Lecture 4 Model Answers to Problems

Transcription

1 Self-Study Problems / Exam Preparation Construct and annotate a valence MO diagram for H 2 CN -. Use your diagram to explain why the neutral radical is more stable than the anion. (this is an old exam question) o use the MO diagram check-list!! o shape is known, point group is C 2v o axial system and symmetry operations are shown in Figure 1 o Use your diagram to explain why the neutral radical is more stable than the anion. The neutral radical has lost one electron from an antibonding MO, this could be expected to stabilise the total energy of the molecule. Comments from the exam: a large portion of students forgot to answer the interpretative question after drawing the MO diagram!! of those that did answer this question a great many were confused over the "neutral radical". H 2 CN - has 12e (it has one negative charge and is a singlet), the neutral radical REMOVES one electron leaving 11e (it is neutral and has one unpaired electron). surprisingly a small number of students tried to evaluate the "bond order", as this molecule has 4 distinct bonds it is irrelevant to evaluate a bond order, this ONLY works for diatomics which have a single bond. also problematic was failure to associate the high energy antibonding orbital as effecting the total energy of the molecule. A logical description is required 6a 1 given this FO energy alignment the b 1 and b 2 interaction should be less than the a 1 because the energy of the FO differs more and both interactions are pi type x z y N C C 2 (z) σ v (xz) H 1 H 2 Figure 1 symmetry elements σ v (yz) 1b 1 3a 1 2b 1 2b 2 b 1 N is more electronegative than C so I estimate that the paos lie deeper than the C nonbonding pao of the CH 2 fragment, but not as deep as the bonding interaction of the 1b 2 FO becuase it is strongly bonding b 2 a 1 1b 1 1b 2 it is difficult to predict the ordering of these two orbitals and a calculation or experimental evidence is required 1b 2 5a 1 The a 1 interaction should be very large because the a 1 FOs are almost degenerate and there is strong overlap of the two fragment pao lobes in a sigma type interaction 4a 1 N is electronegative so I estimate that the sao lies about a deep as the bonding CH 2 orbital 2a 1 the splitting is quite large becuase sfos have the strognest interactions and the energy of the FOs is quite close C N C 3a 1 H H H H 6 electrons 12 electrons 6 electrons Figure 2 MO diagram H 2 CN - N a 1 slightly larger contribution from the lower lying FO (difficult to determine which is lower without calculation) 1a 1 and 2a 1 are the core 1s AOs for N and C respectively 1

2 optimise H 2 CN -, carry out a frequency analysis to confirm you have a minima, then compute the MOs and compare them to your predicted MOs o the computed orbitals are shown in Figure 3 o while the LCAOs provides a good approximation to most of the orbitals, there are some subtle differences o notice the polarisation in the real 5a 1 MO compared to the qualitative LCAO MO o notice also the contributing orbital size difference in the real 1b 2 MO Figure 3 MO diagram H 2 CN - with the computed MOs shown 2

3 construct and annotate a valence MO diagram for I 3 (D h point group). Use as fragments the two terminal I atoms (I I) and a central I atom ( I, assume that the MOs do not undergo mixing (this is an old exam question) antibonding MOs are destabilized by more than bonding MOs are stabilized (0.5 mark) splitting energy is not large because these orbitals are 2 bonds apart and there is minimal overlap and therefore minimal stabilisation or destabilisation(1 mark) 4 2π u 1π g "I 2 " FO contribution is larger because the FO is higher in energy and this is an antibonding MO (0.5 mark) remains non-bonding because there is no FO or the correct symmetry to interact (1 mark) splitting energy is large because the FOs are very close in energy (1 mark) π g π u 5 π u extra electron due to the negative charge I pao D h pi type interactions have a smaller splitting energy than sigma type interactions (1 mark) I I I x y z 3 large sp gap to right of PT (1 mark) 4 1π u FO contributions are more even because the FOs are close in energy (0.5 mark) "I" FO contribution is larger because the AO is lower in energy and closer to the energy of this bonding MO (0.5 mark) 2 I I I I I I 14e 22e 8e Figure 4 MO diagram I 3 3 extra electron due to -1 charge goes to complete octet (0.5 mark) 3

