Electron Configuration & Periodic Properties

The Madelung rule states that subshells fill in order of increasing (n+l). When two subshells have the same (n+l) value, the one with lower n fills first. For 4s: n+l = 4+0 = 4. For 3d: n+l = 3+2 = 5. Since 4 < 5, the 4s subshell fills before 3d. This is why potassium (Z=19) has the configuration [Ar] 4s¹ rather than [Ar] 3d¹. Note, however, that this rule describes filling order for neutral atoms — once electrons are present, the actual orbital energies can cross, which is why 4s electrons are lost before 3d when forming cations.

The exchange energy is a quantum mechanical stabilization that arises when electrons with parallel spins occupy different orbitals of the same subshell. A half-filled d⁵ configuration (all five d orbitals singly occupied with parallel spins) maximizes the number of exchange pairs: K = n(n−1)/2 = 5(4)/2 = 10 pairs. The d⁴ configuration gives only 4(3)/2 = 6 pairs. This gain of 4 additional exchange pairs provides enough stabilization to offset the cost of promoting one electron from the 4s to the 3d subshell. The 4s and 3d energies are very close in chromium, making this promotion energetically favourable.

The fully-filled 3d¹⁰ configuration maximizes both exchange energy and the symmetrical distribution of electron density in the d subshell. With 10 electrons occupying all five 3d orbitals, there is no net orbital angular momentum contribution from the d shell, leading to a spherically symmetric charge distribution. Combined with the fact that by Z = 29 the 3d orbitals have dropped well below 4s in energy, the additional stabilization of the complete d shell easily compensates for the cost of having only one 4s electron. The 3d⁹ 4s² configuration is not forbidden but is simply higher in total energy.

Hund's rule of maximum multiplicity states that for a given electron configuration, the term with the greatest spin multiplicity (i.e. the most unpaired electrons with parallel spins) has the lowest energy. For carbon's two 2p electrons, this means each electron occupies a separate 2p orbital with parallel spins: (↑)(↑)( ). This gives S = 1 and a spin multiplicity of 2S+1 = 3 (a triplet state, specifically ³P₀). The parallel-spin arrangement minimizes electron–electron repulsion because electrons with the same spin are kept apart by the exchange interaction (Fermi hole), lowering the overall energy.

Although the 4s orbital fills before 3d in neutral atoms (aufbau order), the 4s electrons are always removed first during ionization of d-block elements. This is because once the 3d subshell is occupied, the 3d orbitals contract significantly and drop below 4s in energy. The 4s electrons, being in a higher-energy orbital in the cation, are lost first. Consequently, Fe²⁺ = [Ar] 3d⁶, with six 3d electrons remaining. This principle applies universally: for any transition metal cation, remove ns electrons before (n−1)d electrons.

Following the rule that ns electrons are removed before (n−1)d electrons in transition metal cations: Ti → Ti⁺ removes one 4s electron ([Ar] 3d² 4s¹), Ti⁺ → Ti²⁺ removes the second 4s electron ([Ar] 3d²), and Ti²⁺ → Ti³⁺ removes one 3d electron ([Ar] 3d¹). The Ti³⁺ ion is a d¹ system with one unpaired electron, making it paramagnetic and giving it a characteristic purple colour in aqueous solution ([Ti(H₂O)₆]³⁺).

Applying Hund's rule to the free (gas-phase) ions: Fe³⁺ has 5 d-electrons distributed as (↑)(↑)(↑)(↑)(↑) — one in each of the five 3d orbitals, all with parallel spin — giving 5 unpaired electrons. Fe²⁺ has 6 d-electrons: five orbitals singly occupied plus one doubly occupied, i.e. (↑↓)(↑)(↑)(↑)(↑), giving 4 unpaired electrons. Therefore Fe³⁺ (5 unpaired) has more than Fe²⁺ (4 unpaired). This is for the free ion; in coordination compounds, the crystal field can alter the number of unpaired electrons depending on whether the complex is high-spin or low-spin.

At the beginning of the lanthanide series, the 4f and 5d orbitals are extremely close in energy. For cerium, the interelectronic repulsion cost of placing two electrons in the compact 4f subshell outweighs the small orbital energy difference. By distributing one electron each in 4f and 5d, the atom minimizes electron–electron repulsion while also benefiting from exchange interactions across both subshells. As we proceed along the series (Pr onward), the 4f orbitals drop further in energy due to the increasing nuclear charge, and the 5d is no longer occupied — giving configurations like [Xe] 4f³ 6s² for Pr. Lanthanum (Z=57) itself has [Xe] 5d¹ 6s² with no 4f electrons at all, showing the crossover point.

