There is no strictly relationship, but for NON-transition metals (i.e. Non-d-block, non-f-block), over there is one.

Valence electrons are then USUALLY detailed after the noble gas core, within factor (e.g. Tungsten, #"W"#, probably doesn"t have #20# valence electrons, but up come #6# instead).

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For NON-transition metals, the energy level #n# is offered by its duration number, i.e. That row number. Gift on the fourth row the the periodic table way your valence electrons would be the #4s# and #4p# electrons, #n = 4#.

Two examples:

#"Ca"#, with configuration #color(blue)(4s^2)#. This has actually #bb2# valence electrons.

#"P"#, through configuration #color(blue)(3s^2 3p^3)#. This has actually #bb5# valence electrons.

EXCEPTIONS BELOW!

Transition metals on the various other hand have actually easy access to occupied #(n-1)d# orbitals, and also thus have actually the electrons in those orbitals included in their set of valence electrons.

Three examples:

#"Sc"# (scandium), with configuration # color(blue)(3d^1 4s^2)#. This is why #"Sc"# deserve to have a best oxidation state the #+3# (e.g. In #"ScCl"_3#); it has actually #bb3# valence electrons.

#"W"# (tungsten), v configuration #color(red)(4f^14) color(blue)(5d^4 6s^2)# --- keep in mind that the #4f# electrons are hardly used, also though they are detailed after "##", the noble gas core. #"W"# is commonly going to have a #+6# best oxidation state (e.g. In #"WO"_3#), which means it probably has actually #bb6# valence electrons many of the time.

#"Os"# (osmium), v configuration #color(red)(4f^14) color(blue)(5d^6 6s^2)#. It has up to #bb8# valence electrons, e.g. In #"OsO"_4#.

Some heavy #bb(f)#-block metals (mainly lanthanides and actinides) likewise have access to occupied #(n-2)f# orbitals too, and those electron might likewise be had in their collection of valence electrons... They might even not have actually #(n-1)d# valence electron sometimes.

Three examples:

#"Pa"# (protactinium), through configuration #color(blue)(5f^2 6d^1 7s^2)#. This is why #"Pa"# usually has actually a #+5# oxidation state in its compounds (such together #"Pa"_2"O"_5#); it has #bb5# valence electrons.

#"Bk"# (berkelium), through configuration #color(red)(5f^9) color(blue)(7s^2)# --- note that #11# valence electrons would be insane. In berkelium, only some the the #5f# electrons are taken into consideration valence, depending upon context. Because #+4# is the highest possible known easily-accessible oxidation state (e.g. In #"BkO"_2#), #"Bk"# has actually around #bb4# valence electron (not #2#!).

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#"Gd"# (gadolinium), with configuration # color(red)(4f^7) color(blue)(5d^1 6s^2)#. It many reasonably has #bb3# valence electrons, and also indeed, its highest possible oxidation state is generally #+3# (such as in #"Gd"_2"O"_3#).