Vibrational frequencies of water molecule
✅ Paper Type: Free Essay | ✅ Subject: Chemistry |
✅ Wordcount: 1956 words | ✅ Published: 17th May 2017 |
Question 1 |
Normal modes and vibrational frequencies of water molecule
HF/3-21* optimised geometry of the water molecule |
|
H bond length |
0.967 |
HOH bond angle |
107.7˚ |
(ii) Energy of the HF/3-21G optimised water molecule = -75.58596 au
Cycle |
Energy |
Max. Grad. |
Max. Dist. |
1 |
75.58553 |
0.01246 |
0.00304 |
2 |
75.58589 |
0.00324 |
0.00025 |
3 |
75.58596 |
0.00001 |
0.00000 |
Frequency (cm-1) |
Relative Motion |
Stretch or Bend |
Type |
Symmetry (S or A) |
1799.2 |
Bend |
A1 |
S |
|
3812.2 |
Stretch |
A1 |
S |
|
3945.8 |
r |
Stretch |
B1 |
A |
HOD |
||
Energy |
75.58596 au |
|
Geometry |
Bond angle |
107.7 |
Bond length |
0.967 |
|
Vibrational frequencies |
1578.7 |
H moves faster than D symmetric |
2815.3 |
D moves quickly whereas H moves slightly asymmetric |
|
3881.7 |
H moves quickly whereas D moves slightly – asymmetric |
Normal modes and vibrational frequencies of the water dimmer (H2O)2
Hydrogen-Bond Acceptor
Hydrogen bond
Hydrogen Bond Donor
Hydrogen bond length (H””O) = 1.808
Hydrogen bond angle (O-H””O) = 174.9˚
Energy of the F/3.12G optimised water dimer = -151.18902 au
(a) Potential energy calculation:
ΔE = E(dimer) – 2xE(H2O)
= (-396 871.2KJ/mol) – 2x(-198 413.2KJ/mol)
= (-396 871.2) – (-396 826.3)
= – 44.9 KJmol-1
(b)As seen from the surface diagram for H2O, the oxygen has negative charge (δ-) whereas the hydrogens are positively charged (δ+).
In the water dimer molecule, the hydrogen atoms (on the H-bond donor oxygen) are δ+/blue region. The oxygen atom that is bonded to the hydrogen that is the H-bond acceptor has δ- charge/red region. Between in the H-bond, the positive(H) and negative(O) charges combine/green region.
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The hydrogen bond is formed between one of the H atoms and one O, instead between the two oxygens, because the two oxygen atoms are negatively charged, and have δ-, and therefore repulsive interactions are formed between them. So, one H reacts with the O, which donates one of its lone pairs to form the H-bond.
In the structure of the molecule, the H””O bond is almost linear, very close to 180˚ but it is distorted so it is about 175˚. Also, the distortion causes the bond H”’O to become longer.
(c) For the water molecule:
H bond length = 0.967
For the water dimer:
H bond length of H-bond donor = 0.965
H bond length of H-bond acceptor = 0.966, 0.974 (H of H-bond)
The H bond length of the hydrogen of the H-bond is bigger than the other O-H bonds in the molecule. This is because this H is bonded to the oxygen through the H-bond, and it is pulled towards the oxygen, causing its bond with the other oxygen to become a bit longer.
Question 2 |
The water dimer consists of two fragments, the H-bond acceptor (top OH2 group) and the H-bond donor (bottom OH2 group). When a vibration causes both fragments and H-bond to move, then it is considered to be the inter-monomer because it is a vibration between the two molecules. If only one of the fragments vibrates, then the vibration is only in one of the molecules (it is internal) and it is considered to be an intra-monomer.
The vibrational frequencies of the water dimer are the following:
Frequency = 81 cm-1
Type = A’
Bending Mode
Top part of the molecule moving slightly up and down, while the two bottom hydrogens move up and down as well
Inter-monomer: The vibration affects both molecules connected through the hydrogen bond.
Frequency = 133
Type = A”
Bending mode
Top part and bottom part moving right and left.
Inter monomer
Frequency = 172
Type = A”
Bending mode
Middle hydrogen moving right and left and two bottom H atoms moving up and down symmetrically (when one is up, other is down)
Inter Monomer
Frequency = 242
Type = A’
Stretching Mode
Inter monomer
Frequency = 425
Type = A’
Bending Mode
The H-bond acceptor fragment moves to the front and then back, and the H-bond donor fragment moves up and down as well.
