NH3 Molecule

Molecule Information

Optimised NH₃ Molecule
Name of Molecule	Ammonia
Calculation Method	RB3LYP
Basis Set	6-31G(d,p)
Final Energy E(RB3LYP) in atomic units (a.u.)	-56.55776873
RMS Gradient	0.00000485
Point Group	C_3V
N-H Bond Length (Å)	1.01798
H-N-H Bond Angle (°)	105.74115

Optimised Ammonia

The optimisation file is linked to here

"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=-5.986269D-10
 Optimization completed.
    -- Stationary point found.

Animating The Vibrations

"Table of data for each vibration"
Wavenumber (cm-1)	Symmetry	Intensity
1090	A1	145
1694	E	14
1694	E	14
3461	A1	1
3590	E	0
3590	E	0

Questions on Vibrations

From the 3N-6 rule, one would expect to see 6 vibrational modes. The degenerate modes are the modes with 1694 cm ^-1 and 3590 cm ^-1 wavenumbers. The bending vibrational modes are at 1090 cm ^-1 & 1694 cm ^-1. The stretching vibrational modes are at 3590cm ^-1 & 3461cm- ^-1. The highly symmetric modes are 3461 cm ^-1 and 1090 cm ^-1. The umbrella mode is 1090 cm ^-1. In the experimental spectrum of gaseous ammonia, one would expect to see 2 bands.

Picture of Charge Distrubution on Optimised NH₃

The nitrogen would hold a negative charge whilst the hydrogen would be positive. This is because nitrogen has a greater electronegativity, hence it withdraws a greater amount of electron denisty and forms a partial negative charge on the nitrogen. A partial positive charge would resultantly be formed on the hydrogen, thus forming a neutral molecule overall.

N₂ Molecule

Molecule Information

Optimised N₂ Molecule
Name of Molecule	Nitrogen
Calculation Method	RB3LYP
Basis Set	6-31g(D,P)
Final Energy E(RB3LYP) in atomic units (a.u.)	-109.52412868
RMS Gradient	0.02473091
Point Group	D_∞h
Nitrogen Bond Length (Å)	1.11
Nitrogen Bond Angle (°)	No bond angle present

Optimised Nitrogen

The optimisation file is linked to here

"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=-3.401074D-13
 Optimization completed.
    -- Stationary point found.

Animating The Vibrations

"Table of data for each vibration"
Wavenumber (cm-1)	Symmetry	Intensity
2457	SGG	0

Questions on Vibrations

N₂ has one vibrational mode which is a symmetric mode. This vibrational mode is not IR active as there is no change in overall dipole moment.

Picture of Charge Distrubution on Optimised N₂

H₂ Molecule

Molecule Information

Optimised H₂ Molecule
Name of Molecule	Hydrogen
Calculation Method	RB3LYP
Basis Set	6-31G(d,p)
Final Energy E(RB3LYP) in atomic units	-1.17853936
RMS Gradient	0.00000017
Point Group	D_∞h
H-H Bond Length (Å)	0.74279
H-H Bond Angle (°)	No bond angle Present

Optimised Hydrogen

The optimisation file is linked to here

"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.

Animating The Vibrations

"Table of data for each vibration"
Wavenumber (cm-1)	Symmetry	Intensity
4466	SGG	0

Questions on Vibrations

H₂ has one vibrational mode which is a symmetric mode. This vibrational mode is not IR active as there is no change in overall dipole moment.

Picture of Charge Distrubution on Optimised H₂

Structure and Reactivity

Unique Identifier for the Complex: ABASUR

Link to Complex in CCDC

H-H bond length (Å): 1.00

Difference in H-H Bond Length between the crystal Structure and computational value (Å): 0.25721

The bond length of the H-H bond in the transition metal complex is 1.00 Å. The bond length is shorter in the gaseous H₂ molecule than in the transition metal complex (Dihydrogen-(dihydrogen bis(3,5-bis(trifluoromethyl)pyrazolyl)-borate-H,N)-bis(tri-isopropylphosphine)-ruthenium ). This is because in the transition metal complex, Ru-H bonds are also formed with each hydrogen. The d_z² orbital of ruthenium overlaps with the bonding orbital of the H₂ bond, thus withdrawing electron density from the bonding orbital to increase the bond length of the H-H bond in the complex. The d_xy orbital of ruthenium also overlaps with the anti-bonding orbital to partially weaken the bond and thus further increase the bond length of the H-H bond as well. Hence the H-H bond length in the transition metal complex is longer than in the gaseous molecule.

It is also important to note that because we are using computational methods to analyse our molecular structures, there are going to be error which occur due to the limitations of this computational method. Using a better computational method which takes more parameters into consideration would yield a greater accuracy.

