Question

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

Answer 1

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...

![http://www.ptable.com/]()

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.