Too Few Electrons

3 Too Few Electrons

At first glance, the structure of diborane, B2H6, would seem unusual. Why shouldn’t the molecule assume the same geometry as ethane,

C2H6, which after all has the same number of heavy atoms and the same number of hydrogens?

H H H H H H B B CC H H H H H H

diborane ethane The important difference between the two molecules is that diborane has two fewer electrons than ethane and is not able to make the same number of bonds. In fact, it is ethylene which shares the same number of electrons to which diborane is structurally related.

H CCH HH ethylene This activity explores isoelectronic (equal electron) relationships, and shows how they can be employed to anticipate the structures of molecules which have too few electrons to make the “required” bonds. We will begin with the relationship between ethylene and diborane (both of which structures are well known) and then try to predict the geometry of the (unknown) borane analogue of acetylene. 1. Build ethylene and diborane, and put them into the same document.

Construct diborane using the inorganic model kit. Select boron from the Periodic Table and the five coordinate trigonal-bipyramid structure from the list of atomic hybrids. Notice that an icon of fragment is displayed at the top of the model kit. Identify the axial and equatorial positions in this fragment,

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equatorial B equatorial

axial and click on one of the equatorial positions. In response a yellow circle will move to this position. Next click anywhere on screen, and orient the fragment such that you can clearly identify axial and equatorial positions. Click on an equatorial position to make a two- boron fragment.

Next select hydrogen from the Periodic Table and two coordinate linear from the list of hybrids. Click on the upper axial free valence on one boron and then on the lower axial free valence on the other boron. You are left with the structure.

B B

Select Make Bond from the Build menu, and click on the free valence of the upper hydrogen and then on the axial free valence on the adjacent boron. Repeat for the lower hydrogen.

H B B H Select Minimize from the Build menu to produce a “reasonable” geometry for diborane.

2. Obtain equilibrium geometries for ethylene and diborane using the HF/3-21G model. Also request that all six valence molecular orbitals for the two molecules be drawn. After the calculations have completed, sketch the two sets of orbitals. As best you can, associate each valence orbital in ethylene with its counterpart in diborane. Focus on similarities in the structure of the orbitals and not on their “position” in the lists of orbitals. To what orbital in

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3. Build acetylene, C2H2, and optimize its geometry using the HF/ 3-21G model. Request all valence molecular orbitals be drawn. On the basis of the orbital shapes (for acetylene) and on your experience with diborane, suggest at least two “plausible”

geometries for the hypothetical molecule B2H4. Build each, (use techniques similar to those you used to build diborane) calculate its equilibrium geometry using the HF/3-21G model and request all valence molecular orbitals. Also request calculation of the infrared spectrum. Of the possible geometries you suggested, which is favored energetically? Is this structure actually a minimum on the energy surface? To tell, examine the infrared spectrum for the existence of imaginary frequencies (see the essay “Potential Energy Surfaces”). Are any or all of your other selections also energy

minima? Sketch the molecular orbitals of your preferred B2H4 geometry alongside those of acetylene and, as best you can, pair them up. Pay particular attention to the orbitals relating to the two π bonds in acetylene.

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