Energies of the Haber-Bosch Process

E(NH3)= -56.55777 a.u.

2*E(NH3)= -113.11554 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)]=-0.05579 a.u.

ΔE= -146.47848 kJ/Mol

As this is an exothermic reaction, the product formed is more energetically stable than the reactant. Therefore, the ammonia product is 146.48 kJ/Mol (2dp) more stable than the gaseous reactants (N2 and H2)

CO Molecule

Molecule Information

Optimised CO Molecule
Name of Molecule	Carbon Monoxide
Calculation Method	RB3LYP
Basis Set	6-31G(d,p)
Final Energy E(RB3LYP) in atomic units	-113.30824232
RMS Gradient	0.03370421
Point Group	C_∞V
C-O Bond Length (Å)	1.13794
C-O Bond Angle (°)	No Bond Angle present

Optimised Carbon Monoxide

The optimisation file is linked to here

"Item Table"


 Item               Value     Threshold  Converged?
 Maximum Force            0.000007     0.000450     YES
 RMS     Force            0.000007     0.000300     YES
 Maximum Displacement     0.000003     0.001800     YES
 RMS     Displacement     0.000004     0.001200     YES
 Predicted change in Energy=-2.221225D-11
 Optimization completed.
    -- Stationary point found.

Animating The Vibrations

"Table of data for each vibration"
Wavenumber (cm-1)	Symmetry	Intensity
2209.01	SG	67.96

Questions on Vibrations

CO has one stretching vibrational mode which is IR active as it results in a change in overall dipole moment. Hence one band should appear in its IR spectrum.

Picture of Charge Distrubution on Optimised CO

The oxygen has a greater electronegativity than the carbon, thus withdrawing a greater amount of electron density and resulting in a partial negative charge forming on the oxygen. The carbon would hold a partial positive charge.

Molecular Orbitals of CO

Analysis of chosen Molecular Orbitals of CO
	MO 1	MO 2	MO 3	MO 4	MO 5
Image of Molecular Orbital
Analysis of Molecular Orbital	This is a non- bonding orbital which is very deep in energy and occupied. The 1S atomic orbital of the oxygen contributes mainly to this molecular orbital. However, this isn't involved in bonding due to being tightly held in by the oxygen atom.	This is a bonding orbital which is partially deep in energy. The contribution mainly occurs from the 2s orbital on the carbon and the 2p orbital on the oxygen. Being occupied, it also resultantly has an effect on the overall bonding of the molecule.	Here, the majority of the contribution to the molecular orbital is from the 2p orbital on each atom, forming a pi bond. It forms an occupied bonding orbital which isn't very deep in energy.	This is the LUMO and therefore high in energy. The contribution to this molecular orbital mainly occurs from the 2p orbitals on each atom which are out of phase to form an anti-bonding orbital. This is unoccupied (hence not having any contribution towards the bonding of the molecule).	This is the HOMO and therefore high in energy. The contributions mainly occur from the two p-orbitals of each atom (evident considering there are nodes on each atom in this bonding orbital). The orbitals are in phase and thus overlap to form a bonding orbital which is occupied.

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

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 4.5/5

Have you completed the calculation and included all relevant information?

YES

Have you added information about MOs and charges on atoms?

YES, overall very good explanations. For MO2 you could have been more clear that the effect on the overall bonding of the molecule is to increase the bonding. (You say it has an effect but you leave the reader to infer what the effect is, it may seem obvious but you should state exactly what you mean in scientific writing.)

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 found.

NH3 Molecule

Molecule Information

Animating The Vibrations

Questions on Vibrations

Picture of Charge Distrubution on Optimised NH3

N2 Molecule

Molecule Information

Animating The Vibrations

Questions on Vibrations

Picture of Charge Distrubution on Optimised N2

H2 Molecule

Molecule Information

Animating The Vibrations

Questions on Vibrations

Picture of Charge Distrubution on Optimised H2

Structure and Reactivity

Energies of the Haber-Bosch Process

CO Molecule

Molecule Information

Animating The Vibrations

Questions on Vibrations

Picture of Charge Distrubution on Optimised CO

Molecular Orbitals of CO

Marking

Wiki structure and presentation 1/1

NH3 1/1

N2 and H2 0.5/0.5

Crystal structure comparison 0.5/0.5

Haber-Bosch reaction energy calculation 1/1

Your choice of small molecule 4.5/5

Independence 0/1

Picture of Charge Distrubution on Optimised NH₃

N₂ Molecule

Picture of Charge Distrubution on Optimised N₂

H₂ Molecule

Picture of Charge Distrubution on Optimised H₂