View source for Rep:Title=Mod:ChlorineDJN16

==ammonia==

===optimisation===



<u>'''optimised information'''</u>

Calculation Method   =   RB3LYP

Basis Set            =   6-31G(d,p)

E(RB3LYP)            =   -56.55776873 a.u.

RMS Gradient Norm    =   0.00000485 a.u.

Point Group          =   C3V

bond length (N-H)    =   1.01798Å

bond angle  (H-N-H)  =   105.741° 

<pre>

         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=-5.986273D-10
 Optimization completed.
    -- Stationary point found.
</pre>

[[File:Sniperoo1.PNG]]

<i>optimised NH<sub>3</sub></i>


<jmol><jmolApplet>
  <script>frame 1.16</script>
<title>optimised NH3</title>
  <color>black</color>
  <size>300</size>
  <uploadedFileContents>NH3 TEST1 DJN.LOG</uploadedFileContents>
</jmolApplet></jmol>

optimisation file for NH<sub>3</sub> is linked [[media:NH3 TEST1 DJN.LOG| here]]



''Vibrational modes''

[[File:Vibrational modes.PNG]]


Questions:
1. How many modes of vibration are expected from the 3N-6 rule?
2. Which modes are degenerate?
3. Which modes are bending? which are stretching?
4. Which modes are highly symmetric?
5. Which mode is described as 'umbrella' mode?
6. How many main peaks should you expect to see in a spectrum for gaseous ammonia?



Ammonia can absorb IR because it has a dipole moment. 
Ammonia should show six vibrational modes. 
The bends 2&3 are degenerate and the stretches 5&6 are degenerate because the movements are symmetrical but the protons that are bending/stretching are different. 
Mode 4 is highly symmetric. Mode 1 is 'umbrella'. 
The two main peaks in the spectrum are due to mode 1 modes 2/3 (there are also two minuscule peaks due to modes 4 and 5/6)


{| class="wikitable" border="1"
|+ types of vibrational mode
! number of mode !! description
|-
| 1 || umbrella, symmetric mode, bend
|-
| 2 || bend
|-
| 3 || bend
|-
| 4 || symmetric, stretch
|-
| 5 || stretch
|-
| 6 || stretch
|} 


''Charge distribution''

[[File:Chargedist NH3.png| thumb|none | Nitrogen has a higher charge density than the Hydrogen's because it is more electronegative. This gives the Nitrogen a partial negative charge and the Hydrogen's a partially positive charge ]]

==Nitrogen==

===optimisation===

<u>'''optimised information'''</u>

Calculation Method   =   RB3LYP

Basis Set            =   6-31G(d,p)

E(RB3LYP)            =   -109.52412868

RMS Gradient Norm    =   0.00000060 a.u.

Point Group          =   D*H

bond length (N-N)    =   1.10550Å

bond angle  (N-N)    =   180°


<pre>
         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=-3.383662D-13
 Optimization completed.
    -- Stationary point found.
</pre>

[[File:VibrationalN2.PNG]]

<i>optimised N<sub>2</sub></i>


There are no IR peaks for N<sub>2</sub> because there is no dipole moment.


<jmol><jmolApplet>
  <script>frame 1.10</script>
<title>optimised N2</title>
  <color>black</color>
  <size>300</size>
  <uploadedFileContents>N2 DJN.LOG</uploadedFileContents>
</jmolApplet></jmol>


optimisation file for N<sub>2</sub> is linked [[media:N2 DJN.LOG| here]]


[[File:Chargedist N2.PNG |thumb|none | in a diatomic molecule there is no relative charge on either atom.]]


==Hydrogen==

===optimisation===

Calculation Method   =   RB3LYP

Basis Set            =   6-31G(d,p)

E(RB3LYP)            =   -1.17853936 a.u.

RMS Gradient Norm    =   0.00000017 a.u.

Point Group          =   D*H

bond length (H-H)    =   0.74279Å

bond angle  (H-H)    =   180°  

<pre>
         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
 Predicted change in Energy=-1.167770D-13
 Optimization completed.
    -- Stationary point found.
</pre>

There are no IR peaks for H<sub>2</sub> because there is no dipole moment.


