Rep:Mod:HJY9591
NH3 Molecule
NH3 Molecule Summary
Calculation Type: FREQ
Calculation Method: RB3LYP
Basis Set: 6-31G(d,p)
Charge: 0
Spin: Singlet
E(RB3LYP): -56.55776873 a.u.
RMS Gradient Norm: 0.00000485 a.u.
Imaginary Freq: 0
Dipole Moment: 1.8466 Debye
Point Group: C3V
Optimised N-H Bond Distance: 1.01798 Å
Optimised H-N-H Bond Angle: 105.741 degrees
NH3 Optimisation
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=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
The optimisation file is linked to here
NH3 Molecule |
NH3 Vibrations Analysis
All the modes are active at positive frequencies, which means that the optimisation of NH3 molecule is completed.
6 vibrational modes are expected from 3N-6 rule. An ammonia molecule has 4 atoms, N=4 thus 3*4-6 = 6 vibrational modes.
Based on the vibration frequencies in the results, Modes 2 and 3 are degenerate and Modes 5 and 6 are degenerate because they are active at the same frequencies respectively at 1693.95 and 3589.82 cm^(-1).
Bond bending: Modes 1(wagging), 2(scissoring) and 3(scissoring)
Bond stretching: Modes 4, 5 and 6 (higher frequencies)
Highly symmetric stretching: Mode 4
Umbrella mode: Mode 1
According to the vibration analysis result, there are 4 non-degenerate vibrational modes of NH3. Mode 1 (wagging) results in net change of dipole moment. Modes 2 and 3 are bending modes with net change of dipole moment. Modes 5 and 6 (stretching) also have net change of dipole moment. Mode 4 is highly symmetrical stretching, where the dipole effects cancel out causing zero net change of dipole moment. Hence, there are 3 of them that are IR active (i.e. with net change of dipole moment). Three peaks are expected to be found in an experimental spectrum of gaseous ammonia.
NH3 Charge Analysis
Nitrogen: -1.125
Hydrogens: +0.375
As nitrogen is more electronegative than hydrogen, nitrogen pulls the electrons of hydrogens towards itself in order to attain a larger share on electrons. Thus, the unequal sharing of electrons results in a higher electron density on nitrogen and thus a negative charge on nitrogen and positive charges on the hydrogens.
H2 Molecule
H2 Molecule Summary
Calculation Type: FREQ
Calculation Method: RB3LYP
Basis Set: 6-31G(d,p)
Charge: 0
Spin: Singlet
E(RB3LYP): -1.17853935 a.u.
RMS Gradient Norm: 0.00003809 a.u.
Imaginary Freq: 0
Dipole Moment: 0.0000 Debye
Point Group: Dinf*H
H2 Optimisation
Item Value Threshold Converged?
Maximum Force 0.000066 0.000450 YES
RMS Force 0.000066 0.000300 YES
Maximum Displacement 0.000087 0.001800 YES
RMS Displacement 0.000123 0.001200 YES
Predicted change in Energy=-5.726834D-09
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7429 -DE/DX = -0.0001 !
--------------------------------------------------------------------------------
The optimisation file is linked to here
H2 Molecule |
H2 Bond Length
H-H bond length is 0.74289 Å.
H2 Vibrational Frequency
Mode 1: 4464.36 cm^(-1)
H2 only has one vibration mode because it is a simple diatomic molecule. It has no bond angles or torsional angles to consider other modes of vibrations. As expected from 3N-5 rule (linear molecule), number of vibrational mode is equal to 3*2-5=1. This calculation conforms to the frequency analysis result. There is no negative frequency found suggesting that the optimisation is completed successfully.
N2 Molecule
N2 Molecule Summary
Calculation Type: FREQ
Calculation Method: RB3LYP
Basis Set: 6-31G(d,p)
Charge: 0
Spin: Singlet
E(RB3LYP): -109.52412868 a.u.
RMS Gradient Norm: 0.00000001 a.u.
Imaginary Freq: 0
Dipole Moment: 0.0000 Debye
Point Group: Dinf*H
N2 Optimisation
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 0.000000 0.001200 YES
Predicted change in Energy=-8.006836D-17
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
The optimisation file is linked to here
N2 Molecule |
N2 Bond Length
N-N bond length is 1.10550 Å.
