Extrinsic semiconductors are created by doping an intrinsic semiconductor; typically silicon is doped with phosphorus to create free electrons or boron to create "holes".

In the case of phosphorus doping, phosphorus takes the place of a silicon atom in the silicon lattice, and its fifth electron becomes a free electron.

Here is what I am struggling to understand with my basic knowledge. Phosphorus has the configuration $\ce{[Ne] 3s^2 3p_x^1 3p_y^1 3p_z^1}$.

This page gives an description of the promotion of a 3s electron to a 3d shell, allowing five bonds to form in the case of $\ce{PCl5}$. I would assume that a similar process results in four bonds forming in the silicon lattice and leaving a free electron.

But that same page later on mentions a 2007 paper which essentially says that that model is not accurate, because such a state is less stable than other ways of modeling the bond (assuming I'm understanding correctly. The problem is described here, I have not read the actual paper.

What causes phosphorus to form four bonds with silicon and leave a free electron rather than, say, forming three bonds and leaving a silicon with an unpaired electron in the lattice, or bonding with another "free" electron? Is it promoting an electron to form the fourth bond and freeing the last?

In short: Why does phosphorus form four Si bonds and leave a free electron rather than forming three Si bonds when doping silicon?


1 Answer 1


First, don't forget that the phosphorus is sitting on a diamond cubic silicon lattice site, which pretty much defines the symmetry of the potential around it - the P cannot change that, but must respond to it. So, you should not think in terms of atomic orbitals.

Second, the fifth electron is not initially a free electron - it occupies a state that happens to be in the band gap. Now, that state is really close to the conduction band, only 45 meV below it at 300 K, so at room temperature there is a really good chance that it will jump into and roam about the conduction band. Nor is that the only state in the gap for P - there is another state 72 meV above the valence band (i.e. it is slightly easier to ionize a 4-fold coordinated P than a Si).

Why doesn't the P only bond with 3 Si? Well, that would leave one unhappy Si atom, and one not much happier P atom. Again, this is occurring in a crystal, not an isolated molecule. The P will occupy the lowest free energy state. Note that there are other impurities that preferentially occupy various interstitial sites. It is just that P can be coerced into pretending to be pretty much a Si atom - peer pressure, perhaps...


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