[[File:H2 vibrational.PNG]]

<i>optimised H2</i>


<jmol><jmolApplet>
  <script>frame 1.10</script>
<title>optimised H2</title>
  <color>black</color>
  <size>300</size>
  <uploadedFileContents>H2.LOG</uploadedFileContents>
</jmolApplet></jmol>



[[File:H2 Chargedist.PNG |thumb|none | in a diatomic molecule there is no relative charge on either atom.]]

optimisation file for H<sub>2</sub> is linked [[media:H2.LOG| here]]


==Haber-Bosch process==

N<sub>2</sub> + 3H<sub>2</sub> -> 2NH<sub>3</sub>  

<pre>
E(NH3)= -56.55776873 a.u.
2*E(NH3)= -113.1155375 a.u.
E(N2)= -109.52412868 a.u.
E(H2)= -1.17853936 a.u.
3*E(H2)= -3.53561808 a.u.
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579074 a.u.
ΔE= -146.478599028kJ/mol
</pre>

literature value for haber bosch process: ΔH = -45.7kJ/mol. <ref name="literature value for H-B process"/>

The value of ΔE for the computed values is lower than the literature value

==Fluorine==


<u>'''optimised information'''</u>

Calculation Method   =   RB3LYP

Basis Set            =   6-31G(d,p)

E(RB3LYP)            =   -199.49825218 a.u.

RMS Gradient Norm    =   0.00007365 a.u.

Point Group          =   D*H

bond length (F-F)    =   1.40281Å

bond angle  (F-F)    =   180° 



<pre>
         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
 Predicted change in Energy=-1.995024D-08
 Optimization completed.
    -- Stationary point found.
</pre>

[[File:F2 vibrational.PNG]]


<jmol><jmolApplet>
<title>optimised F2</title>
  <color>black</color>
  <size>300</size>
  <uploadedFileContents>F2 OPTIMISATION.LOG</uploadedFileContents>
</jmolApplet></jmol>


optimisation file for F<sub>2</sub> is linked [[media:F2 OPTIMISATION.LOG| here]]

=='''MOs for F<sub>2</sub>'''==



[[File:MOs F2 sigma 1s.PNG|300px|thumb|left| This is the molecular orbital (MO) that corresponds to the σ<sub>1s</sub> MO of the F<sub>2</sub> molecule. The MO has energy of -24.79730 a.u. An energy this deep results in no overlap of the orbitals ad shown in the diagram.]]



[[File:MOs F2 sigma 2s.PNG|300px|thumb|right| This is the molecular orbital that corresponds to the σ<sub>2s</sub> MO of the F<sub>2</sub> molecule. The MO has energy of -1.33659 a.u. This is a very symmetrical and stable MO formed  by the overlap of electrons the 2s atomic orbitals of each Fluorine atom.]]



[[File:MOs F2 sigma star2s.PNG|300px|thumb|left| This is the molecular orbital that corresponds to the σ<sup>*</sup><sub>2s</sub> MO of the F<sub>2</sub> molecule. The MO has energy of -1.09047 a.u. This MO is also formed  by the overlap of electrons the 2s atomic orbitals of each Fluorine atom but this time they are in opposite phase. This σ<sup>*</sup><sub>2s</sub> MO means that the σ<sub>2s</sub> MO will not contribute the the bonding in the F<sub>2</sub> molecule as the antibonding orbital cancels the bonding orbital out.]]



[[File:MOs F2 sigma z 2p.PNG|300px|thumb|right| This is the molecular orbital that corresponds to the σ<sub>z 2p</sub> MO of the F<sub>2</sub> molecule. The MO has energy of -0.58753 a.u. This MO is formed by the overlap of the 2p<sub>z</sub> electrons. It has the lowest energy MO of the MOs formed by the overlap of the 2p atomic orbitals. It's antibonding MO is unfilled therefore the F<sub>2</sub> can exist, but F<sub>2</sub> is a very volatile molecule and exists easily in radical form.]]



[[File:MOs F2 PI 2px.PNG|thumb|left| This is the molecular orbital that corresponds to the π<sub>2 px</sub> MO of the F<sub>2</sub> molecule. The MO has energy of -0.52332 a.u. This MO is formed by the overlap of the 2p<sub>z</sub> electrons. The π bond is formed by the overlap of the 2p electrons.]]


References 

<references>
<ref name="literature value for H-B process">Modak, J. M. (2002). Haber Process for Ammonia Synthesis Jayant M Modak. Resonance, 7(8), 69–77.</ref>