# Why does phosphorus give a free electron in silicon doping?

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?