l2 tutorial problems - hunt research group · 3 has 8e (5 from b and 3 from h 3) which fill this...

Lecture 2 Model Answers to Problems

1

Self-Study Problems / Exam Preparation • revise your computational chemistry workshop from last year:

http://www.huntresearchgroup.org.uk/teaching/year1_lab_start.html o make sure you have checked that the molecule is optimised o carry out the frequency analysis and ensure that all the vibrational modes are positive o then perform the calculation which stores all the molecular orbital information pop=(full,nbo)

• use qualitative MO theory to predict if BeH2 will be bent or linear. o assume that we can apply the same MO diagram as generated for H2O o note that because Be is less electronegative than H the orbital coefficients for the MOs will

change, Figure 1 o the occupation is different!

Figure 1 Correlation diagram for BeH2

o BeH2 will be linear o the stabilisation of the 3a1 MO which occurs in H2O because of mixing does not occur in

BeH2 because the the 1πu orbitals are not occupied and hence significant mixing cannot occur o in addition it is the stability or instability of the 2a1 and 1b1 under the distortion which

determine the structure o in the diagram above I have shown a slight mixing of the 3a1 and 4a1, this is because all

orbitals of the same symmetry actually interact slightly, thus on any distortion a slight amount of stablisation may occur, however in this case this should be tiny.


2

o on moving away from linear the 2a1 is stabilised by overlap of the H 1sAO contributions, but the 1b1 is destabilised by more due to the antibonding overlap of the H1sAO and reduction in overlap with the pAO on the Be.

• optimise BeH2 using the B3LYP method and 6-311G(d,p) basis set, confirm you have a minimum structure. Compute the MOs using pop=(full,nbo), visualise orbitals 1-7 and compare them to your qualitative MO diagram o the orbitals for the optimised linear structure are shown on the MO diagram above, Figure 1

• draw and annotate the MO diagram for linear water HOH, using this diagram explain if linear

CH2 is expected to diagmagnetic or paramagnetic o the MO diagram for linear water HOH is shown in, Figure 2a o note that I have not annotated this diagram, students are expected to annotate diagrams in the

exam.

Figure 2 MO diagrams for (a) linear H2O and (b) linear CH2.

o C is more electronegative than H and thus the linear H2O MO diagram can be used to understand CH2 fragments (MO diagrams apply to molecular fragments as well as stable molecules), Figure 2b

o in linear CH2 two electrons go into a degenerate set of π orbitals, the electrons will remain un-paired and thus linear CH2, if it existed would be paramagnetic

o however as we have a partially filled degenerate orbital we could expect vibronic coupling to lower the symmetry and then orbital mixing (analogous to that found in H2O) to split the degeneracy. In this case these electrons would now pair in the lower lying MO and bent is CH2 diamagnetic. (bent CH2 MO diagram not given)

σ u+

πu πu

σ g+

σ u+

2σ g+

2σ u+

2σ g+

σ g+

1πu

1σ u+

z

x

yO HH

z

x

y

OH H

H O H

σ u+

πu πu

σ g+

σ u+

2σ g+

2σ u+

2σ g+

σ g+

1πu

1σ u+

z

x

yC HH

z

x

y

CH H

H C H


3

• (do AFTER the problems class) draw a MO diagram for planar D3h NH3 o The problems class involved the first step in this process, drawing the MO diagram of BH3,

however, B is less electronegative than N and so we need to alter the diagram, Figure 3 o key differences are

² shifting of orbital levels, N valence moves down ² larger sp gap in N ² N has more electrons than B and so now the 1a2” level is filled

o note carefully the changes in orbital composition (ie the largest contributing FO to each MO), the shift in energies and that in the D3h structure the HOMO and LUMO cannot mix because they are of different symmetry

Figure 3 MO diagram for D3h NH3

′a1

′e

′e

′a1

N

H H

H

H H

H

N

H H

H

H H

H

H H

H

pz (px,py)′′a2

′e

′a1

′′a2

′a1y

x

z

NH H

H

′e

in-phase overlap=> bonding MO

FOs are close in energy hence roughly equal

contribution to the MO

out-of-phase overlap=>antibonding MO

significantly larger contribution from N becuase this FO is

much closer in energy

antibonding MO is destabilised more than

bonding MO is stablised

2 ′a1

1 ′e

1 ′′a2

3 ′a1

2 ′e

non-bonding MO

the large sp gap and electronegative nature of N mean

the 1s is relatively deep

larger contribution from H3 becuase this FO is closer in

energy


4

• (do AFTER the problems class) draw the correlation diagram for the distortion of NH3 from D3h to C3v o The above question involved the first step, drawing the high symmetry MO diagram o Then we have two options:

1. determine the MO diagram for trigonal pyramidal AH3 from scratch 2. use the MO diagram we have constructed and evaluate changes to MOs

o Option 2 is often more useful when forming a correlation diagram, Figure 4 o the far right column of Figure 4 shows the centre of the MO diagram for trigonal planar AH3,

A more electronegative than H o remember when the shape of a molecule is distorted the AO components move with their

atoms, they do not change direction or orientation.

Figure 4 Correlation diagram for pyramidalisation

o first the shift in the energy levels can be determined solely by evaluating changes that occur in overlap and overall bonding character of a MO on distortion. ² the 2a1' MO is slightly stabilised due to an increase in the (through space) H sAO bonding

overlap, this is a result of the H atoms getting slightly closer together as the molecule distorts. There is also a slight decrease in the bonding overlap of the central element sAO with the H sAOs and so overall the stabilisation is not large.

² the 1e' MOs are destabilised due to a reduction in the bonding overlap of the central element pAO with the H sAOs, and an increase in the through space H sAO antibonding interaction, both interactions contribute to destabilisation and hence the shift in energy is larger than for the 2a1' MO. The converse is true for the 2e' MOs.

