Title

NH3 molecule

Key information
Calculation method	RB3LYP
Basis set	6-31G(d.p)
Final energy, E(RB3LYP) (au)	-56.5577687
RMS gradient (au)	0.00000485
Point group	C3V

N-H bond distance = 1.02Å (2 d.p)

H-N-H bond angle = 106° (0 d.p)

         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

Project molecule

NH3 molecule

The optimisation file is liked to here

Mode	Wavenumber (cm-1)	Symmetry	Intensity (arbitrary)	Type of vibration
1	1090	A1	145	Bend
2	1694	E	14	Bend
3	1694	E	14	Bend
4	3461	A1	1	Stretch
5	3590	E	0	Stretch
6	3590	E	0	Stretch

In the NH3 molecule, Nitrogen atom has a charge of -1.125 and all the Hydrogen atoms have a charge of 0.375. I would expect the Nitrogen atom's charge to be more negative in comparison. the reason for this is because Nitrogen is more electronegative in comparison to the Hydrogen atoms. Hence it'll have a higher electron density thus making it more negative.

6 modes are expected from the (3N-6) rule. The modes 2 and 3 are degenerate. 5 and 6 are also degenerate to each other. Modes 1, 2 and 3 are bending vibrations whilst modes 4, 5 and 6 are bond stretch vibrations. The mode which is highly symmetrical is 4. The umbrella mode is 1. 2 bands are expected to be seen in an experimental spectrum of gaseous ammonia due to there being two degenerate pairs. In addition, modes 5 and 6 have an intensity of 0.

N2 molecule

Key information
Calculation method	RB3LYP
Basis Set	6-31G(d,p)
Final energy, E(RB3LYP) (au)	-109.5241287
RMS Gradient Norm (au)	0.00000365
Point Group	D*H

N≡N bond distance = 1.11Å (2 d.p)

N2 is a linear molecule.

         Item               Value     Threshold  Converged?
 Maximum Force            0.000006     0.000450     YES
 RMS     Force            0.000006     0.000300     YES
 Maximum Displacement     0.000002     0.001800     YES
 RMS     Displacement     0.000003     0.001200     YES

Project molecule

N2 molecule

The optimisation file is liked to here

Mode	Wavenumber (cm-1)	Symmetry	Intensity (arbitrary)
1	2457	SGG	0

Both Nitrogen molecules have no charge. The reason for this is because they are the same element, hence they have the same electronegativities. So the molecule is non polar as the charge is distributed evenly along both atoms.

H2 molecule

Key information
Calculation method	RB3YLB
Basis set	6-31G(d.p)
Final energy, E(RB3LYP) (au)	-1.1785394
RMS Gradient Norm (au)	0.00000017
Point Group	D*H

H-H bond distance = 0.74Å (2 d.p)

         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

Project molecule

H2 molecule

The optimisation file is liked to here

Mode	Wavenumber (cm-1)	Symmetry	Intensity
1	4466	SGG	0

Both Hydrogen atoms have a charge distribution of 0. This shows it's a non polar molecular as it has no difference in charge.

Structure and reactivity of H2

Unique identifier = CEFCAS

[[1]]

Molecule	H-H bond (Å)
H2	0.74
CEFCAS	1.48

The H-H bond in the H2 molecule is shorter compared to the H-H bond in the metal complex. The reason for this is because the hydrogen atom is bonded to both the transition metal and another hydrogen. The bond density in the H-H bond becomes smaller compared to the bond density in the H-TM (TM = transition metal) bond. This causes the H-H bond to become longer. In addition, the bonds may be different due to computational errors. This can be fixed by using a different method.

Haber-Bosch reaction energy calculation for NH3

E(NH3)= -56.5577687 au

2*E(NH3)= -113.1155374 au

E(N2)= -109.5241287 au

E(H2)= -1.1785394 au

3*E(H2)= -3.5356182 au

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.0557905 au = -146.5(1 d.p) kJ/mol

The exothermic reaction is favoured. The forward reaction is exothermic. Hence there will be a higher production of ammonia in this reaction because ammonia is more stable.

