# 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 Answer
Jul 20, 2017

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 $\text{IIIB" - "VIIIB}$ and $\text{IB - IIB}$ (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 $1 s , 2 s , 2 p , 3 s , 3 p$. The rest depend on the element and the atomic environment...

2)

$\text{Ni}$, a late transition metal, has core-like $3 d$ orbitals due to its ${Z}_{e f f}$ being on the higher side, and we fill its orbitals with the forethought that the $3 d$ orbitals are significantly lower in energy than the $4 s$.

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$ is the core of the previous noble gas, i.e. of $\text{Ar}$.

We represent that as $\left[A r\right]$, and append the remaining outer-core $3 d$ and valence $4 s$ electrons:

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

5)

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

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

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

$\textcolor{red}{\left[R n\right] 5 {f}^{4} 7 {s}^{2}}$ (ALERT! ALERT! ERROR! ERROR!)

That is, however, INCORRECT. The early actinides (and to a smaller extent, the early lanthanides) are exceptions---namely, $\underline{\text{Ac" - "Np}}$.

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

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

Even though the $6 d$ orbitals are higher in energy than the $5 f$ orbitals, they are similar sizes. Yet, the $5 f$ 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 $6 d$).

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

6)

$\text{Pb}$ is a normal, post-transition metal.

We include the $X e$ core, which by now we know how to write. We also include the $6 s$ by inspecting the $s$ block, and since $\text{Pb}$ is a post-transition metal, we have full $\left(n - 2\right) f$ and $\left(n - 1\right) d$ orbitals.

This gives us, then:

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

But of course, no more than $\boldsymbol{4}$ of those electrons are actually valence.