Slater's Rules: Calculating Effective Nuclear Charge Step by Step

The idea of effective nuclear charge is powerful, but it becomes far more useful when you can put numbers on it. In 1930, the American physicist John C. Slater proposed a set of simple empirical rules for estimating the shielding constant S, and hence Zeff = Z − S, for any electron in an atom. They’re approximate, but they capture the main trends well and are widely taught in first-year university chemistry. This article sets out the rules and works through examples.

Step 1: write the configuration in Slater’s groups

Write the electron configuration, grouping subshells in this order:

(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) …

Notice that s and p of the same shell go together, but d and f are each in a separate group.

Step 2: apply the shielding contributions

To find S for a chosen electron, add up contributions from all the other electrons.

For an electron in an (ns, np) group

Other electrons in…	Contribution per electron
groups to the right (higher)	0
the same (ns, np) group	0.35 (but 0.30 if the group is 1s)
the n − 1 shell	0.85
n − 2 shell and lower	1.00

For an electron in an (nd) or (nf) group

Other electrons in…	Contribution per electron
groups to the right	0
the same (nd) or (nf) group	0.35
all groups to the left	1.00

Step 3: calculate Zeff

Zeff = Z − S

Worked example 1: a 2p electron in nitrogen

Nitrogen, Z = 7: (1s²) (2s² 2p³)

For one 2p electron:

• Other electrons in (2s, 2p): 2 + 3 − 1 = 4, each 0.35 → 1.40
• Electrons in (1s), the n − 1 shell: 2, each 0.85 → 1.70
• S = 1.40 + 1.70 = 3.10
• Zeff = 7 − 3.10 = 3.90

Worked example 2: the 3s electron in sodium

Sodium, Z = 11: (1s²) (2s² 2p⁶) (3s¹)

For the 3s electron:

• Same group (3s, 3p): 0 other electrons → 0
• n − 1 shell (2s, 2p): 8 electrons × 0.85 → 6.80
• n − 2 shell (1s): 2 electrons × 1.00 → 2.00
• S = 8.80
• Zeff = 11 − 8.80 = 2.20

The outer electron of sodium feels only about a fifth of the full nuclear charge. That’s why it’s so easily lost, and why sodium is so reactive.

Worked example 3: a 3p electron in chlorine

Chlorine, Z = 17: (1s²) (2s² 2p⁶) (3s² 3p⁵)

For one 3p electron:

• Same group: 2 + 5 − 1 = 6 × 0.35 → 2.10
• n − 1 shell: 8 × 0.85 → 6.80
• n − 2 shell: 2 × 1.00 → 2.00
• S = 10.90
• Zeff = 17 − 10.90 = 6.10

Compare with sodium (2.20): chlorine’s outer electrons feel nearly three times the effective charge. That explains why chlorine is much smaller than sodium, has a much higher ionisation energy, and readily gains electrons. See atomic radius trend.

Worked example 4: across period 3

Using the same method for the outer electron of each period 3 element:

Element	Z	Configuration (outer group)	S	Zeff
Na	11	3s¹	8.80	2.20
Mg	12	3s²	9.15	2.85
Al	13	3s² 3p¹	9.50	3.50
Si	14	3s² 3p²	9.85	4.15
P	15	3s² 3p³	10.20	4.80
S	16	3s² 3p⁴	10.55	5.45
Cl	17	3s² 3p⁵	10.90	6.10
Ar	18	3s² 3p⁶	11.25	6.75

Each step adds 1 to Z but only 0.35 to S, so Zeff rises by 0.65 per element. This steady increase is the engine behind the trends across a period: shrinking radius, rising ionisation energy and rising electronegativity. See periodic trends explained.

Worked example 5: down group 1

Element	Outer electron	Zeff (Slater)
Li	2s	1.30
Na	3s	2.20
K	4s	2.20
Rb	5s	2.20

Zeff stays almost constant down the group, while the outer electron occupies ever larger shells. With similar pull but greater distance, the outer electron is held less tightly: ionisation energy falls and reactivity rises down the group. See alkali metals.

