Rep:Mod:test page AC

NH3 molecule

N-H bond distance= 1.01798 Angstroms

H-N-H bond angle= 105.741 degrees

Calculation method= RB3LYP

Basis set= 6-31G(d.p)

Final energy= -56.55776873 au

RMS gradient= 0.00000485 au

point group= C3V

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

File:AC NH3 OPTF POP.LOG The optimisation file is liked to here

Questions how many modes do you expect from the 3N-6 rule? A:N=4 therefore modes=6

which modes are degenerate (ie have the same energy)? A: modes 6 and 5 and modes 2 and 3 have the same energy(degenerate) as they have the same frequency.

which modes are "bending" vibrations and which are "bond stretch" vibrations? A: mode1: bond bend, mode2: bond stretch, mode3: bond bend, mode4: bond stretch, mode5: bond stretch, mode6:bond stretch

which mode is highly symmetric? A: mode1

one mode is known as the "umbrella" mode, which one is this? A: mode1

how many bands would you expect to see in an experimental spectrum of gaseous ammonia? 5 bands because the 4th mode is a symmetric stretch.

charge on N atom: -1.125, charge on H atom:0.375. The N would be expected to have a negative charge and the hydrogen positive due to Nitrogen being the more electronegative of the two atoms, pulling electrons in the covalent bond towards itself hence having a negative charge.

N2 molecule

Calculation method:RB3LYP

Basis set:6-31G(d.p)

Final energy: -109.52412868 au

RMS gradient: 0.00000365 au

Point group: D∞H

Item               Value     Threshold  Converged?
 Maximum Force            0.000006     0.000450     YES
 RMS     Force            0.000006     0.000300     YES
 Maximum Displacement     0.000002     0.001800     YES
 RMS     Displacement     0.000003     0.001200     YES

N2 molecule

Frequency for N2 molcecule: 2457.31, this confirming there is no negative frequencies. And satisfying that there should be one frequency. (3x2)-5=1

H2 molecule

Calculation method: RB3LYP

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

Final energy: -1.17853936 au

RMS gradient: 0.09719500 au

Point group: D∞H

         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

Determining the energy of the reaction of N2 + 3H2 -> 2NH3

E(NH3)= -56.55776873 au

2*E(NH3)= -113.1155375 au

E(N2)= -109.52412868 au

E(H2)= -1.17853936 au

3*E(H2)= -3.53561808

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.0557904 au -146.47770636 kJ/mol

It can be seen that the product of ammonia is more stable than the gaseous reactants as there is a release of energy (exothermic reaction) showing that product is lower in energy and therefore more stable.

Project molecule- HCl

Calculation method:RB3LYP

Basis set:6-31G(d.p)

Final energy: -460.80077875 au

RMS gradient: 0.00005211 au

Point group: C∞V

Item               Value     Threshold  Converged?
 Maximum Force            0.000090     0.000450     YES
 RMS     Force            0.000090     0.000300     YES
 Maximum Displacement     0.000139     0.001800     YES
 RMS     Displacement     0.000197     0.001200     YES

HCl molecule

As seen by the image above there is only one mode. This is consistent with the expected value (3x2 -5=1) for a linear molecule with two atoms.

It can be seen from the image above that the negative charge in the molecule lies with the chlorine atom. Chlorine is the more electronegative than hydrogen and so this result is expected as chlorine attracts the two electrons in the covalent towards itself and gives the HCl molecule ionic character.

Orbitals

The occupied MO shown above is very low in energy (-9.47437 au), the AO that contributes to this MO is the 2s from the chlorine. The large energy difference between the 1s AO from the hydrogen and the 2s AO from the Cl atom results in no interaction between the two AOs. Therefore forming an MO where there is no contribution from the hydrogen AO.

This MO above is formed by 3s AO from the Cl atom. It can be seen that this orbital is not perfectly spherical and is the first orbital which seems to have contribution from the H 1s AO; all orbitals lower in energy than this are only contributed to by exclusively the Cl AOs. This MO orbital shape could be due to some mixing between the 1s H AO and the 3s Cl AO. This MO has an energy of -0.84773 au.

The screenshot above represents the main bonding orbital in HCl between the 3pz AO from the Cl atom and the 1s AO from the H atom which the forms the sigma bond in the molecule. The orbital has an energy of -0.47433 au.

This a nonbonding MO that has no contribution from the H 1s AO and all contribution is attributed to the 3px Cl orbital, this is due to wrong orientation therefore no interaction is possible. This is the HOMO orbital with an energy of -0.33163 au, there is also another nonbonding orbital which is of a degenerate energy but in a different plane and is caused by the other 3py AO from Cl which has the wrong orientation for interaction with the H 1s AO also.

The MO above represents the antibonding MO in the HCl molecule and is also the LUMO orbital. The 3pz AO from the Cl atom and the H 1s AO interact to form this antibonding MO. This MO gives an energy of 0.01366 au

HBr molecule

Calculation method: RB3LYP

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

Final energy:-2572.30644608 au

RMS gradient: 0.00000078 au

Point group: C∞V

Item               Value     Threshold  Converged?
 Maximum Force            0.000001     0.000450     YES
 RMS     Force            0.000001     0.000300     YES
 Maximum Displacement     0.000003     0.001800     YES
 RMS     Displacement     0.000004     0.001200     YES

HBr molecule

Comparison to HCl

The optimized energy for HCl is -460.80077875 au and for HBr it is -2572.30644608 au, this could be due to the reduced reactivity of HBr relative to HCl. This could be explained by the larger dipole moment that HCl has relative to HBr.