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NH3 molecule

summary

Calculation MethodːRB3LYP

Basis Setː6-31G(d,p)

Final Energy(RB3LYP)ː-56.5577687 a.u.

RMS Gradientː0.0000049 a.u.

Point GroupːC3V

N-H bond distance = 1.02 Å

H-N-H bond angle = 106°

"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
 

The maximum force is below 0.00045 and the converge of the set of values means that the optimisation is completed

JMol of structure

NH3

The optimisation file is linked here

Display vibrations table

Display Vibrations
















Table of vibrations and intensities

wave number cm-1 1090 1694 3461 3590
symmetry A1 E A1 E
intensity arbitrary unit 145 14 1 0
image

The vibrations at 1694 cm-1 and 3590 cm-1 have two degenerate states respectively, 6 modes of vibrations in total

The difference in the intensity is due to the different in the change of dipole during the vibration . The first mode have the highest intensity among the 6 modes because it has the highest change in the dipole (all H-atoms move to the same side). And the two degenerate vibration, modes 5&6, have the lowest change in dipole among the 6 modes hence the lowest intensity.

Question and answer

how many modes do you expect from the 3N-6 rule?

6.

which modes are degenerate (ie have the same energy)?

Modes2&3 , modes5&6.

which modes are "bending" vibrations and which are "bond stretch" vibrations?

Modes1,2,3 are bending. Modes4,5,6 are bond stretch.

which mode is highly symmetric?

Modes 1,4 are highly symmetric.

one mode is known as the "umbrella" mode, which one is this?

Mode 1.

how many bands would you expect to see in an experimental spectrum of gaseous ammonia?

We expect to see four bands, because modes 2&3 and 5&6 are degenerate. But two of the four bands are too low in intensity (1.06&0.27) to be visualise in an real experiment due to the limitation of experimental appliance.

NBO charge

Charge on the N-atom = -1.125

charge on the H-atom = 0.375

The N-atom are expected to have the negative charge due to its high electronegativity

H2 molecule

summary

Calculation MethodːRB3LYP

Basis Setː6-31G(d,p)

Final Energy(RB3LYP)ː-1.1785394 a.u.

RMS Gradientː0.0000002 a.u.

Point GroupːD∞h

H-H bond distance = 0.74 Å

linear structure

"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

 

The maximum force is below 0.00045 and the converge of the set of values means that the optimisation is completed

JMol of structure

H2

The optimisation file is linked here

Display vibrations table

Display Vibrations
















Table of vibrations and intensities

wave number cm-1 4466
symmetry SGG
intensity arbitrary unit 0
image

The intensity of this vibration is zero due to the vibration is symmetric and there's no change in the dipole.

NBO charge

Charge on both of the H-atoms = 0

The two H-atoms have same electronegativity, the charges are cancelled out due to the molecule's symmetry, so the NBO charges are 0.

N2 molecule

summary

Calculation MethodːRB3LYP

Basis Setː6-31G(d,p)

Final Energy(RB3LYP)ː-109.5241287 a.u.

RMS Gradientː0.0000006 a.u.

Point GroupːD∞h

N-N bond distance = 1.11 Å

Linear structure

"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 
 

The maximum force is below 0.00045 and the converge of the set of values means that the optimisation is completed

JMol of structure

N2

The optimisation file is linked here

Display vibrations table

Display Vibrations

















Table of vibrations and intensities

wave number cm-1 2457
symmetry A1
intensity arbitrary unit 0
image

The intensity of this vibration is zero due to the vibration is symmetric and there's no change in the dipole.

NBO charge

Charge on both of the N-atoms = 0

The two N-atoms have same electronegativity, the charges are cancelled out due to the molecule's symmetry, so the NBO charges are 0.

Mono-metallic TM complex that coordinates H2

DECXOY

Sturcture displayed[here]

H-H Bond distance in the structure with identifier DECXOY = 0.9 Å

H-H bond distance gained by computational calculation = 0.74 Å


The computational distances H-H is smaller than that in the structure we gained(DECXOY) which means the H-H bond is shorter and stronger. This can be explained by the difference in the environment of the H-atoms. In the structure, H-atoms are bonded to metal ion and the electronegative TM pull the electron density away form the hydrogens. The electron density diffusion cause the strength of bond between the H-atoms in the structure to be lower, hence longer distance. In addition, in the computational calculation, only parameters are used, while in real experiment, environment should also be taken in to consideration. The effect of this computational failure could be also be an explanation to difference in the H-atoms distance.

