Effective nuclear charge Z_eff — Slater's rules: Worked examples

This page is the case-by-case drill room for Slater's rules. The parent note gave you the recipe; here we hit every kind of atom and every kind of target electron the rules can throw at you — no surprises left. If a symbol below feels new, it was built in the parent note (the recipe box), and we re-anchor it as we go.

The scenario matrix

Every problem Slater's rules can pose falls into one of these cells. The examples below are labelled with the cell(s) they cover, so by the end you've seen all of them.

Cell	What makes it tricky	Example that covers it
A. Lone valence $s$ electron	same-group count is $0$ (nothing to shield you)	Ex 1 (Na 3s)
B. Full-period buildup	same-group electrons pile up at $0.35$ each	Ex 2 (Cl 3p)
C. Innermost / degenerate case	the $1 s$ electron: special $0.30$ , no shells below	Ex 3 (He 1s, C 1s)
D. $d$ -electron target	all left groups screen at $1.00$ , even same- $n$ $s, p$	Ex 4 (Fe 3d)
E. $s$ vs $d$ split-shell	for a $4 s$ target the $3 d$ counts as $n - 1$ at $0.85$	Ex 5 (Fe 4s)
F. Ion (electron removed)	recount population after losing an electron	Ex 6 (Fe²⁺)
G. Real-world word problem	translate a chemistry fact into a $Z_{e f f}$ comparison	Ex 7 (why K loses 4s)
H. Exam twist / trend limit	compare two atoms; predict the $Δ Z_{e f f}$ per step	Ex 8 (Na→Mg trend)

Case A + the anchor: a lone valence electron

Worked example Example 1 — Na, the

$3 s$ valence electron · cell A Sodium, $Z = 11$ . Configuration in Slater groups: $(1 s)^{2} (2 s, 2 p)^{8} (3 s)^{1}$ . Target = that lone $3 s$ electron.

Forecast: Guess before reading — will $Z_{e f f}$ be close to 11, or close to 1? (Hint: 10 inner electrons sit between it and the nucleus.)

Same group $(3 s, 3 p)$ : population is $1$ , subtract itself → $0$ others → $0 \times 0.35 = 0$ . Why this step? An electron never shields itself; we always use (group population $- 1$ ).
$n - 1$ shell $(2 s, 2 p)$ : $8$ electrons $\times 0.85 = 6.80$ . Why? The whole $n = 2$ shell sits just inside the $n = 3$ electron — strong but imperfect screening.
$\leq n - 2$ shells $(1 s)$ : $2 \times 1.00 = 2.00$ . Why? The $1 s$ is buried deep; each electron blocks essentially a full proton.
Add up: $S = 0 + 6.80 + 2.00 = 8.80$ , so $Z_{e f f} = 11 - 8.80 = 2.20$ .

Verify: $S = 8.80 < Z = 11$ ✓ (shielding can't exceed proton count). The lone valence electron feels only $\sim 2.2$ of 11 protons — hence Na loses it easily. See the bucket picture below.

Figure — Effective nuclear charge Z_eff — Slater's rules

Case B: a full-period target (same-group pile-up)

Worked example Example 2 — Cl, a

$3 p$ valence electron · cell B Chlorine, $Z = 17$ . Configuration: $(1 s)^{2} (2 s, 2 p)^{8} (3 s, 3 p)^{7}$ . Target = one $3 p$ electron.

Forecast: Sodium's valence felt $2.20$ . Chlorine has $6$ more protons but they land in the same shell as our target. Will $Z_{e f f}$ jump by a full $6$ , or by less?

Same group $(3 s, 3 p)$ : population $7$ , minus itself → $6 \times 0.35 = 2.10$ . Why? $s$ and $p$ of the same $n$ live in one Slater group; all six others screen at $0.35$ .
$n - 1$ shell $(2 s, 2 p)$ : $8 \times 0.85 = 6.80$ .
$\leq n - 2$ $(1 s)$ : $2 \times 1.00 = 2.00$ .
$S = 2.10 + 6.80 + 2.00 = 10.90$ , so $Z_{e f f} = 17 - 10.90 = 6.10$ .