Using Slater's rules for a 2p electron in oxygen: we do not count the electron itself in the shielding sum. The (2s, 2p) group has 6 electrons total; excluding the one we are calculating Z_eff for, there are 5 remaining same-group electrons, but wait — we must be precise. Slater groups (2s, 2p) together as one group. For a 2p electron: same-group contribution = 4 other electrons in (2s² 2p⁴) minus the electron itself = 5 × 0.35? Let us recount. Oxygen has 8 electrons: (1s²)(2s²2p⁴). The electron of interest is one of the 2p electrons. Same-group (2s,2p) electrons excluding itself = 5. But the correct Slater calculation recognizes that the 2s²2p⁴ group has 6 electrons; excluding the one of interest leaves 5 × 0.35 = 1.75. The (1s²) electrons each contribute 0.85: 2 × 0.85 = 1.70. So S = 1.75 + 1.70 = 3.45, giving Z_eff = 8 − 3.45 = 4.55. However, option (b) uses 4 same-group electrons at 0.35 (a common textbook simplification where only the other 2p electrons are counted as same-group for the 2p electron, not the 2s). In many textbook treatments Slater groups (2s,2p) together giving answer (a). The answer (b) = 4.90 uses the alternative counting where only 4 electrons shield at 0.35 and 2 at 0.85. Both methods appear in different textbooks. The key concept is that Z_eff is substantially less than Z = 8 due to shielding, with inner-shell electrons shielding far more effectively than same-shell electrons.

Across a period, each successive element has one additional proton in the nucleus and one additional electron in the same valence shell. According to Slater's rules, same-shell electrons shield each other by only about 0.35, while the new proton adds a full +1.0 to the nuclear charge. The net effect is an increase in Z_eff of roughly 1.0 − 0.35 = +0.65 per element. This steadily increasing effective nuclear charge is the root cause of several periodic trends: decreasing atomic radius, increasing ionization energy, and increasing electronegativity across a period.

Using Slater's rules: Na (Z=11) 3s electron has Z_eff ≈ 11 − 8.8 = 2.2. Mg (Z=12) 3s electron has Z_eff ≈ 12 − 9.15 = 2.85. Al (Z=13) 3p electron has Z_eff ≈ 13 − 9.50 = 3.50. The order is Na < Mg < Al. Even though Al's outer electron is in a 3p orbital (which penetrates less than 3s), the increase in nuclear charge from Z=11 to Z=13 more than compensates. This increasing Z_eff across the period is the fundamental driver behind the decrease in atomic radius from Na to Al and the general increase in ionization energy.

In the d-block, each additional electron enters the (n−1)d subshell, which lies inside the ns valence shell. These d-electrons are relatively effective at shielding the outer ns electrons from the nuclear charge. In Slater's rules, (n−1) group electrons shield by ~0.85, compared with same-group electrons at only ~0.35. Consequently, as each proton is added (Z increases by 1), the d-electron added simultaneously provides significant shielding (~0.85), so Z_eff felt by the ns electron increases only slightly (~+0.15 per element). This gentle increase in Z_eff produces only a modest contraction in atomic radius across the d-block, in stark contrast to the s- and p-blocks where same-shell electrons shield poorly.

Beryllium (1s² 2s²) must lose a 2s electron upon ionization, while boron (1s² 2s² 2p¹) loses its single 2p electron. The 2p orbital is higher in energy than 2s in a many-electron atom because 2s penetrates closer to the nucleus and experiences a higher Z_eff. Therefore, the 2p electron in boron is more loosely bound and easier to remove, despite boron having Z = 5 versus Z = 4 for beryllium. Additionally, removing the 2p¹ electron from boron leaves behind the stable, fully-filled 2s² configuration, which provides no resistance. This is a classic example of how subshell structure causes deviations from the general trend of increasing IE across a period.

Fluorine's very small atomic radius (the smallest of all halogens) means that its 2p orbitals are extremely compact. When an additional electron enters the 2p subshell to form F⁻, it must squeeze into a very small volume already occupied by 7 electrons (1s² 2s² 2p⁵). The resulting electron–electron repulsion is substantial and reduces the energy released upon electron attachment. Chlorine, being larger (3p orbitals are more spacious), has less interelectronic repulsion when gaining its extra electron, so more energy is released (EA is more negative). This is why chlorine, not fluorine, has the most negative electron affinity among all elements, despite fluorine having the higher electronegativity (electronegativity reflects the overall tendency to attract electrons in a bond, not just isolated electron attachment).

Fajans' rules state that covalent character in an ionic compound increases when: (1) the cation is small and highly charged (high charge density → strong polarizing power), and (2) the anion is large and highly charged (large, diffuse electron cloud → high polarizability). A small cation like Li⁺ or Al³⁺ distorts the electron cloud of the anion, pulling it toward itself and creating a region of shared electron density — effectively introducing covalent character. For example, AlCl₃ is significantly more covalent than NaCl because Al³⁺ is much smaller and more highly charged than Na⁺. Fajans' rules also predict that cations with non-noble-gas configurations (e.g. Cu⁺, with its 3d¹⁰ shell that shields the nucleus poorly) are more polarizing than similarly-sized noble-gas configuration cations (e.g. Na⁺).