. There are only 2 molecules.

Calculation Method	RB3LYP
Basis Set	6-31G(d,p)
E(RB3LYP)	-1.17853935 a.u.
Dipole Moment      0.0000 Debye
Point Group	D*H

Vibration analysis


wavenumber cm^-1	4464
symmetry	SGG
intensity(arbitary units)	0
Image

Vibrational modes expected from 3N-5 = 1
Degenerate modes = n/a
bending vibrations = n/a
bond stretch vibrations = 4464
highly symmetric stretch = 4464
umbrella mode = n/a
how many bands= zero; no change of dipole in the stretch

Charge output

There is no charge on the atoms, as the electronegativities are equal.

The Haber Process

E(NH3)=-56.55776873 a.u
2*E(NH3)= -113.11553746 a.u
E(N2)=-109.52413 a.u
E(H2)=-1.17854 a.u
3*E(H2)= -3.53562 a.u
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= 0.05578746 a.u
This is equal to -146.5 kJ/mol
The process of converting N2 and 3H2 to 2NH3 is exothermic, and this implies that the more stable is the NH3 product over the reactants.

Project molecule (ClF3)

test molecule (ClF3)

General

Item               Value     Threshold  Converged?
 Maximum Force            0.000050     0.000450     YES
 RMS     Force            0.000028     0.000300     YES
 Maximum Displacement     0.000204     0.001800     YES
 RMS     Displacement     0.000134     0.001200     YES
 Predicted change in Energy==-1.250231D-08

Cl-F(1) bond length= 1.65 a.u
Cl-F(4) bond length= 1.73 a.u
F4-Cl-F2 bond angle= 174
F4-Cl-F3 bond angle= 87

Calculation Method	RB3LYP
Basis Set	6-31G(d,p)
E(RB3LYP)	-759.46531688 a.u.
Dipole Moment      0.8386 Debye
Point Group	C2V

Vibration analysis


wavenumber cm^-1	305	309	401	541	736	752
symmetry	A1	B1	B2	B2	A1	A1
intensity(arbitary units)	14	18	1	541	736	752
Image

Vibrational modes expected from 3N-6= 6
Degenerate modes = n/a
bending vibrations = 305,309,401
bond stretch vibrations = 541,736,752
highly symmetric stretch = 541
umbrella mode = 309
how many bands= 5
There are 5 different vibrational modes that result in a change in dipole moment, so would cause 5 bands to appear in gaseous sample. The condensed state would not show this many, as there is more disorder to account for, and this would result in overlap between the closely lying bands.

Molecular Orbital Analysis


	A Gaussview of the [3] MO of ClF3, showing the 1s electrons of the F2 and F4, which are degenerate in energy.	A Gaussview of the [2] Molecular orbital of ClF3. This shows the 1s orbital of the F3 molecule, and is lower in energy than the adjacent F2 and F4 molecules' 1s MO energies (which are degenerate). This is interesting as you would expect the opposite when you increase the size of the orbitals, as the increased electron density of the region would push electrons away from the molecule in larger orbitals. Here, this is causing the electrons to go closer to the nucleus, thus reducing the overall energy of the electrons in that orbital.
	A Gaussview of the [9] MO of ClF3. This shows the first case of a molecular orbital in this molecule both spatially and energetically large enough to span the whole molecule; the 3pz orbitals here overlap in a pi- bonding manner, for in-phase orbitals for all the individual atomic orbitals of the molecule.	A Gaussview of the [8] MO of ClF3. This is shows the 3py atomic orbital of the Cl atom, and is significantly lower in energy than the [9] MO, with an energy difference of almost 6.2 a.u.
	A Gaussview of the [22] MO of ClF3. This is the Highest Occupied Molecular Orbital (HOMO) in the molecule, and is a clear pi antibonding orbital across all the atoms. There are 3 nodes across the individual atomic orbital of Cl. In addition to this, the F3 molecule, the middle one, is significantly smaller than the F2 and F4 molecules, as there is repulsion from the high electron density in this region. Thus the orbital is smaller due to less electrons being present in this region. This is the reverse effect to the [2] MO of ClF3, which is held more tightly and is lower in energy. ^[1]	A Gaussview of the [23] MO of ClF3. This is the Lowest Unoccupied Molecular Orbital (LUMO, and it is very interesting! There is both sigma and pi antibonding occuring here, as well as some degree of pi and sigma bonding. This is due to the spatial arrangement of the p orbitals of the F2 and F4 atoms, which are arranged in a way that they can make these interactions. The antibonding is clear to see; there is sigma antibonding due to the clear node defined between F2 ( and F4), and the Cl atoms. The pi antibonding is between the F3 and Cl atoms. However, there is also some angled pi bonding occurring between the same phases of the F2F4 atoms and the Cl; a very interesting angle of approach. In addition to this, as we increase the isovalue, a measure of the electron density per region, we encounter some angled sigma bonding between the p molecules of the F atoms. It is an incredibly interesting molecular orbital, so therefore it is a real shame that it contributes nothing to the energy of the bonding of the molecule; it is Unoccupied.

Charge output

The F atoms are more electronegative than Cl, so have the negative charge. Cl has a charge of 1.225, and the individual F atoms have different charges based on position. The F3 middle atom is slightly less charged, with a charge of -0.315 as there is more repulsion of electrons from here due to increased electron density in that region. The outer F atoms have a charge of -0.465.

↑ This is based on works from this article by Henry Rzepa in the Winnower.https://thewinnower.com/papers/vsepr-theory-a-closer-look-at-chlorine-trifluoride-clf3

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 0/1

Is your wiki page clear and easy to follow, with consistent formatting?

YES, however you have written lots of text in the "pre" boxes, and you have not inserted line breaks leaving the reader to scroll sideways to read your work. You don't need to use the boxes the default formatting is much better.

Do you effectively use tables, figures and subheadings to communicate your work?

Lots of your tables are not formatted nicely and have extra rows/columns or no lines. Many of your figures are not cropped and one has been crudely coloured around in paint instead.

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 3.5/5

Have you completed the calculation and included all relevant information?

YES

Have you added information about MOs and charges on atoms?

YES, good detailed explanations, well done. MO9 is made up of s type AOs not p type and the bonding is sigma bonding.

Independence 0/1

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

No independent work.

[ClF3-1] This is based on works from this article by Henry Rzepa in the Winnower.https://thewinnower.com/papers/vsepr-theory-a-closer-look-at-chlorine-trifluoride-clf3

[1]