Verify: From Na ( $2.20$ ) to Cl ( $6.10$ ) is $+ 3.90$ across $6$ protons — that's $0.65$ per proton, exactly the " $1 - 0.35$ " rate the parent note promised. ✓ Higher $Z_{e f f}$ ⇒ Cl is smaller and grabs electrons — consistent with high Electronegativity.

Case C: the innermost electron (the degenerate/edge case)

Worked example Example 3 — the

$1 s$ electrons of He and of C · cell C Two sub-cases, because $1 s$ is special: same-group contribution is $0.30$ (not $0.35$ ), and there is no shell below it.

(a) Helium, $Z = 2$ , config $(1 s)^{2}$ , target a $1 s$ electron. Forecast: one partner electron sitting right beside it — will it feel almost 2, or noticeably less?

Same group $(1 s)$ : $1$ other $\times 0.30 = 0.30$ . Why $0.30$ ? The $1 s$ group is the one exception to the $0.35$ rule.
No lower shells exist → nothing else.
$S = 0.30$ , so $Z_{e f f} = 2 - 0.30 = 1.70$ . Verify: $0 < 1.70 < 2$ ✓ — one companion electron steals a little pull, not a whole proton.

(b) Carbon, $Z = 6$ , config $(1 s)^{2} (2 s, 2 p)^{4}$ , target a $1 s$ electron.

Same group $(1 s)$ : $1$ other $\times 0.30 = 0.30$ .
Outer electrons $(2 s, 2 p)$ are above the $1 s$ target → they contribute nothing. Why? Slater's rules only count electrons in the same group or below; electrons further out don't screen inward.
$S = 0.30$ , so $Z_{e f f} = 6 - 0.30 = 5.70$ . Verify: Nearly the full $6$ — a $1 s$ electron sits so deep it feels almost the bare nucleus. This is why core electrons are ferociously bound.

Case D + E: the split shell — $3 d$ vs $4 s$ in the same atom

The most error-prone scenario. Same atom, two targets, different rules for the $3 d$ shell depending on which target we pick. We use Iron.

Worked example Example 4 — Fe, a

$3 d$ electron · cell D Iron, $Z = 26$ . Slater grouping: $(1 s)^{2} (2 s, 2 p)^{8} (3 s, 3 p)^{8} (3 d)^{6} (4 s)^{2}$ . Target = one $3 d$ electron.

Forecast: Will the $3 s, 3 p$ electrons (same $n = 3$ !) screen at $0.85$ or at $1.00$ ? This is the whole trap.

Same $(3 d)$ group: population $6$ , minus itself → $5 \times 0.35 = 1.75$ . Why? Same-group is still $0.35$ even for a $d$ target.
Everything to the LEFT $(1 s)^{2} + (2 s, 2 p)^{8} + (3 s, 3 p)^{8} = 18$ electrons $\times 1.00 = 18.00$ . Why $1.00$ , not $0.85$ ? For a $d$ (or $f$ ) target, all groups to its left screen fully — the $d$ orbital sits outside them, so even the same- $n$ $s, p$ count as a full $1.00$ . (See Shielding and Penetration.)
$4 s$ electrons: to the right of $3 d$ → contribute $0$ . Why? Only groups at or below the target count.
$S = 1.75 + 18.00 = 19.75$ , so $Z_{e f f} = 26 - 19.75 = 6.25$ .

Verify: $S = 19.75 < 26$ ✓. Note the $18$ left-group electrons all counted at $1.00$ — contrast this sharply with Example 5.

Worked example Example 5 — Fe, a

$4 s$ electron · cell E Same atom, same config. Target = one $4 s$ electron, group $(4 s, 4 p)$ .

Forecast: Now the $3 d$ electrons are below our $4 s$ target. Do they screen at $0.85$ (as an $n - 1$ shell) or at $1.00$ ? Different rule than Example 4!

Same group $(4 s, 4 p)$ : population $2$ , minus itself → $1 \times 0.35 = 0.35$ .
$n - 1$ shell = all $n = 3$ : $(3 s, 3 p)^{8} + (3 d)^{6} = 14$ electrons $\times 0.85 = 11.90$ . Why $0.85$ for the $3 d$ here? For an $s / p$ target, the rule is " $n - 1$ shell at $0.85$ " — and the $3 d$ has $n = 3$ , one below the $4 s$ . So it counts at $0.85$ , not the $1.00$ it got in Example 4.
$\leq n - 2$ shells: $(1 s)^{2} + (2 s, 2 p)^{8} = 10 \times 1.00 = 10.00$ .
$S = 0.35 + 11.90 + 10.00 = 22.25$ , so $Z_{e f f} = 26 - 22.25 = 3.75$ .