N2 Vibrational Frequency
Mode 1: 2457.33 cm^(-1)
Comparing to the vibrational frequency of H2, the vibrational mode of N2 is active at a lower frequency. The grater the masses of the atoms the lower the frequency. H2 has a shorter bond length and thus a stronger bond than N2, which results in a higher vibrational frequency - it takes more energy for the H-H bond to stretch. There is no negative frequency found thus the optimisation is completed.
Haber Process Reaction Energy
E(NH3)= -56.55776873 a.u.
2*E(NH3)= -113.11553750 a.u.
E(N2)= -109.52412868 a.u.
E(H2)= -1.17853935 a.u.
3*E(H2)= -3.53561805 a.u.
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579077 a.u. = -146.47867779 kJ/mol
The energy for converting hydrogen and nitrogen gas into ammonia gas is -146.47867779 kJ/mol and thus the reaction is exothermic. The ammonia product is more stable as it possesses less energy.
CO Molecule
CO Molecule Summary
Calculation Type: FREQ
Calculation Method: RB3LYP
Basis Set: 6-31G(d,p)
Charge: 0
Spin: Singlet
E(RB3LYP): -113.30945314 a.u.
RMS Gradient Norm: 0.00000035 a.u.
Imaginary Freq: 0
Dipole Moment: 0.0599 Debye
Point Group: C*V
CO Optimisation
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=-1.454068D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1379 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
The optimisation file is linked to here
CO Molecule |
CO Bond Length
Optimised C-O Bond Distance: 1.13794 Å
This is a very short and strong bond.
CO Vibrational Frequency
Mode 1: 2209.03 cm^(-1)
CO Charge Analysis
C: +0.506
O: -0.506
Oxygen is more electronegative than carbon thus oxygen pulls electrons towards itself to gain a larger share on electrons leading to the negative charge. Electrons on the covalent bond are closer to the oxygen leaving a positive charge on the carbon.
Molecular Orbitals of CO
Figure 1 shows the occupied bonding combination of two 2s AOs of C and O atoms. These two AOs overlap strongly leading to one extended surface. The overlap takes place along the internuclear axis indicating that a σ bond is formed. The energy level of this MO is -1.15790 au, which is much higher in energy than those two below it but still in the low energy region. This is a valence orbital that contributes to the overall chemical bonding of the molecule.
Figure 2 shows the occupied antibonding MO of two 2s AOs of C and O atoms in the low energy region at -0.57004 au. As mentioned above, these two AOs overlap strongly thus this occupied MO contributes to the chemical bonding. The energy difference between the bonding and antibonding MOs of 2s AOs is large due to the fact that the they overlap to a large extent. This is also a valence orbital that contributes to the overall chemical bonding.
Figure 3 shows the occupied bonding MO of two 2p AOs that are perpendicular to the internuclear axis. This is a sideways overlap creating a π MO in the low energy region at -0.46743 au. It can be seen from the figure that there is another MO of exactly the same energy as this one. Because there are two p AOs on each atom that are perpendicular to the internuclear axis, two π MOs at the same energy level are formed, which are called degenerate. These two MOs contribute to the overall chemical bonding.
Figure 4 illustrates the combination of two 2p AOs along the internuclear axis, which forms a σ bonding MO at -0.37145 au. This is an occupied MO in the HOMO region, which contributes to the overall chemical bonding. The mixing effects of two MOs cause the distorted shape of this MO.
Figure 5 illustrates the antibonding combination of the perpendicular 2p AOs. This is an unoccupied π* MO at -0.02178 au in the LUMO region, which does not contribute a lot to the overall chemical bonding. As shown in the figure, the next MO is at exactly the same energy as this one thus these two antibonding MOs are degenerate.
Figure 6 shows the unoccupied antibonding combination of the 2p AOs along the internuclear axis. This σ* MO is so high in energy that it has a positive energy +0.26241 au. This MO is not highly involved with the overall chemical bonding.
All three antibonding MOs of 2p AOs in CO molecule are not occupied, which makes the overall structure very stable.
Reference
Senseaircom. 2017. Senseaircom. [Online]. [24 February 2017]. Available from: http://www.senseair.com/senseair/gases-applications/ammonia/