Question

Spartan Lab Questions:

1. Discuss the trend in homonuclear double bond energies going down Group 14. Use the HOMO‑LUMO gaps that you have determined to facilitate your discussion. Are there significant differences in the MO diagrams for ethene and digermene?

2. Compare the HOMOs and LUMOs of ethane and digermene. What does the distribution of electron density tell you about the orbitals?

3. Simple alkenes are colorless, but disilenes and digermenes are intensely colored. Explain.

4. All isolated disilenes, digermenes and distannenes have very large substituents, such as t-butyl or TMS groups.^[1] How do large substituents lend 4. greater stability to these compounds?

5. What evidence do the authors in reference 6 give that demonstrates just how weak the “double” bonds in heavier analogs of ethane are?

6. How does the symmetric stretching frequency relate to bond strength?

7. Recently, the first ever compound with a Si-Si triple bond was structurally characterized.^[2] Contrary to what is observed for alkynes, the compound displays a nonlinear R-Si-Si-R geometry. Based on your findings in lab, is this a surprise? Rationalize the bonding using a simple valence bond model.

please answer any that you can!

Answer 1

1) The group 14 elements are C(carbon),Si(silicon),Ge(germanium),Sn(tin),Pb(lead),Fl (flerovium) and the respective bond energies in kJ/mol are 603,274,195,169,164 (standars data from SCI literature). Homonuclear bond is the covalent bond between two identical molecules.If two HOMO orbitals react then two new MO’s will be obtained (one lower and one higher in energy) and The only energy gain is when a HOMO from one molecule reacts with a LUMO from the other molecule. The comparison of bond energies shows the level of decrease on moving down its due to the strength of HOMO energy excitation to LUMO, for example C-C bond is always higher, so only diamond is much stronger C on comparing with other atoms/molecules.

Digermene (H₄Ge₂) and Ethene (C₂H₄), yes there is significant difference for the both in HOMO-LUMO energy gap the below figure represents the difference,

3) It all about the metal's electrons and also about their d orbitals.Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level.

However when the metal starts bonding with other ligands, this changes. Due to the different symmetries of the d orbitals and the inductive effects of the ligands on the electrons, the d orbitals split apart and become non-degenerate (have different energy levels).

This forms the basis of Crystal Field Theory. How these d orbitals split depend on the geometry of the compound that is formed. For example if an octahedral metal complex is formed, the energy of the d orbitals will look like this:

As you can see, previously the d orbitals were of the same energy, but now 2 of the orbitals are higher in energy. Now what does this have to do with its colour?

Well, electrons are able to absorb certain frequencies of electromagnetic radiation to get promoted to higher energy orbitals. These frequencies have a certain energy which correspond to the energy difference between different orbitals. Now most substances are only able to absorb frequencies of radiation which are outside the visible light spectrum, for example they might be able to absorb radiation which has a frequency of 300

GhZ (that is infrared radiation). This means that it reflects all other types of radiation, including the full spectrum of visible light. So our eyes see a mixture of all the colours; red, green, blue, violet, etc. This is seen as white (this is why several organic compounds are white).

However transition metals are special in that the energy difference between the non-degenerate d orbitals correspond to the energy of radiation of the visible light spectrum. This means that when we look at the metal complex, we don't see the entire visible light spectrum, but only a part of it.

So for example, if the electrons in an octahedral metal complex are able to absorb green light and get promoted from the dyz

orbital to the dz2

orbital, the compound will reflect all other colours except for green. Therefore by using the colour wheel, we can find the complementary colour of green which will be the colour of the compound, which is magneta.

This explains why not all transition metal complexes are colourful. For example copper sulfate is a bright blue compound, however zinc sulfate on the hand is a white compound despite being a transition metal. The reason behind this is because zinc's d orbitals are completely filled up with electrons, meaning that it is not possible for any electron to make a d-> d transition as they are all filled up. Hence you might sometimes see zinc referred as not being a transition metal.

6) The below animation shows the symmetry stretching, it stretches to the maximum depends on bond stregth, on comparing C=C and C-C, C=C stretching will be high on compared to C-C due to its pi bond energy level.