Rep:Mod:Vladimir

Molecular Optimisation

Different molecular analysis

Molecule: NH3

Bond angle = 109.471

Bond lenght = 1.3 angstrom

Calculation method: RB3LYP

Basis Set: 6-31G(d,p)

Final Energy (au):-56.55776871

RMS Gradient: 0.05399560

Point Group: C3V

Charge on N atom = -1.125 Charge on H atoms = +0.375

I would expect a negative charge on N because it is a more electronegative element than H.

 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

Ammonia

Link to the Files

From the 3N-6 rule I would expect 3*4-6=12 vibrations in these molecule. As seen from the list of vibrations above this rule is followed. Modes 2-3 and 5-6 are degenerate. They vibrate with the same frequency and thus the same energy. Modes 1,4,5,6 are stretching. Modes 2 and 3 are bending. Mode 4 is highly symmetric. Mode 1 is called the umbrella vibration mode. In the experimental spectrum of gaseous ammonia you would expect to see only 2 modes, because modes 4,5 and 6 have a very low intensity and as a result of noise on the spectrum these would not be seen. Also, modes 2 and 3 are degenerate so only one signal would be seen for these two as the spectrometer cannot resolve modes of the same frequency.

Molecule: H2

Bond angle = N/A

Bond lenght = 0.74279 angstrom

Calculation method: RB3LYP

Basis Set: 6-31G(d,p)

Final Energy (au):-1.17853936

RMS Gradient: 0.00000017

Point Group: Dinfh

Charge on H atoms = no charge, neutral 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

Hydrogen

Link to the Files

Only one vibration is observed which is IR inactive because there is no net change in dipole.

Molecule: N2

Bond angle = N/A

Bond lenght = 1.10550 angstrom

Calculation method: RB3LYP

Basis Set: 6-31G(d,p)

Final Energy (au):-109.52412868

RMS Gradient: 0.00000060

Point Group: Dinfh

Charge on H atoms = no charge, neutral 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

Nitrogen

Link to the Files

Only one vibration is observed which is IR inactive because there is no net change in dipole.

Total Energy Calculation

N2 + 3H2 -> 2NH3

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]=

2*E(NH3)= -113.11553742 E(N2)= -109.52412868 3*E(H2)= -3.53561808

ΔE=-113.11553742-[-109.52412868+-3.53561808]= -0.05579066 au

ΔE=-148.12 kJmol^-1

According to an online resource the reaction is indeed exothermic, but releases only 92.4 kJ mol^-1 of energy

This is good as the value confirms that the reaction is exothermic; however, the calculations account only for the energy of the reactants and products. In actual fact the environment where the reaction was carried out matters and the way the molecules interact.

Molecule: Cl2

Bond angle = N/A

Bond lenght = 2.04101 angstrom

Calculation method: RB3LYP

Basis Set: 6-31G(d,p)

Final Energy (au):-920.34987875

RMS Gradient: 0.00011755

Point Group: Dinfh

Charge on Cl atoms: Neutral molecule thus no charge

        Item               Value     Threshold  Converged?
 Maximum Force            0.000204     0.000450     YES
 RMS     Force            0.000204     0.000300     YES
 Maximum Displacement     0.000566     0.001800     YES
 RMS     Displacement     0.000801     0.001200     YES

Chlorine

Link to the Files

Similarly to N2 only one vibration is observed which is also IR inactive because there is no net change in dipole.

Molecular Orbitals of Cl2

1)

In this example two 1s AOs contribute to the MO. However, their energy is so small that no overlap of orbitals is seen. Thus this orbital is non-bonding.

2)

In this example two 3py orbitals overlap in phase in order to form a pi orbital. Such configuration is stable and thus has a low energy. This orbital is bonding.

3)

In this example two 3py orbitals overlap out of phase to form a pi orbital which is of a higher energy than in the previous example where the overlap was in phase. This orbital is anti-bonding.

4)

In this example two 3pz orbitals overlap in phase to form a pi orbital of low energy. The geometry of the overlap is different to the overlap of 3py orbitals in phase and in fact the overlap in this example produces a configuration of a lower energy. This orbital is bonding.

5)

Finally, here 3pz orbitals overlap out of phase forming the LUMO with a high energy. It is the only molecular orbital which does not have electrons and has the highest energy at the same time. This orbital is anti-bonding.