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For example for 3d, would an element like Mn have 2 or 7 valence electrons? Would Cu have 1 or 11 valence electron? I'm pretty sure that Ga has 3 valence electrons right? Can someone explain when 3d is included in valence electrons?
The 3d shell is a continuation of the n=3 shell, so it does not play any role in affecting the number of valence electrons. The number of valence electrons is determined by the number of electrons in the outermost shell. The outermost shell for period 4 elements is the n=4 shell, therefore these transition metals will all have 2 valence electrons (in the 4s orbital). I suppose for the exception of Cr and Cu, their d shells are more stable with half-full 3d shells / full 3d shells respectively, so in these cases they would both have one valence electron (according to their electron configuration, in which their 4s shells are 4s1 instead of 4s2).
Valence electrons in the d block are usually determined by their group number, or the number of electrons in their valence shell. In general, the transitional metals have 2 valence electrons.
From my understanding, the answer to that depends on which definition of valence electrons is being used. For instance, if valence electrons are being defined as the number of electrons in the outermost energy level, then manganese would have 2 valence electrons. However, if valence electrons are defined as the electrons that participate in chemical reactions/bonds, then manganese would have 7 valence electrons in an example such as MnO4-, where the oxidation state of Mn is +7. I would consider the first definition (the number of electrons in the outermost energy level) the most common one, so generally, transition metals would have 2 valence electrons under this definition. Hope that helps.
The textbook problem 2A.1c states that Mn has 7 valence electrons, but it is very clear about "including d electrons."
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