Definition and physical meaning
Effective nuclear charge is the net positive charge felt by a particular electron in a multi-electron atom. Inner (core) electrons partially shield or screen the nucleus, so an outer (valence) electron does not feel the full nuclear charge \(Z\).
The standard model is: \[ Z_{\mathrm{eff}} = Z - S \] where \(Z\) is the atomic number and \(S\) is the shielding (screening) constant contributed by other electrons.
How to estimate shielding with Slater’s rules (ns/np electron)
For a valence electron in an ns or np orbital, Slater’s rules group electrons by shells and assign the following shielding contributions:
| Electrons counted toward \(S\) | Shielding per electron | Applies when the electron of interest is |
|---|---|---|
| Other electrons in the same principal shell \(n\) (same ns/np group) | \(0.35\) each (except \(1s\): \(0.30\) each) | \(ns\) or \(np\) |
| Electrons in shell \(n-1\) | \(0.85\) each | \(ns\) or \(np\) |
| Electrons in shell \(n-2\) or lower | \(1.00\) each | \(ns\) or \(np\) |
Worked example: effective nuclear charge for a 3p electron in chlorine
Chlorine has atomic number \(Z = 17\) and electron configuration 1s2 2s2 2p6 3s2 3p5. A 3p valence electron is an \(ns/np\) case with \(n=3\).
Step 1: Compute shielding \(S\) from the same shell \(n=3\).
In the \(3s,3p\) group there are \(3s^2 3p^5\) electrons. For one chosen 3p electron, the “other electrons in the same shell” count is \(2 + 4 = 6\). \[ S_{n} = 6 \cdot 0.35 = 2.10 \]
Step 2: Add shielding from shell \(n-1 = 2\).
Shell 2 contains \(2s^2 2p^6\), i.e. \(8\) electrons. \[ S_{n-1} = 8 \cdot 0.85 = 6.80 \]
Step 3: Add shielding from shells \(n-2\) and lower (shell 1).
Shell 1 contains \(1s^2\), i.e. \(2\) electrons. \[ S_{n-2\ \text{and lower}} = 2 \cdot 1.00 = 2.00 \]
Step 4: Total \(S\) and compute \(Z_{\mathrm{eff}}\).
\[ S = 2.10 + 6.80 + 2.00 = 10.90 \] \[ Z_{\mathrm{eff}} = 17 - 10.90 = 6.10 \] So a 3p valence electron in chlorine experiences an effective nuclear charge of approximately \(Z_{\mathrm{eff}} \approx 6.10\) by Slater’s rules.
Visualization: what “net attraction” means
Check: trend across Period 3 (Slater estimates for a valence electron)
Slater’s rules predict that \(Z_{\mathrm{eff}}\) increases from left to right across a period, which helps explain decreasing atomic radius and increasing first ionization energy across Period 3.
| Element | \(Z\) | Valence electron considered | Shielding \(S\) (Slater) | \(Z_{\mathrm{eff}} = Z - S\) |
|---|---|---|---|---|
| Na | 11 | 3s | \(0 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 8.80\) | \(2.20\) |
| Mg | 12 | 3s | \(1 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 9.15\) | \(2.85\) |
| Al | 13 | 3p | \(2 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 9.50\) | \(3.50\) |
| Si | 14 | 3p | \(3 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 9.85\) | \(4.15\) |
| P | 15 | 3p | \(4 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 10.20\) | \(4.80\) |
| S | 16 | 3p | \(5 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 10.55\) | \(5.45\) |
| Cl | 17 | 3p | \(6 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 10.90\) | \(6.10\) |
| Ar | 18 | 3p | \(7 \cdot 0.35 + 8 \cdot 0.85 + 2 \cdot 1.00 = 11.25\) | \(6.75\) |
Key takeaway
Effective nuclear charge summarizes the competition between nuclear attraction and electron–electron shielding; numerically \(Z_{\mathrm{eff}} = Z - S\). For chlorine’s 3p electron, Slater’s rules give \(Z_{\mathrm{eff}} \approx 6.10\), consistent with stronger net attraction as the period advances.