Inter-monomer
Frequency = 826
Type = A”
Bending mode
The H of the H-bond (middle H) is moving to the right and left, causing the rest of the molecule to move in that way as well
Inter-monomer
Frequency = 1782
Type = A’
Bending Mode
The hydrogen atoms on the H-bond donor fragment move up and down to the sides going further away and then coming closer.
Intra-monomer
Frequency = 1854
Type = A’
Bending Mode
The hydrogen atoms on the H-bond acceptor fragment separate and go further away and then come closer together again.
Intra-monomer
Frequency = 3724
Type = A’
Stretching mode
The hydrogen forming the H-bond moves closer to the oxygen of the H-bond and then further from it, causing the O-H bond to come smaller and the H””’O bond to become bigger, and the opposite.
Intra-monomer
Frequency = 3849
Type = A’
Stretching mode
The hydrogen atoms move symmetrically so that their bonds with the O of the H-bond donor are becoming bigger (stretch out) and then smaller.
Intra-monomer
Frequency = 3907
Type = A’
Stretching mode
The O-H bond of the H not involved in the H-bond acceptor fragment is stretching out, causing the bond to become longer, while the bond of the oxygen with the other H, which is involved in the H-bond, becomes shorter.
Intra-monomer
Frequency = 3982
Type = A”
Stretching mode
It is an unsymmetrical movement, where one O-H bond in the H-bond donor fragment becomes shorter and the other longer.
Intra-monomer
Question 3 |
Isotopic substitution in the water dimer
Free Energy (H-TS) = 37.8
ΔΗ Total = 127.5
Free Energy (H-TS) = 39.7
ΔΗ Total = 126.5
ΔG = G(B) – G(A) = 39.7KJmol-1 – 37.8KJmol-1 = 1.9 KJ/mol
K = e(-ΔG/RT) = exp(-1.9×10-3Jmol-1/8.314JK-1mol-1x298K) = 1.00000077
Deuterium prefers the position shown in B (connected to the oxygen of the H-bond acceptor fragment, but doesn’t take part directly in the H-bond) because the molecule has higher free energy for this arrangement.
Question 4 |
Interconversion of water dimer structures
Frequency = i302 Type = B1 |
|
Frequency = 105 Type = B2 |
|
Frequency = 208 Type = A1 |
|
Frequency = 225 Type = B1 |
|
Frequency = 256 Type = A2 |
|
Frequency = 591 Type = B2 |
|
Frequency = 1785 Type = A1 |
|
Frequency = 1831 Type = A1 |
|
Frequency = 3829 Type = A1 |
|
Frequency = 3862 Type = A1 |
|
Frequency = 3952 Type = B1 |
|
Frequency = 3961 Type = B2 |
Acyclic water dimer Cyclic water dimer
- The acyclic water molecule energy is 3.969×10-5 KJmol-1 whereas the energy of the cyclic one is —-. The cyclic molecule is less stable than the acyclic one because its ability to move around is effectively reduced compared to the acyclic one, due to the two bonds formed between the oxygen of one molecule and the two H of the other molecule.
- The imaginary frequency has the value of i306.9. One of the middle hydrogens moves up while the other moves down, in an unsymmetrical movement as shown in the pictures above.
- For the acyclic water dimer there are no imaginary frequencies and it corresponds to the valley. This shows that it is very stable and this structure is preferred.
- The cyclic molecule contains one vibrational frequency and this suggests that it is not as stable as the acyclic one. It corresponds to the hilltop of molecule-mountain.
- If a molecule has more than one vibrational frequency it corresponds to the mountain passes and it is a very unstable and unfavoured structure for the molecule to be at, which most probably does not exist.
- The cyclic structure is not very stable, and therefore it is not preferred over the acyclic one.
Question 5 |
Syn-butane:
No imaginary frequencies à Valley à stable structure, highly favoured
Boat cyclohexane:
ne imaginary frequency à Hilltop à fairly unstable, exists but not preferred
All-syn cyclohexane:
More than one imaginary frequencies à Mountain Pass à does not exist, very unstable
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