As @Custer explained previously,
it would be wrong to directly link any atomic energy levels to solid state energy levels. That mapping just does not work.
Here, I will try to elucidate it with an intuitive example.
One form of elemental carbon is diamond, where every tetrahedral carbon center combines with 4 other carbon atoms as shown below.
A total of 8 electrons are involved in the above bonding (4 from central $C$ atom and 1 each from the $C$ atoms in the corners). Likewise, 8 atomic orbitals are involved in the process, giving rise to 8 new molecular orbitals.
The 8 electrons occupy the molecular orbitals in the order of increasing energy levels. Thus, 4 bonding orbitals are filled with electrons and the 4 anti-bonding orbitals are empty. Also note, there is a big difference in energy between the bonding and anti-bonding orbitals.
Just as one central $C$ atom and 4 corner $C$ atoms gave rise to 4 $\phi$ and 4 $\phi^{*}$ molecular orbitals at 4 different energy levels, increasing the number of $C$ atoms in the material would form multitude of $\phi$ and $\phi^{*}$ molecular orbitals at varied energy levels, as shown below. In a group of $n$ carbon atoms bonded together, each atom will contribute 4 orbitals and 4 electrons. There will be 2n bonding molecular orbitals, each filled with 2 electrons, and 2n empty anti-bonding orbitals. When the number of carbon atoms in the material approaches infinity ($n → ∞$), the energy differences between bonding molecular orbitals becomes vanishingly small, essentially becoming a continuum. The same is true for the energy differences between the anti-bonding orbitals. Thus, the filled molecular orbitals and the empty molecular orbitals become two, separated bands with a gap in between.
Note that $1s$ atomic orbitals of $C$ also form bonding and anti-bonding orbitals during covalent bonding. Their bands, which are not shown above, occur below the filled band. Also note, since, $C-C$ bond has a considerable bond strength, the band gap between the filled and empty bands shown above is considerably large. Silicon, whose crystalline form also assumes diamond like molecular configuration, has a relatively weaker $Si-Si$ bond due to which its band gap is smaller than diamond's.
I hope this gives you an intuition on why one-to-one mapping of atomic energy levels with solid state energy levels would be wrong. For e.g., carbon's $2s, 2p_{x}, 2p_{y}$, and $2p_{z}$ atomic orbitals, all contribute to both of its filled and empty bands, i.e., the same set of atomic orbitals contribute to two different bands. Likewise, it might also be possible that bands contributed by two different sets of orbitals overlap, in which case it technically is a single band.
In the case of metals, although no such covalent $\sigma$ bonds occur between the atoms as in diamond or silicon, the core concept of band formation remains essentially the same.
Density of states
We know each band is composed of multitude of states, where no two electrons can occupy the same state. Also for some value of energy in a band, there exists more states than the other. The density of states function $g(E)$ is a probability distribution function such that,
$$\int_{E_1}^{E_2}g(E).dE$$ gives the number of states in the energy range $E_1$ to $E_2$ per unit volume of material. It should come no surprise that for energies inside a band gap, $g(E) = 0$.
Filling of electrons in bands
At thermodynamic equilibrium, the probability that a state of energy $E$ is filled with an electron is given by the Fermi-Dirac distribution
$$f(E)=\frac {1}{1+ e^{(E-\mu)/(k_{B}T)}}$$
The number of electrons in the energy range $E_1$ to $E_2$ per unit volume of material should be
$$\int_{E_1}^{E_2}g(E).f(E)dE$$
And, the density of electrons in the material should be
$$\frac{N}{V} = \int _{-\infty }^{\infty }g(E).f(E)\,dE$$
The function $f(E)$ looks as follows at $T=0K$, i.e., when the electrons in the solid assume their ground states due to no possible thermal excitation.
for $E>\mu$, $f(E) = 0$
for $E<\mu$, $f(E) = 1$
for $E=\mu$, $f(E) = indeterminate$
This value of $\mu$ is called Fermi level, below which all electronic states must be filled at $0K$ and above which all states must be empty.
N.B. No states are allowed in band gaps, i.e., the gaps are always empty.
When $T ≠ 0K$, at Fermi level, function $f(E)$ returns $\frac{1}{2}$, implying that if there are electronic states allowed at Fermi level, half of the states must be occupied at all temperatures.
If the Fermi level lies in a band gap, the band immediately below it is called the valence band. Valence in the sense that it is the highest energy band electrons occupy at ground state of the solid. Similar terminologies are valence shell, the highest energy shell occupied by electrons in an atom at ground state, and valence orbitals, the highest energy orbitals occupied by electrons. Likewise, if the Fermi level lies within a band, then this band would be the valence band.
Please note, it would be wrong to blindly map the valence orbitals of an atom to the valence band of the solid formed of those atoms as discussed previously.
The basis of conductivity
We talked about band as essentially being a continuum of energy eigenstates. It can also be talked about as an almost continuous collection of momentum eigenstates (i.e., just as energy levels, only certain momenta are allowed for an electron). For every eigenstate there is an another with an equal and opposite momentum, i.e., if we populate every momentum eigenstate in a band, the net momentum of electrons will be zero.
If a band is not that populated, then under the influence of electric field, the electrons can redistribute themselves to appropriate unoccupied energy/momentum eigenstates within that band. Whereas, if the band is densely populated, there are not much free states available for redistribution. The more free states are available within a band, the more conductive it becomes.
Note that electrons can also lose their energy/momenta as thermal losses due to collisions with kernels in the lattice. This causes them to shift to a lower energy state within the band, and they can drift under the field once again. Likewise, electrons whose momenta are directed such that they decelerate under the field also will have their energy/momenta lost.
All in all, this net motion of electrons (drift velocity) is what's needed to sustain an electric current.
What determines the conductivity of a band?
In metals, the fermi level $(E_F)$ passes through the valence band. As $T$ becomes greater than $0$, the electrons just below $E_F$ surpass the level and shift to higher energy states within the band. As there are a lot of unoccupied eigenstates within the band allowing redistribution under an electric field, the valence band is the conduction band for metals.
In semi-metals, $E_F$ passes through the trough of the valence band which has bimodal density of states. So, they need to be at somewhat a higher temperature than metals to be conductive (so that the electrons under $E_F$ can shift to energy levels with greater density of states allowing them to be more conductive). Technically, this bimodal valence band is also the conduction band.
In semi-conductors, $E_F$ lies in the band gap. All states in the band below $E_F$ (the valence band) are filled at $0K$ and the band is not conductive. At higher temperatures, however, the electrons can have enough energy to surpass the band gap and shift to the immediately higher unoccupied band. The electrons in this higher band, now become conductive, and so the band is called the conduction band.
In insulators (e.g., diamond), the $E_F$ passes through a very large band gap. The valence band is filled and non-conductive, and the occupancy of electrons in the band immediately higher than $E_F$, even at higher temperature, is almost null because of the very large band gap that needs to be surpassed. This higher band is what we refer to as conduction band, although no appreciable conduction takes place here. This should be an example to the fact that just as extremely populated bands, extremely non-populated bands are non-conductive too.
