Why is an n-type semiconductor different from an n-type?

Question

Electronics

If a 5 valence impurity is added to silicon, then 1 electron will remain without a covalent bond and will be free, but why when we add a 3 valence impurity, then 1 electron from silicon does not remain.
Or if you look from the other side: we add a 3 valence impurity to silicon, then the neighboring silicon molecule gives up 1 electron to create a covalent bond, while violating its energy balance and becoming positive, but when we add a 5 valence impurity, then the extra electron of the impurity is not tries to detach from the molecule and create a covalent bond, thereby making a "hole".

Answer 1

It is calculated by the excess of electrons or holes. Ionized impurity and silicon atoms are of no interest, since nailed to the crystal lattice and do not participate (in the first approximation) in the current flow.

Answer 2

In contrast to the clear title (albeit ignoring the Latin alphabet), the description of the details of your question is very vague. It is noticeable that you have not mastered the concept of a hole as a virtual carrier of a positive charge. Therefore, I will try to answer only the title, remembering the remnants of the FTT, preserved in my head from university lectures many decades ago.
So, what is the difference between N and P semiconductors? By and large, only in the mobility of carriers (ceteris paribus) are electrons more mobile than holes. Everything else follows from this, including the difference in the parameters of pairs of transistors that are considered "complementary".