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NH3 Molecule

Type of Molecule NH3
Calculation Method RB3LYP
Basis set 6-31G(d.p)
Final energy (au) -56.55776873
RMS gradient 0.00000323
Point group C3V
NH bond length (A) 1.01798
HNH bond angle 105.745

A literature value for the HNH bond angle in NH3 is 106.7. This shows that the optimization value is very accurate.[1] A literature value for the NH bond length in NH3 is 1.012. This also shows that the optimization value is very accurate.[1]

  Item               Value     Threshold  Converged?
 Maximum Force            0.000006     0.000450     YES
 RMS     Force            0.000004     0.000300     YES
 Maximum Displacement     0.000014     0.001800     YES
 RMS     Displacement     0.000009     0.001200     YES

Ammonia

Media:MA6516PHUNT_NH3_OPTF_POP.LOG


Table showing the vibrations of a molecule of ammonia

How many modes do you expect from the 3N-6 rule? 6 Which modes are degenerate (ie have the same energy)? (2 and 3), (5 and 6) Which modes are "bending" vibrations and which are "bond stretch" vibrations? bending (1, 2, 3) stretching(4, 5, 6) Which mode is highly symmetric? 4 One mode is known as the "umbrella" mode, which one is this? 1 How many bands would you expect to see in an experimental spectrum of gaseous ammonia? 2. This is due to mode 1 having a strong intensity. Modes 2 and 3 are degenerate and also have relatively strong intensities. The other 3 modes have such small intensities they are probably not visible on the spectrum.

A Gaussview image of the charge distribution in a molecule of ammonia

When carrying out calculations it is always good to check your results against your expectations. Write a sentence saying what charge (positive or negative) you would expect for N and H and why. I would expect the hydrogen atoms in ammonia to carry a more positive charge and the nitrogen atom to carry a more negative charge as nitrogen has a lone pair of electrons attached whereas the hydrogen atoms are just positively charged nuclei as they have each shared an electron to form sigma bonds with the nitrogen. Also nitrogen is more electronegative than hydrogen so it is expected that it carries the negative charge. As expected the hydrogen atoms are positively charged and the nitrogen atom is negatively charged.

H2 Molecule

Type of Molecule H2
Calculation Method RB3LYP
Basis set 6-31G(d.p)
Final energy (au) -1.17853936
RMS gradient 0.00000017
Point group D*H
HH bond length (A) 0.60000

A literature value for the bond length in H2 is 0.74 angstrom. This shows that the optimization value is relatively accurate.[2] There is no overall charge in hydrogen as it is a diatomic homonuclear molecule, so no overall charge or polarity arises within the molecule.

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
Mode 1 shows a symmetrical stretch of the HH bond in H2. As H2 is a symmetrical molecule, a symmetrical stretch will produce no overall dipole, therefore it will not produce an intensity or a band in an IR spectrum.


Hyrdogen

Media:MA6516-H2_OPTF_POP.LOG

N2 Molecule

Type of Molecule N2
Calculation Method RB3LYP
Basis set 6-31G(d.p)
Final energy (au) -109.52412868
RMS gradient 0.00000060
Point group D*H
NH bond length (A) 1.09200

A literature value for the bond length in N2 is 1.10 angstrom. This shows that the optimization value is relatively accurate.[2] There is no overall charge in nitrogen as there it is a diatomic homonuclear molecule, so there is no overall charge or polarity within the molecule.

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
Mode 1 shows a symmetrical stretch of the NN bond in N2. As N2 is a symmetrical molecule, a symmetrical stretch will create no overall dipole, therefore it will register no overall intensity on an IR spectrum.
Nitrogen

Media:MA6516_N2_OPTF_POP.LOG

Haber-Bosch process

N2 + 3H2 -> 2NH3

Energies of gaseous reactants and products and overall energy change during the reaction
E(NH3) -56.55
2*E(NH3) -113.11
E(N2) -109.52
E(H2) -1.18
3*E(H2) -3.54
ΔE=2*E(NH3)-[E(N2)+3*E(H2)] -0.06

Values above are in Hartrees. Value for ΔE in kJ/mol is -146.48. Identify which is more stable the gaseous reactants or the ammonia product? The ammonia product is more stable as it is overall lower in energy than the gaseous reactants

Project Molecule F2

Charge distribution in an F2 molecule. As it is a diatomic homonuclear molecule, there is no overall charge of polarity in the molecule as expected.
Type of Molecule F2
Calculation Method RB3LYP
Basis set 6-31G(d.p)
Final energy (au) -199.49825218
RMS gradient 0.00007365
Point group D*H
FF bond length (A) 1.40281

A literature value for the bond length in F2 is 1.42 angstrom. This shows that the optimization value is relatively accurate.[2]


         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

Mode 1 shows a symmetrical stretch of the FF bond. The intensity registers as zero as the molecule is symmetrical so a symmetrical stretch will not create a dipole within the molecule.
Fluorine

