Electrons involved in bonding cannot be involved in conduction?

Question

I am currently studying the textbook Physics of Photonics Devices, Second Edition, by Shun Lien Chuang. In a section discussing the basic concepts of semiconductor band and bonding diagrams, the author says the following:

The As atom has an atomic number of 33 with an $[\text{Ar}]3d^{10}4s^24p^3$ configuration or five valence electrons in the outermost shell ($4s$ and $4p$ states). For a simplified view, we show a planar bonding diagram [2, 3] in Fig. 1.2a, where each bond between two nearby atoms is indicated with two dots representing two valence electrons. These valence electrons are contributed by either Ga or As atoms. The bonding diagram shows that each atom such as Ga is connected to four nearby As atoms by four valence bonds or eight valence electrons. If we assume that none of the bonds is broken, then all of the electrons are in the valence band, and no free electrons are in the conduction band. The energy band diagram as a function of position is shown in Fig. 1.2b, where $E_c$ is the band edge of the conduction band and $E_v$ is the band edge of the valence band. When a photon with an optical energy $hv$ above the band-gap energy $E_g$ is incident on the semiconductor, optical absorption is significant. Here $h$ is the Planck constant and $v$ is the frequency of the photon,

$$hv = \dfrac{hc}{\lambda} = \dfrac{1.24}{\lambda} \text{(eV)} \tag{1.1.1}$$

where $c$ is the speed of light in free space, and $\lambda$ is wavelength in micrometers ($\mu m$). The absorption of a photon may break a valence bond and create an electron-hole pair, shown in Fig. 1.2c, where an empty position in the bond is represented by a hole. The same concept in the energy band diagram is illustrated in Fig. 1.2d, where the free electron propagating in the crystal is represented by a dot in the conduction band. It is equivalent to acquiring an energy larger than the band gap of the semiconductor, and the kinetic energy of the electron is that amount above the conduction-band edge. The reverse process can also occur if an electron in the conduction band recombines with a hole in the valence band; this excess energy may emerge as a photon, and the process is called spontaneous emission. In the presence of a photon propagating in the semiconductor with electrons in the conduction band and holes in the valence band, the photon may stimulate the downward transition of the electron from the conduction band to the valence band and emit another photon of the same wavelength and polarization, which is called a stimulated emission process. Above the conduction-band edge or below the valence-band edge, we have to know the energy versus momentum relation for the electrons or holes. These relations provide important information about the number of available states in the conduction band and in the valence band. By measuring the optical absorption spectrum as a function of the optical wavelength, we can map out the number of states per energy interval. This concept of joint density of states, which is discussed further in the following chapters, plays an important role in the optical absorption and gain processes in semiconductors.

Thanks to Jon Custer's comments in this question, most of my confusion regarding this matter has been clarified. However, there is still one aspect of this textbook's explanation that I would like clarified.

If I'm understanding this correctly, the electrons involved in bonding cannot be involved in conduction. This is because bonding electrons are located in the valence band, whereas conduction requires free electrons, which are located in the conduction band. Is my understanding here correct?

I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

When one is talking about solids in quantum mechanical language, one has to realize that these are models designed to describe the quantum mechanical behavior of about $10^{23}$ atoms/molecules per mole, let alone the number of electrons. The band theory of solids has been one such successful model.

See how in conductors how close in energy levels the conduction band and the valence are. That is why the comments to the question say one cannot be absolute about individual electrons, since everything is quantum mechanical probabilities at the atomic level.

Now the images you copied show another model of solid state, the lattices with positive charge surrounded be the orbitals of valence negative electrons and the orbitals over the whole lattice of the conduction band.

If you look at molecular orbitals, you can see that the negative charges leave in space holes where the positive charges of the nucleus can dominate. Thus, LEGO like there are bonds created.

after this preamble:

If I'm understanding this correctly, the electrons involved in bonding cannot be involved in conduction. This is because bonding electrons are located in the valence band, whereas conduction requires free electrons, which are located in the conduction band. Is my understanding here correct?

Would be correct for a classical model. For a quantum mechanical model one should rephrase:

The probability of electrons involved in bonding to be found in the conduction band is small. The energy levels of the valence band have small overlap to the energy levels of the conduction band due to the ubiquitous Heisenberg uncertainty principle, there will be a probability of electrons changing places.

Answer 2

A large amount of energy is required to move electrons from the valence to the conduction band. Only in the conduction band can electrons move easily and conduct electricity. In insulators there are practically no electrons in the conduction band. The amount of energy electrons can gain from thermal agitation is not enough to lift them from the valence band into the conduction band. The energy gap is too large. No current can flow.

http://electron6.phys.utk.edu/phys250/modules/module%204/conduction_in_solids.htm

In a semiconductor the mobility of electrons (referring to ‘conduction electrons’ or ‘free-electrons’) is greater than that of a holes (indirectly referring to ‘valence electrons’) because of different band structure and scattering mechanisms of these two carrier types. Conduction electrons (free-electrons) travel in the conduction band and valence electrons (holes) travel in the valence band. In an applied electric field, valence electrons cannot move as freely as the free electrons because their movement is restricted. The mobility of a particle in a semiconductor is larger if its effective mass is smaller and the time between scattering events is larger. Holes are created by the elevation of electrons from innermost shells to higher shells or shells with higher energy levels. Since holes are subjected to the stronger atomic force pulled by the nucleus than the electrons residing in the higher shells or farther shells, holes have a lower mobility.

In an intrinsic silicon, at temperature $300 K$:

Electron mobility $= 1500 cm^2/(V \cdot s)$

Hole mobility $= 475 cm^2/(V \cdot s)$

Answer by Nebi Caka: https://www.researchgate.net/post/Why_is_the_mobility_of_holes_different_from_that_of_electrons