CH4 molecule

Key information
Calculation method	RB3LYP
Basis set	6-31G(d.p)
Final energy, E(RB3LYP) (au)	-40.5240140
RMS Gradient Norm (au)	0.00003263
Point Group	TD

C-H bond distance = 1.07Å (2 d.p)

H-C-H bond angle = 109° (0 d.p)

         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

Project molecule

CH4 molecule

The optimisation file is liked to here

Mode	Wavenumber (cm-1)	Symmetry	Intensity (arbitrary)	Type of vibration
1	1356	T2	14	Bend
2	1356	T2	14	Bend
3	1356	T2	14	Bend
4	1579	E	0	Bend
5	1579	E	0	Bend
6	3046	A1	0	Stretch
7	3162	T2	25	Stretch
8	3162	T2	25	Stretch
9	3162	T2	25	Stretch

The carbon atom has a charge distribution of -0.930 and all the hydrogen atoms have a charge distribution of 0.233. The reason why the carbon atom is more negative is because it has a higher electronegative value compared to hydrogen. hence the electrons in the carbon-hydrogen bond will be more attracted to the carbon. Therefore, the carbon atom has a higher electron density compared to the hydrogens atoms. Thus it is more negative.

9 modes are expected from the (3N-6) rule. The modes 1, 2 and 3 are degenerate which each other. Mode 4 and 5 are degenerate to each other. Modes 7, 8 and 9 are also degenerate to each other. Mode 6 is highly symmetrical. 2 bands are expected to see in an experimental spectrum of gaseous methane.

Haber-Bosch reaction energy calculation for CH4

C + 2H2 → CH4

E(H2)= -1.1785394 au

2*E(H2)= -2.3570788 au

E(C)= -37.77600769 au

E(CH4)= -40.5240140 au

ΔE=E(CH4)-[E(C)+2*E(H2)]= -0.39092751 au = -1026.4 kJ/mol

[[2]]

Molecular orbitals of CH4
MO 1	MO 2	MO 3	MO 5	MO 6

This MO is occupied. This MO shows contribution from a 1s orbital on the central carbon. This orbital is not involved in bonding, this is because the energy of the MO is very deep -10.16707 a.u (5dp).	This MO is occupied. This MO is the 2s orbital on carbon bonding with the 1s orbitals on all four hydrogen atoms. This is not antibonding. There is a sigma interaction, no pi interactions.	This MO is occupied. This MO is a 2px atomic orbital on the central carbon atom, bonding to all the 1s hydrogen atoms. However, hydrogen atoms 2 and 5 are in the same phase bonding to each other. This is true for hydrogens 3 and 4. In addition to this, hydrogen 2 and 5 are antibonding to 3 and 4 s orbitals. MO's 3,4 and 5 are degenerate (same energies), simply because they either have a px, py or pz interacting with the hydrogens 1s orbital.	this MO is occupied. This MO has interactions similar to previous MO, just it uses a 2pz orbital on carbon interacting with hydrogens 1s. This MO is the HOMO, it is lower in energy than the LUMO. This is involved in bonding with sigma interactions.	This MO is the LUMO. It is higher in energy than HOMO. It is not involved in bonding because it is unoccupied. There is anti bonding interaction between the 3s orbital on the carbon to the surrounding hydrogens 1s orbitals. We see two different phases. In addition to this, the 1s orbitals on all the hydrogen are in the same phase and they have a bonding interaction. This MO is not involved in bonding.

Independence - O2

Key information
Calculation method	RB3LYP
Basis set	6-31G(d.p)
Final energy, E(RB3LYP) (au)	-150.2574243
RMS Gradient Norm (au)	0.00007502
Point group	D*H

O=O bond distance = 1.22Å (2 d.p)

         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

Project molecule

O2 molecule

The optimisation file is liked to here

Mode	Wavenumber (cm-1)	Symmetry	Intensity (arbitrary)	Type of vibration
1	1643	SGG	0	Stretch

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 0.5/1

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, overall it is good, however you didn't use the built in headings which allow the wiki to auto-generate a table of contents, this is very useful to the reader of a long wiki page with lots of data.

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 4.5/5

Have you completed the calculation and included all relevant information?

YES

Have you added information about MOs and charges on atoms?

YES, very good explanations well done. Note that MO6 is actually the antibonding counterpart to the bonding MO2.

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 did an extra calculation on O2 well done!