4 marks: 1 fragments and molecule on diagram 1 for axis in correct alignment (-1 z-axis in wrong direction) 2 depicting, labelling FOs complex due to the many problems in general, if made reasonable attempt would give 1 or worse attempt would give 0.5 even if all of below were present -0.5 for capital symbols -0.5 if wrong symmetry -0.5 for pi orbitals not looking like pi orbitals -0.5 for bonding/anti-bonding in wrong position -0.5 for missing out orbitals -0.5 for "stretching bonds" 2 for energy of the FOs roughly correct (complex due to the many problems in general, was harsher than in the above, but 0.5 if there was something to work with even if all of the below were present -0.5 splitting too large -0.5 for I AOs not in the middle -0.5 for sp gap not large enough -0.5 for bonding/anti-bonding in wrong position -0.5 for missing out orbitals 2 for basic shapes of MOs (reasonably generous here) 1 for getting relative size of AO contributions correct (-0.5 if comments and actual pictures do not match up) 4 for a selection of annotations related to MOs (see model answer) 1 correct symmetry labels MOs (very generous here, carry through errors accepted) 2 MO energies roughly right (reasonably generous here, accepted interaction with the lower for full marks) 1 correct electronic configuration (-0.5 for missing out the negative charge) total In general this question was answered well overall, however there is a subset of people with some problems. 1. I- -I is isoelectronic with F 2 which was covered in year 1 and in tutorials and in online notes this year. It is the simplified version of the hetero-nuclear diatomic, a homonuclear diatomic. It has the basic splitting pattern of saos below the paos which split into pi and sigma (sigma split more than pi). Key problems included -missing the 2 level out completely! -including only the bonding FOs -putting in only the bonding π u and anti-bonding -making errors in the symmetry label of the and orbitals 2. As the separation of the I- -I fragment is two bonds apart any overlap is weak and hence energies of stabilisation and destabilisation are small, a small splitting was expected, but many students had a large splitting for the I- -I fragment orbital energy levels. 3. The point in 2 also impacts on the "reference" for the MO diagram. In the model answer this is the non-bonding paos of the I fragment. Some students placed the paos below the weakly bonding I- -I FOs, the non-bonding FO must be higher than the weakly bonding FOs. 4. I lies to the right of the periodic table and although it is lower down still has sufficient s-p separation that the two manifolds do not overlap. Many students however had the s- 4

5 p gap too small. No marks were deducted for making this mistake and the MO diagram was marked on the basis of the position of the FOs as given in the student s diagram. 5. The electronic configuration of I is [Kr]4d 10 5s 2 5p 5. The valence shell is the n=5 shell with 7 valence electrons. A number of students tried to include the 4d electrons. As this shell is completely full it is not included in the valence MO diagram. If you were making a Lewis structure with I you would not include the daos, but count only the 7 valence electrons. No mark penalty was applied for including 4d electrons (this section was ignored if present). 6. Some students wrote the symbol labels as capitals, ie Π and not π or Σ and not σ! (it was repeated often in lectures and is in the notes, that MOs are referred to by noncapitals, capitals are reserved for labelling vibrations or electronic states) 7. Everyone put in the place holders for the molecular fragments, well done! However (despite specifically mentioning this in lectures, and in the notes for last years exam) place holders for the atoms where not present for the FOs. This led to expanding bonds! (see below) 8. Some students placed the 1π g MOs above the 2π u MOs, however the 1π g MOs are essentially non-bonding while the 2π u MOs are very anti-bonding. This indicates students did not examine the bonding character of MOs when determining their energy ordering. 9. Some students combined the FO of I- -I with the orbital of I. As mentioned above some students didn't include the bonding FO of I- -I at all, and placed the anti-bonding FO where the bonding FO should have been. Some students interacted the orbital of I with the s based orbital of I- -I, this was accepted. 10. Some students forgot the -1 charge! (given in the question) 11. In MO theory the charge is not "assigned" to an atom, this is because MOs are spread out over the whole molecule. The extra electron is added in when filling up the energy levels. If you really want to "assign" it, look at the highest occupied MO, in this case the 1π g MOs. These MOs are spread evenly over all 3 atoms, and so the charge is really very delocalised. If the the atoms do "expand-a-bond" NOT move, they provide the underlying "structure" for the orbitals always draw the "basic structure" then put the FOs or MOs on it HOMO of a molecule has a significantly larger contribution from a particular atom AO, then we may consider the electron to "spend more time" around that atom, but this is not the case for I There were also some odd references to the electronegativity of orbitals and of atoms. As all the atoms are identical there will be no electronegativity difference. 1π g YES slighly antibonding very close to non-bonding NO Figure 5 LCAOs Figure 6 LCAOs π u anti-bonding Figure 7 σ-orbital interactions 5

6 Orbitals are not electronegative, atoms are electronegative, I suspect students were trying to refer to the relative energy of the FO energy levels?? 13. Some students STILL did not put annotations down. And in some cases annotations repeated the same information, repetitions are not marked. Not every annotation on the model answer is required, just a selection up to 4 marks worth. Basic annotations get 0.5 mark and those referring to more complex concepts get 1 mark. Annotations related to determining the symmetry of orbitals were also counted. o what are the other possible fragments? Why are the suggested fragments the best fragments to use? You could use an I 2 I fragments: (I I ) and ( I), however this would be problematic because the fragments will not have the σ h mirror plane. o with reference to your MO diagram explain if I 3 is linear or bent and why I 3 will be bent. Removing two electrons from the degenerate HOMO will leave a partially occupied degenerate level. On a reduction in symmetry to C 2v the HOMO will become a 1 and b 1. The "unoccupied" a 1 MO can now mix with the occupied nonbonding 5 MO which also becomes a 1, the energy of the 5 MO is stabilised, stabilising the whole molecule. The HOMO will remain essentially unchanged except for the redefinition of the symmetry label. compute the MOs for I 3 and compare them to your predicted MOs. o note that the 6s AOs lie very close to the 5s and 5p (closer than the 5d!) and could potentially mix, however from the real MOs there is little mixing evident Figure 8 Real MOs 6