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Rep:Mod:le1517

From ChemWiki

Wiki Report for computational models by Lazarus Esayas

NH3

What is the molecule?

NH3

What is the calculation method?

RB3LYP

What is the basis set?

6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)?

-56.44397188

What is the RMS gradient?

0.05399560

What is the point group of your molecule?

C3V

Optimised bond length: 1.01798 Angstroms

Optimised bond angle: 105.741


 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

Optimised NH3

NH3

How many modes do you expect from the 3N-6 rule?

6

Which modes are degenerate (ie have the same energy)?

2:3, 5:6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

Bending modes = 1, 2, 3

Stretching modes = 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

Charge for N = -1.1.25 Charge for H = 0.375

This is due to high difference in electronegativity between nitrogen and hydrogen and so a negative charge is expected for the hydrogen.

N2

What is the molecule?

N2

What is the calculation method?

RB3LYP

What is the basis set?

6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)?

-109.52412868

What is the RMS gradient?

0.00000060

What is the point group of your molecule?

D*H

Optimised bond length: 1.10550 Angstroms

Optimised bond angle: 180



 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

List the frequencies and confirm that there are no negative frequencies

2457.33, therefore no negative frequencies


Optimised N2

Nitrogen


H2

What is the molecule?

H2

What is the calculation method?

RB3LYP

What is the basis set?

6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)?

-1.17853936

What is the RMS gradient?

0.00000017

What is the point group of your molecule?

D*H

Optimised bond length: 0.74279 Angstroms

Optimised bond angle: 180

4465.68cm-1 , therefore no negative frequencies


 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

Optimised H2

Hydrogen

Energy Calculations of Reaction

E(NH3)=

-56.44397188

2*E(NH3)=

-112.88794376

E(N2)=

-109.52412868

E(H2)=

-1.17853936

3*E(H2)=

-3.53561808

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]=

-112.88794376 - (-109.52412868 + -3.53561808) = 0.171803

= 451.068811 kJ mol -1

Since energy of reaction is positive, the reactants are less stable than the product of NH3


CF4

What is the molecule?

CF4

What is the calculation method?

RB3LYP

What is the basis set?

6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)?

-437.47627267

What is the RMS gradient?

0.00004048

What is the point group of your molecule?

TD

Optimised bond length: 1.32939 Angstroms

Optimised bond angle: 109.471


Charge on F = -0.352 Charge on H = 1.407

This is due to the highly electronegative fluorine atoms withdrawing electron density from the carbon atom.



Item               Value     Threshold  Converged?
 Maximum Force            0.000078     0.000450     YES
 RMS     Force            0.000042     0.000300     YES
 Maximum Displacement     0.000133     0.001800     YES
 RMS     Displacement     0.000071     0.001200     YES

Optimised CF4


CF4

This molecular orbital is a result of the overlap between the π orbitals of each of the fluorine atoms and is around the LUMO region.

Node along each of the bonds and so it can be seen that these are anti-bonding orbitals coming from the fluorine p orbitals.

Another π orbital, anti-bonding orbital with contributions from fluorine atoms, overlap between non bonding atoms.

Bonding orbital using the πz orbital from the fluorine atom and the σ orbital from the carbon which is low in energy

Overlap of σ orbitals from all 5 atoms in the molecule and so can be judged to be a bonding orbital and is deeper in energy than the rest of the bonding orbitals.