Worked example 6: why potassium’s 19th electron goes into 4s

Potassium, Z = 19. Compare two possibilities for the last electron.

Option A: 4s¹, configuration (1s²)(2s² 2p⁶)(3s² 3p⁶)(4s¹)

• n − 1 shell (3s, 3p): 8 × 0.85 = 6.80
• n − 2 and lower: 10 × 1.00 = 10.00
• S = 16.80 → Zeff = 2.20

Option B: 3d¹, configuration (1s²)(2s² 2p⁶)(3s² 3p⁶)(3d¹)

• For a d electron, all 18 electrons to the left contribute 1.00 each: S = 18.00 → Zeff = 1.00

The 4s electron feels more than twice the effective nuclear charge of a 3d electron, so it’s more strongly bound and lower in energy. Slater’s rules reproduce the observed filling order. See the Aufbau principle.

Worked example 7: a 3d electron in zinc

Zinc, Z = 30: (1s²)(2s² 2p⁶)(3s² 3p⁶)(3d¹⁰)(4s²)

For one 3d electron:

• Groups to the right (4s): 0
• Same (3d) group: 9 × 0.35 = 3.15
• All groups to the left: 18 × 1.00 = 18.00
• S = 21.15 → Zeff = 8.85

For a 4s electron in zinc:

• Same group (4s, 4p): 1 × 0.35 = 0.35
• n − 1 shell (3s, 3p and 3d, all with n = 3): 18 × 0.85 = 15.30
• n − 2 and lower: 10 × 1.00 = 10.00
• S = 25.65 → Zeff = 4.35

The 4s electrons feel much less effective charge than the 3d electrons, so they’re lost first when zinc forms Zn²⁺. See electron configurations of ions.

Using Zeff to estimate relative size

Because Zeff tells you how strongly the outer electrons are pulled in, the ratio of shell number to Zeff gives a rough guide to relative size. Chlorine (n = 3, Zeff = 6.10) has a much smaller ratio than sodium (n = 3, Zeff = 2.20), consistent with chlorine atoms being much smaller. Such estimates aren’t precise, but they’re a useful way to check that an explanation of a periodic trend makes sense.

Limitations of Slater’s rules

• They’re empirical approximations, designed to fit data, not derived exactly.
• They treat s and p in the same shell identically, ignoring the extra penetration of s electrons.
• Values for heavy elements are less reliable.
• More accurate Zeff values come from quantum-mechanical calculations (for example, those published by Clementi and Raimondi in the 1960s), which give somewhat different numbers but the same trends.

Use Slater’s rules for trends and comparisons, not precise values.

Practice questions

1. Calculate Zeff for a 2p electron in fluorine (Z = 9).
2. Calculate Zeff for the 4s electron in calcium (Z = 20).
3. Calculate Zeff for a 2s electron in carbon (Z = 6).

Answers:

1. Same group: 6 × 0.35 = 2.10; 1s: 2 × 0.85 = 1.70; S = 3.80; Zeff = 5.20
2. Same group: 1 × 0.35 = 0.35; n − 1 (3s, 3p): 8 × 0.85 = 6.80; lower: 10 × 1.00 = 10.00; S = 17.15; Zeff = 2.85
3. Same group (2s, 2p): 3 × 0.35 = 1.05; 1s: 2 × 0.85 = 1.70; S = 2.75; Zeff = 3.25

Key takeaways

• Group orbitals as (1s)(2s,2p)(3s,3p)(3d)(4s,4p)…
• For s/p electrons: same group 0.35 (1s: 0.30), n − 1 shell 0.85, lower shells 1.00.
• For d/f electrons: same group 0.35, everything to the left 1.00.
• Zeff rises by about 0.65 per element across a period but stays nearly constant down a group.
• Slater’s rules explain why 4s fills before 3d and why 4s electrons are lost first in ions.