Media:MA6516_F2_OPTF_POP.LOG

Molecular Orbitals F2

(1) 1s bonding orbital. MO made up of core 1s AOs from each F atom. This orbital is very deep in energy and there is very little overlap. This MO and its respective antibonding MO contribute very little to the overall bonding in F2.
Energies of MOs in F2. The 1s bonding (1) and antibonding(2) MOs are very deep in energy compared to the other MOs. As a result they are not heavily involved in the chemical bonding of fluorine. The next two orbitals comprise of the bonding (3) and antibonding (4) MOs formed from valence 2sAOs. Due to better overlap between the AOs the energy difference between these two MOs is much greater than the energy difference between the 1s MOs. The following six MOs are formed from the valence 2pAOs from each F atom. 5 of the 6 orbitals are occupied with only one antibonding (10) MO being left unoccupied. This orbital is the LUMO.
(3)2s bonding orbital. MO formed by the overlap of the two valence 2s AOs on each F atom. This MO has a much higher energy than that of the 1s MO. They overlap to a much greater extent, however the bonding character of this orbital is cancelled out due to the corresponding antibonding orbital also being occuppied.
(4) 2s antibonding orbital. This MO is formed from valence 2s AOs from each F atom. As this MO is occuppied it cancels out the bonding character of the corresponding bonding 2s MO.
(5) 2p bonding orbital occupied. As this orbital is occupied and there is overlap between the two F atoms that is characteristic of a sigma bond. As the corresponding antibonding MO is unoccupied, this MO contributes a sigma bond to F2
(6) 2p bonding orbital occupied. This MO is formed from 2pAOs from each F atom. As the overlap is above and below the plane, it shows pi bonding character. As the corresponding antibonding MO is occupied, this cancels out the pi bonding character between the two atoms and so a pi bond does not form between the two F atoms in F2
(9) 2p antibonding orbital. This is the HOMO. This MO is formed of pAOs from each F atom. As this antibonding orbital is occupied, the pi bonding character from the corresponding MO is cancelled out, leaving only a sigma bond between the two F atoms.
(10) 2p antibonding orbital unoccupied. This is the LUMO and an antibonding orbital. The fact it is unoccupied and that its corresponding bonding orbital is occupied means that there is only one sigma bond present in the bonding in F2. The LUMO is considerably higher in energy than the HOMO.



Reaction of fluorine with water

2F2 + 2H2O ==> O2 + 4HF

Energies of reactants and products and overall energy change during the reaction
E(H2O) -76.42
E(F2) -199.50
E(O2) -150.27
E(HF) -100.43
2*E(H2O) -152.84
4*E(HF) -401.71
ΔE=[4*E(HF) + E(O2)] - [2*E(F2) + 2*E(H2O)] -0.14

Values above given in Hartrees. The value for ΔE in kJ/mol is -367.34.

H2O

As oxygen is more electronegative than hydrogen, it is expected that the oxygen atom carries a negative charge and the hydrogen atoms carry a positive charge. This is as expected.
Type of Molecule H2O
Calculation Method RB3LYP
Basis set 6-31G(d.p)
Final energy (au) -76.41973740
RMS gradient 0.00006276
Point group D2V
OH bond length (A) 0.96522
HOH bond angle 103.745

A literature value for the bond length in H2O is 0.96 angstrom. A literature value for the HOH bond angle is 104.5. This shows that the optimization is relatively accurate.[3]




Item               Value     Threshold  Converged?
 Maximum Force            0.000099     0.000450     YES
 RMS     Force            0.000081     0.000300     YES
 Maximum Displacement     0.000115     0.001800     YES
 RMS     Displacement     0.000120     0.001200     YES
Mode 1 shows a symmetric bend of the OH bonds. Mode 2 shows the OH bonds symmetrically stretching. Mode 3 shows the OH bonds asymmetrically stretching. The IR spectrum for water should show two bands of strong intensity as mode 2 records a very weak intensity.


Water

Media:MA6516_H2O_OPTF_POP.LOG

O2

Type of Molecule O2
Calculation Method RB3LYP
Basis set 6-31G(d.p)
Final energy (au) -150.26605044
RMS gradient 0.00007560
Point group D2V
OO bond length (A) 1.16

A literature value for the OO bond length is 1.12 angstrom. This shows that the optimization is relatively accurate.[4]. As oxygen is a diatomic homonuclear molecule, there is no overall charge or dipole within the molecule.

 Item               Value     Threshold  Converged?
 Maximum Force            0.000131     0.000450     YES
 RMS     Force            0.000131     0.000300     YES
 Maximum Displacement     0.000081     0.001800     YES
 RMS     Displacement     0.000115     0.001200     YES
Mode 1 shows a symmetrical stretching of the OO bond. As oxygen is a homonuclear diatomic molecule, no overall diploe is induced so no intensities will be produced on the IR spectrum
Oxygen

Media:MA6516_O2_OPTF_POP.LOG

HF

As fluorine is more electronegative than hydrogen, it is expected that the hydrogen has a positive charge and the fluorine has a negative charge within the molecule. This is as expected
Type of Molecule HF
Calculation Method RB3LYP
Basis set 6-31G(d.p)
Final energy (au) -100.42746153
RMS gradient 0.00001092
Point group C*V
HF bond length (A) 0.88000

A literature value for the bond length in HF is 0.92 angstrom. This shows that the optimization is relatively accurate.[3]

Item               Value     Threshold  Converged?
 Maximum Force            0.000019     0.000450     YES
 RMS     Force            0.000019     0.000300     YES
 Maximum Displacement     0.000016     0.001800     YES
 RMS     Displacement     0.000022     0.001200     YES
As HF is a polar molecule, it registers an intensity in IR spectroscopy. The one vibrational mode is the stretching of the HF bond.
Hydrogen Fluoride

Media:MA6516_HF_OPTF_POP.LOG

References

  1. 1.0 1.1 CRC Handbook of Chemistry and Physics, 94th ed. http://www.hbcpnetbase.com. Page 9-26.Retrieved 18 June 2013. via https://en.wikipedia.org/wiki/Ammonia_(data_page)
  2. 2.0 2.1 2.2 Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958; B. deB. Darwent, "National Standard Reference Data Series," National Bureau of Standards, No. 31, Washington, DC, 1970; S.W. Benson, J. Chem. Educ., 42, 502 (1965).
  3. 3.0 3.1 http://www.chegg.com/homework-help/questions-and-answers/o-h-bond-lengths-water-molecule-h20-096-h-o-h-angle-1045o-dipolemoment-water-molecule-185--q279785
  4. http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html