Effective Nuclear Charge and Shielding
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You should be comfortable with the quantum numbers that define atomic orbitals and understand that electrons in multi-electron atoms do not experience the full nuclear charge due to repulsion from other electrons.
Why this matters
Effective nuclear charge explains why the periodic table trends the way it does.
Without the concept of shielding, we could not explain why sodium is much larger than fluorine despite having only a few more protons, or why ionisation energies do not increase linearly with atomic number. By quantifying how much of the nucleus is "felt" by a valence electron, we gain a predictive tool for atomic size, reactivity, and bonding behaviour across the entire periodic table.
The Core Idea
Imagine the nucleus as a magnet and an electron as a piece of iron. In a hydrogen atom, the electron feels the full pull of the single proton. In a multi-electron atom, however, the other electrons act like a screen between the nucleus and the electron of interest. They repel the electron of interest, partially cancelling out the attractive force of the positive nucleus. The net attraction the electron feels is therefore less than the full nuclear charge. This net attraction is called the effective nuclear charge. It is the key variable that determines how tightly an atom holds its outer electrons and how large the atom becomes.
Slater's Rules for Shielding
To calculate the shielding constant \(S\), we use Slater's rules, which provide a semi-empirical method for estimating how much each electron shields the nucleus. For an electron in an \(ns\) or \(np\) orbital, the rules are specific. First, every other electron in the same \(n\) shell (the same \(ns, np\) group) contributes \(0.35\) to the shielding constant. Second, every electron in the \((n-1)\) shell contributes \(0.85\). Third, every electron in shells \((n-2)\) and lower contributes \(1.00\). This hierarchy reflects the fact that electrons in inner shells are closer to the nucleus and thus shield more effectively, while electrons in the same shell are on average further apart and shield less effectively.
Once \(S\) is determined, the effective nuclear charge is simply the difference between the actual nuclear charge \(Z\) and the shielding constant \(S\). This value represents the net positive charge experienced by the electron in question.
Trends Across a Period
As we move from left to right across a period, the atomic number \(Z\) increases by one for each element. However, the added electrons enter the same principal shell \(n\). According to Slater's rules, these new electrons only contribute \(0.35\) to the shielding constant for the valence electron. Therefore, the increase in \(Z\) is not fully compensated by the increase in \(S\). The result is a steady increase in \(Z_{\text{eff}}\). Because the valence electrons are pulled more strongly toward the nucleus, the atomic radius decreases. This stronger attraction also makes it harder to remove an electron, leading to an increase in the first ionisation energy. This trend is consistent with the observed periodic properties of elements.
Trends Down a Group
When moving down a group, the principal quantum number \(n\) increases. The valence electron is now in a higher energy level, further from the nucleus. Although \(Z\) increases significantly, the number of inner shell electrons also increases substantially. These inner electrons provide strong shielding, contributing \(0.85\) or \(1.00\) to \(S\). Consequently, the increase in \(S\) almost matches the increase in \(Z\), leaving \(Z_{\text{eff}}\) roughly constant. However, because the valence electron is in a larger orbital (higher \(n\)), it is physically further from the nucleus. The combination of constant effective charge and increased distance results in a larger atomic radius and a lower first ionisation energy.
The result
Periodic trends
Define the effective nuclear charge felt by an electron as \(Z_{\text{eff}}=Z-S\), where \(S\) (the shielding constant) is computed from Slater's rules: for an electron in an \(ns\) or \(np\) orbital, sum \(0.35\) for every other electron in the same \(n s,np\) group, \(0.85\) for every electron in the \(n{-}1\) shell, and \(1.00\) for every electron in shells \(n{-}2\) and below. Then, moving left to right across a period, \(Z_{\text{eff}}\) for the valence electron increases (atomic radius decreases, first ionisation energy increases); moving down a group, \(Z_{\text{eff}}\) stays roughly constant while the valence principal quantum number \(n\) increases (atomic radius increases, first ionisation energy decreases).
A common misconception is that shielding is a simple, uniform reduction of nuclear charge. In reality, the distribution of electron density is complex, and Slater's rules are an approximation. For example, \(d\) and \(f\) electrons shield less effectively than \(s\) and \(p\) electrons due to their radial distribution. While Slater's rules work well for main-group elements, they are less accurate once \(d\) and \(f\) electrons are involved, as in the transition metals. Always remember that \(Z_{\text{eff}}\) is a model-dependent value, not a directly measurable physical constant. C
Worked example
Problem. Calculate the effective nuclear charge \(Z_{\text{eff}}\) experienced by a valence electron in a phosphorus atom (\(Z=15\)). The electron configuration of phosphorus is \(1s^2 2s^2 2p^6 3s^2 3p^3\). Consider a \(3p\) electron.
- Identify the electron of interest: a \(3p\) electron. The relevant shell is \(n=3\).
- Count electrons in the same \(n=3\) shell: There are \(2\) electrons in \(3s\) and \(2\) other electrons in \(3p\) (since one is the electron of interest). Total same-shell electrons = \(4\). Contribution: \(4 \times 0.35 = 1.40\).
- Count electrons in the \(n-1=2\) shell: There are \(2\) in \(2s\) and \(6\) in \(2p\). Total = \(8\). Contribution: \(8 \times 0.85 = 6.80\).
- Count electrons in shells \(n-2\) and below: There are \(2\) electrons in \(1s\). Contribution: \(2 \times 1.00 = 2.00\).
- Sum the shielding constant \(S\): \(S = 1.40 + 6.80 + 2.00 = 10.20\).
- Calculate \(Z_{\text{eff}}\): \(Z_{\text{eff}} = Z - S = 15 - 10.20 = 4.80\).
Units check. \(Z_{\text{eff}}\) is a dimensionless quantity representing the net nuclear charge in units of elementary charge \(e\).
Three quick tests
- Why does the atomic radius decrease across a period even though the number of electrons increases?
Answer
The increase in nuclear charge \(Z\) outpaces the increase in shielding \(S\), leading to a higher \(Z_{\text{eff}}\) that pulls the electron cloud closer to the nucleus. - Calculate \(Z_{\text{eff}}\) for a \(2p\) electron in nitrogen (\(Z=7\), config \(1s^2 2s^2 2p^3\)).
Answer
Same shell (\(n=2\)): \(2\) in \(2s\) + \(2\) other in \(2p\) = \(4\) electrons. \(4 \times 0.35 = 1.40\). Inner shell (\(n=1\)): \(2\) electrons. \(2 \times 1.00 = 2.00\). \(S = 3.40\). \(Z_{\text{eff}} = 7 - 3.40 = 3.60\). - Why is the first ionisation energy of potassium lower than that of sodium?
Answer
Potassium's valence electron is in the \(n=4\) shell, which is further from the nucleus and more shielded by inner electrons than sodium's \(n=3\) electron, despite similar \(Z_{\text{eff}}\).
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.