Why doesn't the classical model of electrical conduction work with iron, cadmium and bismuth?

My textbook says this:

In most metals, the charge carriers are electrons and the charge – carrier density determined from Hall – effect measurements is in good agreement with calculated values for such metals as lithium (Li), sodium (Na), copper (Cu), and silver (Ag), whose atoms each give up one electron to act as a current carrier. In this case $n_v$ is approximately equal to the number of conducting electrons per volume. This classical model however is not valid for metals like iron (Fe), bismuth (Bi) and Cadmium (Cd) or for semiconductors. These discrepancies can only be explained by using a model based on the quantum nature of solids.

(The context of $n_v$ here is that used in the measurement of Hall voltage): $$Delta V_H = frac{IBd}{n_vqA}$$

In an example where the Hall voltage was determined for a copper strip, the author mentioned that a semiconductor’s Hall voltage would be much larger than that for metals because $n_v$ is much smaller than it is in metals that contribute one electron per atom.

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My first question is how exactly does a semiconductor have a value of $n_v$ lower than one per atom? I read that the charge carriers increase in density with temperature in semi-conductors. Could this property be related to it?

(I found out that semiconductors follow electrical conduction based on band-theory and this explains how semiconductors have a lower $n_v$ than one per atom. The second question still puzzles me.)

I don’t see why metals like iron and bismuth don’t behave in the same way as copper and silver. How are they any different from the other metals and why does a discrepancy occur? They share similar properties with metals like magnesium (Mg) and zinc (Zn), so I really can’t find any reason for their unusual behaviour. Could someone please suggest a possible reason?

Any help would be appreciated!