Effective nuclear charge Z_eff — Slater's rules: Exercises

Before drilling, pin down the two words the whole page leans on.

Intuition WHY the weights are

$0.35$ , $0.85$ , $1.00$ (not black boxes) Picture yourself as the chosen electron looking toward the nucleus. An electron blocks the nucleus best when it sits between you and it.

A deep inner electron ( $\leq n - 2$ ) is a tight little cloud hugging the nucleus — it sits almost entirely between you and the charge, so it cancels nearly a whole proton: weight $\approx 1.00$ .
An electron in the shell just inside you ( $n - 1$ ) is mostly between you and the nucleus, but its cloud is puffy enough to sometimes drift outside you — so it blocks strongly but not fully: weight $0.85$ .
A same-shell neighbour sits beside you at about the same distance, not between you and the nucleus, so it blocks poorly: weight $0.35$ .
The $1 s$ -pair special case ( $0.30$ ) is a tiny tweak: the two $1 s$ electrons are so tightly squeezed together that each screens its partner slightly less than a normal same-group $0.35$ . It only ever applies when the target is itself a $1 s$ electron.

Quick reference for the whole page (from the parent):

Level 1 — Recognition

(Can you read a configuration and place electrons into Slater groups and buckets?)

L1.1

Write the electron configuration of fluorine ( $Z = 9$ ) split into Slater groups.

Recall Solution L1.1

Fluorine has 9 electrons: $1 s^{2} 2 s^{2} 2 p^{5}$ . Grouped the Slater way ( $s$ and $p$ of the same $n$ together): $(1 s)^{2} (2 s, 2 p)^{7}$ The second group holds $2 + 5 = 7$ electrons. That is the entire configuration in Slater form.

L1.2

For a $3 p$ electron of phosphorus ( $Z = 15$ , config $(1 s)^{2} (2 s, 2 p)^{8} (3 s, 3 p)^{5}$ ), sort every electron into the three buckets same group / $n - 1$ / $\leq n - 2$ . (Just count — don't multiply yet.)

Recall Solution L1.2

Target is in Slater group $(3 s, 3 p)$ , so $n = 3$ .

Same group $(3 s, 3 p)^{5}$ : 5 electrons total, but we exclude the target itself → $5 - 1 = 4$ others.
$n - 1$ shell $(2 s, 2 p)^{8}$ : 8 electrons.
$\leq n - 2$ $(1 s)^{2}$ : 2 electrons. Check the count: $4 + 8 + 2 = 14$ = all electrons except the one target. ✓

Level 2 — Application

(Turn the buckets into a number.)

L2.1

Compute $S$ and $Z_{e f f}$ for a $2 p$ electron of oxygen ( $Z = 8$ ).

Recall Solution L2.1

Config: $(1 s)^{2} (2 s, 2 p)^{6}$ . Target in $(2 s, 2 p)$ , so $n = 2$ .

Same group: $6 - 1 = 5$ others $\times 0.35 = 1.75$ (beside-you neighbours, weak block).
$n - 1$ shell = $(1 s)$ : $2 \times 0.85 = 1.70$ (shell just inside, strong-but-not-full block).
$\leq n - 2$ : none. $S = 1.75 + 1.70 = 3.45, Z_{e f f} = 8 - 3.45 = 4.55$

L2.2

Compute $Z_{e f f}$ for a $3 s$ electron of magnesium ( $Z = 12$ ).

Recall Solution L2.2

Config: $(1 s)^{2} (2 s, 2 p)^{8} (3 s)^{2}$ . Target in $(3 s, 3 p)$ , $n = 3$ .

Same group: $2 - 1 = 1$ other $\times 0.35 = 0.35$ .
$n - 1$ shell $(2 s, 2 p)^{8}$ : $8 \times 0.85 = 6.80$ .
$\leq n - 2$ $(1 s)^{2}$ : $2 \times 1.00 = 2.00$ (deep hugging cloud, near-full block). $S = 0.35 + 6.80 + 2.00 = 9.15, Z_{e f f} = 12 - 9.15 = 2.85$ Compare with Na's valence $Z_{e f f} = 2.20$ (parent): Mg's valence electron feels a stronger pull — consistent with Mg being smaller and harder to ionize.

