Question #9def5

You would look at the number of electrons, and try to point to the correct "block" of the periodic table is designated for that element. Then you can find the column, provided you pointed to the correct block...

That is the most foolproof way, without caring about which electrons are actually valence electrons.

In general, there are a few cases:

• Main group elements
• Transition Metals
• Lanthanides/Actinides

Noting that there ARE "exceptions" amongst the normal ones (which I consider to be the ones that actually contain electrons in their proper designated block), here are examples of "normal" cases:

• #[Ar] color(blue)(4s^2)#

#overbrace(18)^(Ar) + overbrace(2)^"valence" = 20# electrons, corresponding to neutral #"Ca"#, group #"2A"#. It is in the #s#-block.

• #[Ar] 3d^10 4s^2 color(blue)(4p^2)#:

#overbrace(18)^(Ar) + overbrace(10)^("outer core") + overbrace(2 + 2)^"valence" = 32# electrons, corresponding to neutral #"Ge"#, group #"4A"#. It is in the #p#-block.

• #[Kr] color(blue)(4d^5) 5s^2#:

#overbrace(36)^(Kr) + overbrace(5)^("outer core") + overbrace(2)^"valence" = 43# electrons, corresponding to neutral #"Tc"#, group #"7B"#. It is in the #d#-block.

• #[Xe] 4f^14 color(blue)(5d^9) 6s^1#:

#overbrace(54)^(Xe) + overbrace(14 + 9)^("outer core") + overbrace(1)^"valence" = 78# electrons, corresponding to neutral #"Pt"#, column #3# within group #"8B"#, in the #d#-block.

• #[Xe] color(blue)(4f^5) 6s^2#:

#overbrace(54)^(Xe) + overbrace(5)^("outer core") + overbrace(2)^"valence" = 61# electrons, corresponding to neutral #"Pm"#, column #5# within the Lanthanide period, in the #f#-block.

Lanthanide and actinides don't really have "group names".

As an example of what I mean by an "exception":

• #[Rn] color(blue)(5f^0) color(red)(6d^2) 7s^2#:

#overbrace(86)^(Rn) + overbrace(0 + 2)^("outer core") + overbrace(2)^"valence" = 90# electrons, corresponding to neutral #"Th"#, column #2# within the Actinide period.

As you can see, thorium is still in the #f#-block, not the #d#-block.