Determining the energy for Haber-Bosch process

E(NH3)= -56.5577687 a.u.

2*E(NH3)= 3*-1.17853936 = -113.1155375 a.u.

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= 3*-1.17853936 = -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]

=-113.1155375-[-109.52412868+-3.53561808]

=-0.05579064 a.u.

=-146.5 KJ/mol


The energy change have negative sign, so the ammonia product is more stable.

O2 molecule

summary

Calculation MethodːRB3LYP

Basis Setː6-31G(d,p)

Final Energy(RB3LYP)ː-150.2574243 a.u.

RMS Gradientː0.0000750 a.u.

Point GroupːD∞h

O=O bond distance = 1.22 Å

linear structure

"Item" table

 
         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

 

The maximum force is below 0.00045 and the converge of the set of values means that the optimisation is completed

JMol of structure

O2

The optimisation file is liked to here

Display vibrations table

Display Vibrations















Table of vibrations and intensities

wave number cm-1 1643
symmetry SGG
intensity arbitrary unit 0
image

The intensity of this vibration is zero due to the vibration is symmetric and there's no change in the dipole.

NBO charge

Charge on both of the O-atoms = 0

The two O-atoms have same electronegativity, the charges are cancelled out due to the molecule's symmetry, so the NBO charges are 0.



MOs

1s-1s σ bonding orbital This is non-bonding MO formed by combination of two 1s AO of the oxygen atoms with the same phase. It is deep in energy with the lowest energy level in the structure of oxygen molecule. This MO is occupied by two pair-spin electron and it does not contribute to the bond forming
1s-1s σ* orbital This is an non-bonding MO formed by combination of two 1s AO that out of phase. It is deep in energy with the second lowest energy level in the structure of oxygen molecule. This MO is occupied by two pair-spin electron and it does not contribute to the bond forming
2py-2py π* orbital This is an anti-bonding MO formed by overlapping of two parallel py orbital that out of phase. It is in the HOMO reigion and occupied by two pair-spin electron in the calculated MOs structure. But in theory it is only occupied by one self-spin electron. This difference is due to the failure of calculation.This MO destabilise the O-O bonding
2pz-2pz π* orbital This is an non-bonding MO formed by overlapping of two parallel pz orbital that out of phase. It is in the LUMO reigion and unoccupied with no contribution to the bond forming in the calculated MOs structure. But in theory it is in the HOMO region and is an anti-bonding orbital with the same energy as the other π2p orbital(degenerate). It also have one self-spin electron in it and destabilise the O-O bonding in theory. This difference is due to the failure of calculation.
2px-2px σ* orbital This is an non-bonding MO formed by overlapping of two head-to-head px orbital that out of phase. It is higher in energy and unoccupied in the calculated MOs structure. But in theory it is in the LUMO region. This difference is due to the failure of calculation.This MO do not contribute to the O-O bond forming.

IndependenceːCO2 molecule

summary

Calculation MethodːRB3LYP

Basis Setː6-31G(d,p)

Final Energy(RB3LYP)ː-188.5809395 a.u.

RMS Gradientː 0.0000115 a.u.

Point GroupːD∞h

C=O bond distance = 1.17 Å

O=C=O bond angle = 180 °

linear structure

"Item" table

 
         Item               Value     Threshold  Converged?
          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

 

The maximum force is below 0.00045 and the converge of the set of values means that the optimisation is completed

JMol of structure

CO2

The optimisation file is liked to here

Display vibrations table

Display Vibrations

















Table of vibrations and intensities

wave number cm-1 640 1372 2436
symmetry PIU SGG SGU
intensity arbitrary unit 31 0 545
image

The vibration at 640 cm-1 have two degenerate vibration states.

The intensity of second mode of vibration is zero due to the vibration is symmetric and there's no change in the dipole. The 4th mode of vibration has higher intensity than the 1st and 2nd ones' because the 4th mode vibration have higher change in the dipole.

NBO charge

Charges on both of the O-atoms and C-atom = 0

The two O-atoms have same electronegativity, the charges are cancelled out due to the molecule's symmetry as the three-atom molecule have linear shape, so the NBO charges are 0.

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

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 overall, well done!

One area I would improve is when a molecule is unoccupied you call it a non-bonding MO. It is clearer to say for e.g. the LUMO: This MO has antibonding interations between the O atoms, but does not reduce the bonding in the molecule because it is unnoccupied.

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

You did an extra calculation, well done.

Do some deeper analysis on your results so far