Level 3 — Analysis

(Compare two electrons or two atoms and explain the difference.)

L3.1

In potassium ( $Z = 19$ , config $(1 s)^{2} (2 s, 2 p)^{8} (3 s, 3 p)^{8} (4 s)^{1}$ ), compute $Z_{e f f}$ for the $4 s$ electron. Then say in one sentence why $K$ readily forms $K^{+}$ .

Recall Solution L3.1

Target $4 s$ , group $(4 s, 4 p)$ , $n = 4$ .

Same group: $1 - 1 = 0$ others → $0$ .
$n - 1$ shell = all $n = 3$ = $(3 s, 3 p)^{8}$ (there is no $3 d$ in K): $8 \times 0.85 = 6.80$ .
$\leq n - 2$ = $(1 s)^{2} + (2 s, 2 p)^{8} = 10 \times 1.00 = 10.00$ . $S = 0 + 6.80 + 10.00 = 16.80, Z_{e f f} = 19 - 16.80 = 2.20$ That lone $4 s$ electron feels only $\approx 2.2$ of the 19 protons → very loosely held → $K$ easily loses it to form $K^{+}$ .

L3.2

For argon ( $Z = 18$ ), compute $Z_{e f f}$ on a $3 p$ electron. Then compare to sodium's valence $Z_{e f f} = 2.20$ and state which atom is smaller and why. See the bar picture below.

Figure — Effective nuclear charge Z_eff — Slater's rules

Recall Solution L3.2

Ar config: $(1 s)^{2} (2 s, 2 p)^{8} (3 s, 3 p)^{8}$ . Target $3 p$ , $n = 3$ .

Same group $(3 s, 3 p)^{8}$ : $8 - 1 = 7$ others $\times 0.35 = 2.45$ .
$n - 1$ shell $(2 s, 2 p)^{8}$ : $8 \times 0.85 = 6.80$ .
$\leq n - 2$ $(1 s)^{2}$ : $2 \times 1.00 = 2.00$ . $S = 2.45 + 6.80 + 2.00 = 11.25, Z_{e f f} = 18 - 11.25 = 6.75$ Sodium's valence electron ( $Z_{e f f} = 2.20$ ) and argon's ( $Z_{e f f} = 6.75$ ) sit in the same third shell, but argon's is pulled in with over 3× the net charge. Stronger pull → the cloud is drawn in tighter → Ar's valence electrons are held far closer than Na's, matching the trend of shrinking Atomic Radius across a period.

Level 4 — Synthesis

(Chain the $4 s$ -vs- $3 d$ logic to explain real chemistry.)

L4.1

For iron ( $Z = 26$ , config $(1 s)^{2} (2 s, 2 p)^{8} (3 s, 3 p)^{8} (3 d)^{6} (4 s)^{2}$ ), compute $Z_{e f f}$ for (a) a $3 d$ electron and (b) a $4 s$ electron. Then state which electrons leave first to make $F e^{2 +}$ and justify with your numbers. See the ladder figure.

Recall Solution L4.1

(a) $3 d$ target (a $d$ electron → same group $0.35$ , everything left $1.00$ ):

Same $3 d$ group: $6 - 1 = 5$ others $\times 0.35 = 1.75$ .
All groups to the left $(1 s)^{2} + (2 s, 2 p)^{8} + (3 s, 3 p)^{8} = 18 \times 1.00 = 18.00$ .
The $4 s$ electrons are to the right of $3 d$ → contribute $0$ . $S = 1.75 + 18.00 = 19.75, Z_{e f f} (3 d) = 26 - 19.75 = 6.25$

(b) $4 s$ target ( $n = 4$ ; note $3 d$ counts as $n - 1$ here!):

Same group $(4 s, 4 p)^{2}$ : $2 - 1 = 1$ other $\times 0.35 = 0.35$ .
$n - 1$ = all $n = 3$ = $(3 s, 3 p)^{8} + (3 d)^{6} = 14 \times 0.85 = 11.90$ .
$\leq n - 2$ = $(1 s)^{2} + (2 s, 2 p)^{8} = 10 \times 1.00 = 10.00$ . $S = 0.35 + 11.90 + 10.00 = 22.25, Z_{e f f} (4 s) = 26 - 22.25 = 3.75$

Conclusion: $Z_{e f f} (3 d) = 6.25 > Z_{e f f} (4 s) = 3.75$ . The $4 s$ electrons feel the weaker net pull, so they are held loosest and leave first when iron ionizes to $F e^{2 +}$ . This is the Slater-rules echo of the Aufbau and 4s vs 3d filling story.

