IMM2js2016
NH3 molecule
Summary information
N-H bond length = 1.01798 Å
Corresponding literature value = 1.002 Å[1]
H-N-H bond angle = 105.741o
Corresponding literature value = 107.2o [1]
Calculation method: B3LYP
Basis set: 6-31G(d,p)
Symmetry: C3v
Final energy E(RB3LYP) = -56.5577687299 au
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
NH3 molecule |
Frequency analysis
Mode # Freq Infrared 1 1089.54 145.3814 2 1693.95 13.5533 3 1693.95 13.5533 4 3461.29 1.0608 5 3589.82 0.2711 6 3589.82 0.2711
From the general equation 3N-6 = number of vibrational modes, ammonia is made up of four atoms (N=4) so 6 vibrational modes are expected. Our calculations fall in line with this prediction. From the table above, you can see that both modes 2 and 3, and 5 and 6 are the degenerate pairs. Modes 1, 2 and 3 are bending vibrations whilst modes 4, 5 and 6 are bond stretching vibrations. Mode 4 has highly symmetric vibrations and mode 1 is the 'umbrella' mode.
4 bands would be expected in an experimental spectrum of ammonia.
Charge on the nitrogen atom is -1.125 and charge on the hydrogen atom is 0.375. Nitrogen is more electronegative than hydrogen so nitrogen pulls the surrounding electrons closer to itself, hence nitrogen should have a negative value for atomic charge whilst the hydrogen atoms should have positive values.
N2 molecule
Summary information
Calculation method: B3LYP
Basis set: 6-31G(d,p)
Symmetry: D∞h
E(RB3LYP)= -109.52359111 au
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
N2 molecule |
Frequency analysis
Mode # Freq Infrared 1 2457.33 0.0000
H2 molecule
Summary information
Calculation method: B3LYP
Basis set: 6-31G(d,p)
Symmetry: D∞h
E(RB3LYP)= -1.15928020 au
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
H2 molecule |
Frequency analysis
Mode # Freq Infrared 1 4465.68 0.0000
Calculations for Haber-Bosch Process
E(NH3)= -56.55776873 au 2*E(NH3)= -113.11553746 au E(N2)= -109.52412868 au E(H2)= -1.17853936 au 3*E(H2)= -3.53561808 au ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.0557907 au = -146.47848285 kJ/mol
Since ΔE is negative, the product in this reaction must have a lower energy than the reactants. Therefore the ammonia product must be more stable than the gaseous reactants.
Project molecule (F2)
Summary information
Calculation method: B3LYP
Basis set: 6-31G(d,p)
Symmetry: D∞h
E(RB3LYP) = -199.42620785 au
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
F2 molecule |
Frequency analysis
Mode # Freq Infrared 1 1065.09 0.0000
Since fluorine molecules are diatomic, the only mode of vibration is a bond stretching vibration. There are no atomic charges on either of the fluorine atoms because two atoms of the same element are covalently bonded together.
Molecular orbitals
| MO | Visualization of MOs | Comments |
|---|---|---|
| 1 and 2 |   |
The visualization of these molecular orbitals show the combination of the 1s atomic orbitals. The molecular orbitals are hardly seen in the images since there is very little overlap between these orbitals and fluorine has such a high electronegativity that the majority of electron density is incredibly close to the nuclei of both fluorine atoms. The bottom image shows the anti-bonding orbital with the anti-phase (green) part of the wavefunction just about visible. Both of these molecular orbitals are filled so there is no overall contribution towards the bond order of the molecule. These molecular orbitals have very low energy values since they are derived from core (1s) atomic orbitals. |
| 3 and 4 |   |
These images display the mixing of the 2s atomic orbitals. Compared to molecular orbitals 1 and 2, there is much more overlap between the fluorine atoms shown in the top image therefore this molecular orbital contributes significantly to the chemical bonding in the fluorine molecule. This also explains significant energy difference ween these two molecular orbitals since the greater overlap deepens the energy of the bonding orbital. The anti-bonding orbital (bottom) has no overlap. Again, both of these molecular orbitals are filled so neither can influence the bond order. |
| 5 |  |
The molecular orbital here is created from the bonding overlap of the p atomic orbitals which are along the axis of the bond. "Mixing" is what gives rise to the shape of this molecular orbital with the positive part of the wavefunction located in-between the fluroine atoms and the negative part pointing outwards. This is the only molecular orbital in this fluorine molecule that contributes to the chemical bonding, hence resulting a fluorine molecule to have an overall bond order of 1. |
| 6 and 7 |   |
These filled molecular orbitals are formed from the combination of the p atomic orbitals which are orthogonal to the bond axis, creating pi bonds. These two molecular orbitals are identical in energy and thus are degenerate. |