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This is a semiconductor-related question, but in order to understand the acceptor level energy state of boron in bilicon, I figured I'd ask here for hints.

Boron is a p-type material when introduced in a semiconductor like silicon. Boron forms three covalent bonds with silicon, leaving one silicon atom frustrated, not forming a bond. This doping process introduces the idea of the hole, that is, absence of electron.

However, my chemistry knowledge is limited, and I don’t understand what kind of bond exists when an adjacent silicon valence electron jumps in this electron vacancy. Is this new electron now shared by both silicon and boron, or does it belong to silicon, since silicon complies to the octet rule, whereas boron does not?

Why does the acceptor level reside above the silicon valence band edge and what does this level exactly represent? Does it represent the energy needed for an adjacent silicon valence electron to jump into this vacancy?

boron in silicon lattice Image source: National Schools' Observatory

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    $\begingroup$ Crystals aren’t molecules. Bloch functions extend throughout the crystal. $\endgroup$ – Jon Custer Dec 20 '20 at 2:57
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    $\begingroup$ well this doesn't really help me further. but I see your point. still, not clear what this acceptor level represents to me $\endgroup$ – pit fermi Dec 20 '20 at 4:21
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Using your diagram, the boron-silicon bond with only the silicon electron has only half of the possible electron content. A "hole" as you said. The hole is not confined to the one $\ce{B-Si}$ bond, but can be in any of the 4 positions. It's not like the boron sits there and the silicon pops its electron into the potential bond; the bond is there and it's only half filled.

Now this isn't a molecule, but you can focus on the boron part of the crystal and say that all four adjacent silicons are bonded to the boron, but you have only 7 electrons where the "molecular" orbitals could contain 8.

Electrons don't really "belong" to one atom or another; they comprise some sort of sea of electrons, and the sea above this boron is a little lower than elsewhere. I think of the sea of electrons in a copper-boron alloy as being very fluid, and having a level surface everywhere, but the sea in a p-type semiconductor is more like mud (quicksand?) and can have some dips that could catch and puddle some water (~electrons). The sea is still controlled mostly by the silicon, just modified by boron (submerged holes).

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  • $\begingroup$ yea, all that makes sense, however, why does the acceptor level reside above the valence band and not below is or inside it? this level is the energy a nonfree electron inside the "disturbed"/imperfect/doped silicon crystal will have , when it is located in one of these 4 incomplete bonds. an energy level that mathematically derives from the the modified hydrogen model? are my assumptions so far correct? $\endgroup$ – pit fermi Dec 21 '20 at 5:08
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The band gap in semiconductors is described in terms of levels. https://en.wikipedia.org/wiki/Band_gap however, levels don't have energy, electrons have energy. The levels are what we think we put them in to indicate their energy.

enter image description here

In the picture, a, toward the left, represents pure silicon. If we remove an electron from one $Si-Si$ bond, there will be a little dip in the filling of the valence band because we have taken one electron away. When we add boron to make the $B-Si$ bond, we get that same little dip in the valence band. However, when we add another electron into that bond (from some electron excitation, or from an electron donor somewhere else in the material), that electron is somewhat stable, but not as tightly bound - it is either in a peak of the valence band or just into the conduction band.

The picture shows the transition from atomic orbitals (on the right), which is what the OP shows in the diagram in the question, to a continuity of structure and a broadening of energy levels (on the left). Drawing individual atoms keeps us at the right end of the picture, whereas discussing semiconductors moves us to the left side of the picture. Holding those two sides of the picture (and the middle, too, which represents metallic conduction) in your mind at the same time is somewhat like wave-particle duality.

So, in answer to the comment on my first answer, the electron in the $B-Si$ bond is lower than the top of the valence band, but the second electron in the bond is more energetic than the top of the $Si-Si$ valence band, and may go into the conduction band, or more easily be excited into it. The levels don't conduct electricity, the electrons (or holes) do.

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  • $\begingroup$ that makes more sense. i am aware of how the bands are formed, but the energy vs atomic radius picture is new to me. in the case of SI-B, this means that the interatomic distance between B and SI is smaller or bigger than from Si-Si? $\endgroup$ – pit fermi Dec 23 '20 at 17:49
  • $\begingroup$ The overall silicon structure remains, so the Si-B-Si distance is pretty much the same (maybe a little shrunken). But the individual electron energy at the "hole" is affected even more by the nucleus substitution (-9) plus the electron shielding (also -9, but directed). Focus on the electron energy, but switch from atomic orbital levels to lattice energy bands and gaps. Imagine the situation at one end of the graph, then at the other end; then try to switch back and forth as smoothly as you can. $\endgroup$ – James Gaidis Dec 24 '20 at 14:09

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