Level 5 — Mastery

(Predict a periodic trend end-to-end from $Z_{e f f}$ — and handle the $1 s$ edge case.)

L5.1

Using Slater's rules, compute the valence $Z_{e f f}$ for Li ( $2 s$ , $Z = 3$ ), Na ( $3 s$ , $Z = 11$ ), and K ( $4 s$ , $Z = 19$ ). Then predict how Atomic Radius and Ionization Energy change going down group 1, and explain whether $Z_{e f f}$ alone accounts for the trend.

Recall Solution L5.1

Li, $2 s$ target $(1 s)^{2} (2 s)^{1}$ :

Same group: $0$ others → $0$ .
$n - 1$ = $(1 s)^{2}$ : $2 \times 0.85 = 1.70$ . $Z_{e f f} = 3 - 1.70 = 1.30$

Na, $3 s$ (from parent): $S = 8 (0.85) + 2 (1.00) = 8.80 \Rightarrow Z_{e f f} = 11 - 8.80 = 2.20$ .

K, $4 s$ (from L3.1): $Z_{e f f} = 19 - 16.80 = 2.20$ .

Prediction & reasoning: The valence $Z_{e f f}$ is roughly flat down the group ( $1.30 \to 2.20 \to 2.20$ ) — it does not collapse. Yet each step down puts the valence electron in a higher shell ( $n = 2 \to 3 \to 4$ ), which sits physically much farther out. With similar net pull but a much larger orbit, the valence electron is farther and looser:

Atomic radius increases down the group.
Ionization energy decreases down the group.

So $Z_{e f f}$ alone does not explain the group trend — the jump in principal quantum number $n$ (shell size) is the dominant factor going down, whereas $Z_{e f f}$ is the dominant factor going across. See the two-effect summary diagram.

L5.2 — the $1 s$ edge case

Compute $Z_{e f f}$ for a $1 s$ electron of helium ( $Z = 2$ ) and of lithium ( $Z = 3$ ). These are the only place the special $0.30$ weight is ever used — show it in action.

Recall Solution L5.2

Helium, $1 s$ target $(1 s)^{2}$ :

The target's only same-group partner is the other $1 s$ electron. Because the target is a $1 s$ electron, this same-group weight is the special $0.30$ , not $0.35$ : $1 \times 0.30 = 0.30$ .
There is no shell inside $1 s$ , so no $n - 1$ or $\leq n - 2$ terms. $S = 0.30, Z_{e f f} = 2 - 0.30 = 1.70$

Lithium, $1 s$ target $(1 s)^{2} (2 s)^{1}$ :

Same group $(1 s)$ : one other $1 s$ electron $\times 0.30 = 0.30$ . (The lone $2 s$ electron is outside the $1 s$ target, so it is to the right — it shields nothing.) $S = 0.30, Z_{e f f} = 3 - 0.30 = 2.70$ Both cases use $0.30$ because the target itself is a $1 s$ electron; you would never use $0.30$ when the target is a $2 s$ , $3 s$ , $3 d$ , … electron.

Recall One-glance answer key

L2.1 O $2 p$ : $Z_{e f f} = 4.55$ · L2.2 Mg $3 s$ : $2.85$ · L3.1 K $4 s$ : $2.20$ · L3.2 Ar $3 p$ : $6.75$ · L4.1 Fe $3 d$ : $6.25$ , $4 s$ : $3.75$ · L5.1 Li: $1.30$ , Na: $2.20$ , K: $2.20$ · L5.2 He $1 s$ : $1.70$ , Li $1 s$ : $2.70$ .