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Rep:MOD:3817fhp

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Molecular Modelling 2

NH3 Molecule

Summary Information

What is the molecule? NH3

What is the calculation method? RB3LYP

What is the base set? 6-31G(d,p)

What is the final energy E(RB3lYP) in atomic units (au)? -56.55776873

What is the RMS gradient (au)? 0.00000485

What is the point group of your molecule? C3V

What is the optimised NH bond distance( Å )? 1.01798

What is the optimised NH bond angle(degrees)? 105.741

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
NH3 molecule

The optimisation file is linked to here

Vibrational Modes

The frequency is in Hz

Description of modes

Mode 1 - N-H wagging (change in dipole moment: active)

Mode 2 - H-N-H scissoring (change in dipole moment: active)

Mode 3 - H-N-H scissoring (change in dipole moment: active)

Mode 4 - N-H symmetric stretching (change in dipole moment: active)

Mode 5- N-H asymmetric stretching (change in dipole moment: active)

Mode 6 - N-H asymmetric stretching (change in dipole moment: active)

How many modes do you expect from the 3N-6 rule? 3(4)-6 = 6

Which modes are degenerate? Using the mode numbers on the Display Vibrations Window, Modes 2 and 3 are degenerate and modes 5 and 6 are degenerate.

Which modes are "bending" vibrations and which are "bond stretch" vibrations? Bending Vibrations:1,2 and 3. Bond Stretch Vibrations = 4,5 and 6.

Which mode is highly symmetric?Mode 4

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? You would expect to see 4 vibration modes for ammonia spectrum. Although there are 6 modes reported, there are two degenerate pairs therefore only 4 modes will be observed as peaks. There are 2 degenerate stretches and two degenerate bends.

Charge Analysis

The charge of Nitrogen atom: -1.125 The charge on the Hydrogen atom: 0.375

You would expect the nitrogen atom to have a negative charge as it is the more electronegative atom. Likewise you would expect the charge on the Hydrogen to be positive as it is more electropositive. Therefore our results fit with what was expected.

N2 Molecule

Summary Information

What is the molecule? N2

What is the calculation method? RB3LYP

What is the base set? 6-31G(d,p)

What is the final energy E(RB3lYP) in atomic units (au)? -109.52412868

What is the RMS gradient(au)? 0.00000060

What is the point group of your molecule? D∞h

What is the optimised NN bond distance(Å)? 1.10550

What is the optimised NN bond angle(degrees)? 180


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
N2 molecule

The optimisation file is linked to here

Frequency Analysis

Vibrational Frequency: 2457.33. This is a positive number therefore the molecule is fully optimised.

H2 molecule

summary Information

What is the molecule? H2

What is the calculation method? RB3LYP

What is the base set? 6-31G(d,p)

What is the final energy E(RB3lYP) in atomic units (au)? -1.17853936

What is the RMS gradient(au)? 0.00000017

What is the point group of your molecule? D∞h

What is the optimised HH bond distance( Å )? 0.74279

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
H2 molecule

The optimisation file is linked to here

Frequency Analysis

Vibrational frequency: 4465.68 This is a positive energy value therefore the molecule i is fully optimised.

Haber-Bosch Reaction

Determining the Energy of Reaction using computer calculated values

E(NH3)=56.55776873

2×E(NH3)=113.1155375

E(N2)=109.52412868

E(H2)=1.17853936

3×E(H2)=3.53561808

ΔE=2×E(NH3)[E(N2)+3×E(H2)]=0.05579074

ΔE(kJmol1)=146.478599028

ΔE=146.479kJmol1

This was converted from hartree to kJmol-1 using an online energy converter

Energy Analysis

The energy change for the reaction is negative therefore energy is given out (exothermic). The product (gaseous ammonia) is more stable than the reactants (gaseous hydrogen and ammonia) as it exists at a lower energy state.

Project Molecule - F2

Summary Information

What is the molecule? F2

What is the calculation method? RB3LYP

What is the base set? 6-31G(d,p)

What is the final energy E(RB3lYP) in atomic units (au)? -199.49825218

What is the RMS gradient (au)?0.00007365

What is the point group of your molecule? D∞h

What is the optimised FF bond distance( Å )? 1.40281

Item Table

        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

The optimisation file is linked to here

Frequency Analysis

Vibrational Frequency: 1065.09. This is a positive value showing that the molecule is fully optimised.

Vibrational information

There is only one vibration for my linear F2 molecule. This is a symmetrical stretch with a frequency of 1065.09.

Charge information

There is no charge on either of my F atoms in the F2 molecule. This is because the atoms are the same therefore, even though Flourine is the most electronegative atom and would usually hold a negative charge, there is no charge interactions between the two atoms. Therefore overall each atom has no charge.

