How do you write the condensed electron configurations for the following atoms, using the appropriate noble-gas core abbreviations?

Well, I will do the three hard ones, and you'll have to figure out the three easy ones. How does that sound?

The hard ones:

#2)" ""Ni": [Ar] 3d^8 4s^2#

#5)" ""U": [Rn] 5f^3 6d^1 7s^2#

#6)" ""Pb": [Xe] 4f^14 5d^10 6s^2 6p^2#

For the easy ones, I'll just put the noble gas core... It's your job to figure out the rest.

#1)" "color(red)("C": [He] 2s^(?) 2p^(?))#

#3)" "color(red)("Se": [Ar] 3d^(?) 4s^(?) 4p^(?))#

#4)" "color(red)("Cd": [Kr] 4d^(?) 5s^(?))#

(No, none of the easy ones are exceptions.)

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Pull out your periodic table...

As we know,

• The first two columns are the so-called #s# block.
• The last six columns are given the label "#p# block".
• The transition metals make up the #d# block, groups #"IIIB" - "VIIIB"# and #"IB - IIB"# (or #3 - 12#).
• The lanthanides and actinides make up the #f# block.

  And we fill the orbitals by energy, followed by considerations of closeness in energy to nearby orbitals if needed (generally more applicable in the #f# block and in the so-called "Aufbau exceptions").

  The first few orbitals in the typical ordering are #1s, 2s, 2p, 3s, 3p#. The rest depend on the element and the atomic environment...

  #2)#

  #"Ni"#, a late transition metal, has core-like #3d# orbitals due to its #Z_(eff)# being on the higher side, and we fill its orbitals with the forethought that the #3d# orbitals are significantly lower in energy than the #4s#.

  #1s^2 2s^2 2p^6 3s^2 3p^6# is the core of the previous noble gas, i.e. of #"Ar"#.

  We represent that as #[Ar]#, and append the remaining outer-core #3d# and valence #4s# electrons:

  #=> color(blue)(barul(|stackrel" "(" "[Ar] 3d^8 4s^2" ")|))#

  #5)#

  #"U"#, an actinide, has an unusual electron configuration. We utilize the previous noble gas for the noble gas core, #"Rn"#, justifiably shorthand for #1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 4f^14 5s^2 5p^6 5d^10 6s^2 6p^6#.

  Then, we consider the outer-core #5f#, valence #6d# and valence #7s# orbitals in the appropriate order.

  Although the #5f# electrons are outer-core for uranium, the #7s# and #6d# are really close in energy, practically degenerate. Since #"U"# is the fourth-column actinide we would expect an Aufbau configuration of:

  #color(red)([Rn] 5f^4 7s^2)# (ALERT! ALERT! ERROR! ERROR!)

  That is, however, INCORRECT. The early actinides (and to a smaller extent, the early lanthanides) are exceptions---namely, #ul("Ac" - "Np")#.

  The electron that we would have liked to believe is in the #5f# orbital is actually placed in the #6d# orbital. Thus, the correct configuration is actually:

  #=> color(blue)(barul|stackrel" "(" "[Rn] 5f^3 6d^1 7s^2" ")|)#

  Even though the #6d# orbitals are higher in energy than the #5f# orbitals, they are similar sizes. Yet, the #5f# is more compact due to its higher angular momentum #l# giving it one more nodal plane (but one less radial node at #n = 5# than in the #6d#).

  So, the #6d# is likely the choice that relieves some electron repulsion.

  #6)#

  #"Pb"# is a normal, post-transition metal.

  We include the #Xe# core, which by now we know how to write. We also include the #6s# by inspecting the #s# block, and since #"Pb"# is a post-transition metal, we have full #(n-2)f# and #(n-1)d# orbitals.

  This gives us, then:

  #=> color(blue)(barul(|stackrel" "(" "[Xe] 4f^14 5d^10 6s^2 6p^2" ")|))#

  But of course, no more than #bb4# of those electrons are actually valence.