The periodic table and electron configuration
The periodic table orders the elements by atomic number, and its rows and columns are no whim: they reflect how the electron shells fill. This topic explains the electron configuration, the blocks of the table and why atomic size, ionization energy and electronegativity change regularly from one side to the other.
Atomic number, groups and periods
The atomic number Z is the number of protons (and of electrons in the neutral atom). Elements in the same group (column) have the same configuration in their outer shell, which is why they are chemically alike; the period (row) says how many shells are occupied. The table is divided into blocks according to the last subshell being filled: s (groups 1 and 2), p (13 to 18), d (3 to 12, the transition metals) and f (lanthanides and actinides).
Electron configuration
Electrons occupy subshells (s, p, d, f, holding 2, 6, 10 and 14) in order of increasing energy. The order is given by Madelung's rule: first the subshell with the lowest (with = 0, 1, 2, 3 for s, p, d, f) and, for equal , the one with the lowest . That is why 4s fills before 3d.
It is abbreviated with the previous noble gas in brackets: iron, Z = 26, is [Ar] 3d⁶ 4s². Within a subshell electrons first spread out one by one with the same spin (Hund's rule) and then pair up. There are exceptions when a d subshell ends up full or half full, which is especially stable: chromium is [Ar] 3d⁵ 4s¹ and copper [Ar] 3d¹⁰ 4s¹, not 3d⁴ 4s² or 3d⁹ 4s².
Periodic trends
The key is the effective nuclear charge: the attraction an outer electron feels after the screening by the inner electrons. Across a period the protons increase but the new electrons enter the same shell and screen little, so the effective charge grows; going down a group a whole shell is added, farther from the nucleus.
| Property | Across a period (→) | Down a group (↓) |
|---|---|---|
| Atomic radius | decreases | increases |
| 1st ionization energy | increases (with exceptions) | decreases |
| Electronegativity | increases | decreases |
| Metallic character | decreases | increases |
The ionization energy is the energy needed to remove the outermost electron from a gaseous atom; electronegativity (Pauling scale) measures how strongly an atom attracts the electrons of a bond. That is why the alkali metals (bottom left) lose electrons easily and the halogens (top right) gain them: fluorine is the most electronegative, 3.98.
Common mistakes
- Filling 3d before 4s (or keeping 4s when removing electrons: Fe²⁺ is [Ar] 3d⁶, with no 4s).
- Forgetting the Cr and Cu exceptions (and silver and gold, in the same column as Cu).
- Confusing the period with the number of valence electrons.
- Expecting the ionization energy to always rise across a period: it drops at groups 13 and 16.
Worked examples
Example 1Configuration of iron
Solution
- Madelung order: 1s² 2s² 2p⁶ 3s² 3p⁶ (18 electrons, argon), 4s² (20) and 6 remain for 3d.
- Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s². By shell: 2, 8, 14, 2.
- The last subshell filled is d → block d. Period 4 (highest shell, n = 4). Group: 2 (from 4s) + 6 (from 3d) = 8.
Solve it in Simuveliq
- In Chemistry → Periodic table, click Fe (or search for it). The card gives the configuration [Ar] 3d⁶ 4s², the electrons per shell 2, 8, 14, 2 and «Group · period · block» 8 · 4 · d.
Example 2The copper exception
Solution
- One electron moves from 4s to 3d and completes it: Cu = [Ar] 3d¹⁰ 4s¹. A full d subshell is more stable than the energy difference between 4s and 3d, which at that point is very small.
- Chromium does the same to have 3d half full (five unpaired electrons): Cr = [Ar] 3d⁵ 4s¹.
- That explains why copper forms Cu⁺ (it loses the 4s¹) as well as Cu²⁺.
Solve it in Simuveliq
- The Cu card gives [Ar] 3d¹⁰ 4s¹, electrons per shell 2, 8, 18, 1 and usual oxidation states +1, +2. The tool applies Madelung and corrects it with the list of known exceptions; open Cr (24) to see [Ar] 3d⁵ 4s¹.
Example 3Chlorine and its ion
Solution
- Cl: [Ne] 3s² 3p⁵: period 3, block p, group 17 (the halogens); it has 7 electrons in the valence shell.
- It is one short of argon's configuration. Its electronegativity is high (3.16) and so is its ionization energy (12.97 eV): losing electrons is hard and it gains one easily → Cl⁻ = [Ne] 3s² 3p⁶ = [Ar].
- With sodium ([Ne] 3s¹, ionization energy 5.14 eV, electronegativity 0.93) it forms NaCl: the electronegativity difference, 2.23, indicates an ionic bond.
Solve it in Simuveliq
- The Cl card gives [Ne] 3s² 3p⁵, 1st ionization energy 12.968 eV, electronegativity 3.16 and usual oxidation state −1 (and +1, +3, +5, +7). Open Na (11) to compare.
Example 4The two irregularities of period 2
Solution
- B: [He] 2s² 2p¹. The electron removed is in 2p, higher in energy and more screened by 2s² than the 2s of Be.
- O: [He] 2s² 2p⁴. The first three 2p electrons go alone (Hund, N has 2p³ half full); the fourth pairs up with another in the same orbital, and their repulsion makes it easier to remove.
- Outside those two points the trend is as expected: from 5.39 eV (Li) to 21.56 eV (Ne).
Solve it in Simuveliq
- In the Periodic table, change Color to «1st ionization energy»: the table is painted as a heat map and the trend shows; the O card gives 13.618 eV and N's, 14.534 eV.
Example 5Going down group 1
Solution
- Covalent radii: Li 128 pm, Na 166, K 203, Rb 220, Cs 244 pm: each period adds a shell.
- Ionization energy: Li 5.39 eV, Cs 3.89 eV. Cesium's 6s¹ electron is much farther from the nucleus and screened by 54 inner electrons.
- Losing that electron is what alkali metals do when they react with water; the easier it is, the faster and more violent the reaction.
Solve it in Simuveliq
- The Cs card gives [Xe] 6s¹, 1st ionization energy 3.8939 eV and covalent radius 244 pm; with Color = «Covalent radius» you see the size grow toward the bottom left.