² the 1a2" MO is non-bonding and so the energy of this MO does not change

A

H

H

H

2 ′a1

1 ′e

1 ′′a2

3 ′a1

2 ′e

AHy

z

xH H

xz

y

2a1

1e

2e

3a1

4a1

D3h

maximium distortion will occur for occupied 3a1 and

empty 4a1

C3v

θ=pyramidalisation angle

θ >90 θ =90


5

o there is also a change in point group on distortion from D3h to C3v, and the symmetry labels of the orbitals change ² there is no significant effect on the e'->e MOs, or in the deep lying 2a1' -> 2a1 MO ² however 1a2" (D3h) MO becomes 3a1 (C3v), and 3a1' (D3h) MO becomes 4a1 (C3v). ² these orbitals now have the same symmetry and can potentially mix, the extent of mixing

is determined by the rules outlined in Lecture 3: o Rules for MO mixing:

² only MOs of the same symmetry can mix ² mixing must stabilise the total energy of the molecule ² mixing tends to be large when at least one of the following criteria are met:

o MOs are close in energy o one of the MOs is non-bonding or unoccupied o MOs are in the HOMO-LUMO region o we evaluate the effects of mixing between the 3a1 and 4a1 (C3v) MOs (Figure 5) and include

them on our diagram (Figure 3).

Figure 5 Mixing of the 3a1 and 4a1 MOs for NH3

o It is important to realise that mixing will be small unless there are electrons in the 4a1 LUMO, this is why NH3 is pyramidal while BH3 is planar!

o BH3 has 6e (3 from B and 3 from H3) which fill this diagram to the 1e' (or 1e) level. On distortion the 1e' level is destabilised, more than the 2a1' is stabilised and hence BH3 prefers to remain planar.

o NH3 has 8e (5 from B and 3 from H3) which fill this diagram to the 3a1 level. While the 3a1 MO is not stabilised by the geometry change, the loss of symmetry means that mixing can now occur between the occupied 3a1 and unoccupied 4a1 MO. This mixing is very strong and stabilises the 3a1 MO substantially and hence NH3 is trigonal pyramidal and not planar.

o comparing the orbitals that mix from BH3 to NH3 it is clear there is a stronger bonding and antibonding interaction between the N and H3 than found between B and H3, Figure 6

Figure 6 Mixing of the 3a1 and 4a1 MOs for BH3

ψ 1 =ψ (3a1) +ψ (4a1)ψ (3a1) ψ (4a1)

−ψ (3a1) ψ 2 =ψ (4a1) −ψ (3a1)ψ (4a1)

+

+

NH3

ψ 1 =ψ (3a1) +ψ (4a1)ψ (3a1) ψ (4a1)

−ψ (3a1) ψ 2 =ψ (4a1) −ψ (3a1)ψ (4a1)

BH3


6

• (do AFTER the problems class) discuss the bonding and structure of the acid-base aduct H3BNH3 o BH3 in the gas phase is planar while NH3 is trigonal pyramidal, on forming the aduct the

overall symmetry of the molecules will be reduced, the symmetry becomes C3v, this will change the symmetry of the contributing orbitals.

Figure 7 Symmetry of H3NBH3

o to a first approximation we can consider H3BNH3 as the MO diagram between two fragments H3B and NH3.

o the reduced local symmetry of the BH3 fragment changes the symmetry labels for the MOs allowing the HOMO and LUMO to mix, nevertheless without any occupation in the 2a2” (now the 3a1’) this is not favoured

o however there will be an interaction between the HOMO of NH3, and the low lying 2a2” LUMO on BH3, electrons will “flow downhill” and the resulting MO that forms can be thought of as partially occupying the empty 2a2” LUMO (now the 3a1’) on BH3

Figure 8 Symmetry of H3NBH3

B N

H

H H

HH

H

C3(z)z

y

x

σ(yz)σv mirror in the plane

with each pair of HBNHσv

σv

σv

2a1

1e

2e

3a1

4a1

N BH

HH

H

H

HB

H

HH

NH

HH

2a1

1e

2e

LUMO

HOMO

4a1

3a1

1a1

1e

2a1

3a1

4a1

5a1

6a1

2e

3e

4e


7

² thus the base has donated electrons into the D3h 1a2” “LUMO” if BH3 and mixing is now favoured leading to distortion of the BH3 molecule. The more electron density transferred into the 1a2" LUMO of BH3 the greater the mixing that can occur and the larger the stabilisation and the greater the structural distortion.

² the electrons removed from the 3a1’ on NH3 may lead to a reduction in mixing and the molecule will become less pyramidal. The extent of distortion of the BH3 and NH3 molecules in an acid-base adduct can be correlated with the basicity of the base! ie the amount of electron density donated from the base into the LUMO of BH3

o the reality of the orbital interactions is more complex than this simple argument would present. All of the a1 MOs can mix, and this leads to a rather complex set of MOs. Moreover the HOMO of NH3 is very stable. One way to rationalise the final computed MOs is to consider the interaction of the NH3 HOMO with the 2a1 MO of BH3. Both of these orbitals are occupied however it is the stabilisation of the “antibonding” pair by MO mixing with the BH3 LUMO that leads to the stabilisation of the molecule, Figure 9

Figure 9 Computed MOs of H3NBH3

2a1

1e

2e

3a1

4a1

N BH

HH

H

H

HB

H

HH

NH

HH

2a1

1e

2e

LUMO

HOMO

3a1

4a1

5a1

6a1

3e

4e

3a1

2e

2a1

mixing

l2 tutorial problems - hunt research group · 3 has 8e (5 from b and 3 from h 3) which fill this...

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