Verify: The same $3 d$ electrons contributed $18, 00 \to$ full in Ex 4 but $0.85$ each here — because the target changed, not the atom. Compare answers: $3 d$ feels $6.25$ , $4 s$ feels only $3.75$ . The $4 s$ is held more loosely. Figure below.

Case F: an ion — recount the population

Worked example Example 6 — Fe²⁺, a

$3 d$ electron · cell F Making Fe²⁺ removes the two $4 s$ electrons first (Ex 5 showed they're the loosest). New config: $(1 s)^{2} (2 s, 2 p)^{8} (3 s, 3 p)^{8} (3 d)^{6}$ . $Z$ is still 26 (protons don't change!). Target = a $3 d$ electron.

Forecast: We removed the $4 s$ electrons — but Example 4 showed the $4 s$ never shielded the $3 d$ anyway. So does $Z_{e f f}$ for the $3 d$ change at all?

Same $(3 d)$ group: still $6$ electrons → $5 \times 0.35 = 1.75$ .
Left groups: still $18 \times 1.00 = 18.00$ .
No $4 s$ to consider (they're gone, and they contributed $0$ anyway).
$S = 19.75$ , so $Z_{e f f} = 26 - 19.75 = 6.25$ — identical to neutral Fe's $3 d$ .

Verify: Same number as Ex 4 ✓. Lesson: removing electrons that never shielded your target leaves its $Z_{e f f}$ unchanged. (The remaining electrons do contract the ion, but Slater's simple recipe doesn't capture that subtlety — it's an estimate.) The tighter grip explains higher Ionization Energy for successive ionizations.

Case G: a real-world word problem

Case H: the exam twist — predict $Δ Z_{e f f}$ across a period

Worked example Example 8 — "By how much does

$Z_{e f f}$ rise from Na to Mg?" · cell H No full calculation allowed — predict, then confirm. Mg, $Z = 12$ , config $(1 s)^{2} (2 s, 2 p)^{8} (3 s)^{2}$ ; target a $3 s$ electron.

Forecast: Adding one proton ( $+ 1$ to $Z$ ) but the new electron goes into the same $3 s$ group as our target. Net change should be $+ 1 - 0.35 = + 0.65$ . Let's check.

Same group $(3 s)$ : now $2$ electrons, minus itself → $1 \times 0.35 = 0.35$ . Why? Mg's second $3 s$ electron is a same-group partner, adding $0.35$ to $S$ .
$n - 1$ $(2 s, 2 p)$ : $8 \times 0.85 = 6.80$ (unchanged from Na).
$\leq n - 2$ $(1 s)$ : $2 \times 1.00 = 2.00$ (unchanged).
$S = 0.35 + 6.80 + 2.00 = 9.15$ , so $Z_{e f f} = 12 - 9.15 = 2.85$ .

Verify: $2.85 - 2.20 = 0.65$ ✓ — exactly the predicted per-step climb. This one number drives the whole Periodic Trends story: rising $Z_{e f f}$ ⇒ shrinking radius, rising ionization energy, rising electronegativity left→right. (The $4 s$ -vs- $3 d$ filling behind these configs is in Aufbau and 4s vs 3d filling.)

Recall Quick self-check on the trap cells

For a $3 d$ target, what do the same- $n$ $3 s, 3 p$ electrons contribute? ::: $1.00$ each (all left groups screen fully for a $d / f$ target). For a $4 s$ target, what do the $3 d$ electrons contribute? ::: $0.85$ each (they're the $n - 1$ shell for an $s$ target). When you strip electrons to make an ion, does $Z$ change? ::: No — $Z$ is the proton count and never changes; only the electron population (and thus $S$ ) does. Which single number tracks Na $\to$ Mg $\to$ Ar and explains the period trend? ::: $Z_{e f f}$ , rising $\approx 0.65$ per step.