How is the energy level of an atom's valence electrons related to its period in the periodic table?
1 Answer
There is no strict relationship, but for NON-transition metals (i.e. non-d-block, non-f-block), there is one.
Valence electrons are then USUALLY listed after the noble gas core, within reason (e.g. tungsten,

For NON-transition metals, the energy level
Two examples:
"Ca" , with configuration[Ar]color(blue)(4s^2) . This hasbb2 valence electrons.
"P" , with configuration[Ne]color(blue)(3s^2 3p^3) . This hasbb5 valence electrons.
EXCEPTIONS BELOW!
Transition metals on the other hand have easy access to occupied
Three examples:
"Sc" (scandium), with configuration[Ar] color(blue)(3d^1 4s^2) . This is why"Sc" can have a maximum oxidation state of+3 (e.g. in"ScCl"_3 ); it hasbb3 valence electrons.
"W" (tungsten), with configuration[Xe]color(red)(4f^14) color(blue)(5d^4 6s^2) --- note that the4f electrons are hardly used, even though they are listed after "[Xe] ", the noble gas core.
"W" is commonly going to have a+6 maximum oxidation state (e.g. in"WO"_3 ), which means it probably hasbb6 valence electrons most of the time.
"Os" (osmium), with configuration[Xe]color(red)(4f^14) color(blue)(5d^6 6s^2) . It has up tobb8 valence electrons, e.g. in"OsO"_4 .
Some heavy
Three examples:
"Pa" (protactinium), with configuration[Rn]color(blue)(5f^2 6d^1 7s^2) . This is why"Pa" usually has a+5 oxidation state in its compounds (such as"Pa"_2"O"_5 ); it hasbb5 valence electrons.
"Bk" (berkelium), with configuration[Rn]color(red)(5f^9) color(blue)(7s^2) --- note that11 valence electrons would be insane. In berkelium, only some of the5f electrons are considered valence, depending on context.
Since+4 is the highest known easily-accessible oxidation state (e.g. in"BkO"_2 ),"Bk" has aroundbb4 valence electrons (not2 !).
"Gd" (gadolinium), with configuration[Xe] color(red)(4f^7) color(blue)(5d^1 6s^2) . It most reasonably hasbb3 valence electrons, and indeed, its highest oxidation state is usually+3 (such as in"Gd"_2"O"_3 ).