How do you write the condensed electron configurations for the following atoms, using the appropriate noble-gas core abbreviations?
1) C
2) Ni
3) Se
4) Cd
5) U
6) Pb
1) C
2) Ni
3) Se
4) Cd
5) U
6) Pb
1 Answer
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.)
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
The first few orbitals in the typical ordering are
#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
#=> color(blue)(barul(|stackrel" "(" "[Ar] 3d^8 4s^2" ")|))#
Then, we consider the outer-core
Although the
#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,
The electron that we would have liked to believe is in the
#=> color(blue)(barul|stackrel" "(" "[Rn] 5f^3 6d^1 7s^2" ")|)#
Even though the
So, the
We include the
This gives us, then:
#=> color(blue)(barul(|stackrel" "(" "[Xe] 4f^14 5d^10 6s^2 6p^2" ")|))#
But of course, no more than