Molecular oribital Information

Molecular Orbital Diagrams

Referring to a standard molecular orbital diagram aids in the understanding and explanation of the MO diagrams. Note: the molecular orbital diagram shown does not contain

Source: P.Atkins and J. de Paula, 'Molecular Structure: Homonuclear Diatomic Molecules', in Atkins Physical Chemistry, Oxford University Press, Oxford UK, 8th edition, 2006, Chapter 11, 366

Axis Definition

The definition of the Cartesian axis also aids in the description of the MO diagrams. Here the bonding axis is defined as the z axis with the x and y axis being orthogonal to this.

1 σ (1s) Bonding Orbital

This orbital is the 1 σ g orbital formed as result of the constructive overlap of the two 1s orbitals of the F atoms. It is difficult to make out the orbital in the image as it is covered by the ball of the F atom. This is because the 1 σ MO's are considered to be core orbitals. They are so deep in energy their involvement in the bonding and interaction of the two atoms is very small. Their relative energy lies at -24.79730 compared to -1.33659 of the 2 σ orbitals

1 σ (2s) Antibonding orbital

This orbital is the 2s σ antibonding orbital. This orbital is formed as a result of the destructive overlap of two 2s orbitals of opposite phase. This antibonding orbital is occupied by 2 electrons. As it is a σ orbital there are no nodes along the bonding (Z) axis, however there is one node orthogonal to the z axis in the middle of the bond. Here the bonding electrons have a high probability of occupying space therefore there is a low probability of finding the antibonding electrons within this area. As an antibonding orbital it has a destabilising effect on the energy of the molecule as the presence of electrons within this bond increases the probability of electrons being found outside of the bonding region. Its energy lies higher than that of its corresponding bonding orbital.

2 σ 2pz g Bonding Orbital

This orbital is the 2 σ g orbital. This orbital is formed as a result of the constructive overlap of two 2p orbitals along the same axis i.e. two 2pz orbitals; one from each Flourine atom. This is a bonding MO orbital, occupied with two electrons. This MO is not very deep in energy at -0.58753 below the relative 0 energy of the molecule and so will have quite a large effect on bonding.

2 π u Bonding Orbital

The Cartesian Axis shown aids in the description of the MO. The orbitals described so far have been sigma orbitals. This is a π orbital. Sigma orbitals have no nodes along the bonding axis. Conversely this π orbital has a node along the bonding axis (shown with the Cartesian axis as the z axis). This is an occupied bonding orbital formed through the combination of two 2px orbitals that lie orthogonal to the bonding axis. This orbital is not very deep in energy at a relative -0.52332 which lies close to the relative 0 energy and so this MO has a strong effect on bonding. There is an additional orbital that of degenerate energy which is formed as a result of the constructive overlap of two 2py orbitals. This degenerate orbital lies orthogonal to the orbital shown in the diagram as another π orbital and exhibits the same shape as this orbital.

1 π g (2px) antibonding orbital

This is the HOMO of the F2 molecule. This is an antibonding orbital occupied by 2 electrons (Note, as before, there is an additional degenerate antibonding orbital occupied by two electrons along the orthogonal y axis exhibiting the same shape). This antibonding orbital is formed as a result of the destructive overlap of the 2px orbitals orthogonal to the bonding (z) axis. This orbital is found in the HOMO/LUMO energy region. As this orbital is located in the HOMO/LUMO region it has a large effect on bonding.

2pz σ u Antibonding Orbital

This is the LUMO in the MO of F2. This is the 2 σ antibonding orbital and is not occupied by any electrons. It is in the HOMO/LUMO region of energy and is formed as a result of the destructive overlap of two 2pz orbitals. As a result of being located in the HOMO/LUMO energy region, this orbital has a large effect on bonding. As the LUMO, if it was to be occupied by electrons, the strong σ bond primarily holding the two F atoms together would break. Therefore this orbital is very important in determining bonding. As a result, this orbital can be considered to have a large destabilising effect on the bonding of the molecule.

Independent section - H2SiO

Summary Information

What is the molecule? H2SiO

What is the calculation method? RB3LYP

What is the base set? 6-31G(d,p)

What is the final energy E(RB3lYP) in atomic units (au)? -365.90001403

What is the RMS gradient (au)?0.00000941

What is the point group of your molecule? Cs

What is the optimised SiH bond distance( Å )? 1.48652

What is the optimised SiO bond distance( Å )? 1.53172

What is the optimised OSiH bond angle( degrees )? 124.156

Item Table

        Item               Value     Threshold  Converged?
Maximum Force            0.000023     0.000450     YES
RMS     Force            0.000009     0.000300     YES
Maximum Displacement     0.000023     0.001800     YES
RMS     Displacement     0.000017     0.001200     YES

Vibrational Analysis

All frequencies here are positive showing that the molecule has been fully optimised.

Charge Distribution

Hydrogen: 0.090

Silicone: 0.681

Oxygen: -0.502

This charge distribution is as expected for this molecule. Oxygen is the most electronegative atom therefore will attract electrons most strongly. Therefore this is expected to have the most negative value.