How configuration decides the position of an element
Electrons fill subshells in order of energy (Aufbau order): 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
Because properties depend mostly on the outer electrons, elements with the same outer arrangement fall in the same group.
Period from configuration
The period number = the highest principal quantum number (n) in the configuration. Na = [Ne] 3s¹ → n = 3 → period 3.
Each period starts with filling a new ns and ends with a full np (noble gas). So the number of elements in a period = number of electrons that fit in the subshells filled in it:
- Period 1: 1s → 2 elements
- Periods 2 and 3: ns, np → 8
- Periods 4 and 5: ns, (n−1)d, np → 18
- Periods 6 and 7: ns, (n−2)f, (n−1)d, np → 32
Group from configuration
- s-block: group = number of ns electrons (1 or 2).
- p-block: group = 10 + (ns + np electrons).
- d-block: group = ns electrons + (n−1)d electrons.
- f-block: all placed in group 3 (shown separately at the bottom).
Special cases
Chromium is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) and copper is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²), because a half-filled or full d subshell is extra stable. The group rule still works: Cr → 1 + 5 = 6, Cu → 1 + 10 = 11.
Helium is 1s² (s-block) but is placed in group 18 because its shell is full and it behaves like a noble gas. Hydrogen (1s¹) sits in group 1 but is a non-metal; it is sometimes discussed separately.
s-block elements
Groups 1 (alkali metals) and 2 (alkaline earth metals). Outer configuration ns¹ and ns².
- Reactive metals with low ionisation enthalpy; they lose outer electrons to form M⁺ and M²⁺ ions.
- Reactivity increases down the group.
- Mostly ionic compounds (except Li and Be, which show some covalent character).
p-block elements
Groups 13 to 18. Outer configuration ns² np¹ to ns² np⁶. Together, s- and p-block elements are called representative or main group elements.
- Contains metals, non-metals and metalloids. Non-metallic character increases across a period and metallic character increases down a group.
- Group 17 (halogens, ns² np⁵) have very negative electron gain enthalpy.
- Group 18 (noble gases, ns² np⁶, He 1s²) have a full outer shell and are very unreactive.
d-block elements (transition elements)
Groups 3 to 12, in the middle of the table. General outer configuration (n−1)d¹⁻¹⁰ ns⁰⁻² (Pd has 4d¹⁰ 5s⁰).
- All are metals, usually hard with high melting points.
- Many form coloured ions, show several oxidation states, are paramagnetic and act as catalysts.
- Zn, Cd and Hg ((n−1)d¹⁰ ns²) do not show most of these features.
- They form a "bridge" between the very reactive s-block metals and the p-block elements, hence the name transition.
f-block elements (inner transition elements)
The two rows at the bottom: lanthanoids (Ce, Z = 58 to Lu, 71) and actinoids (Th, Z = 90 to Lr, 103). General outer configuration (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns².
- All are metals. Within each row the properties are very similar.
- Actinoids are radioactive. Elements after uranium (Z > 92) are called transuranium elements and are mostly made in the lab.
Metals, non-metals and metalloids
More than 78% of elements are metals (left side and centre). Non-metals are at the top right. A zig-zag line from B through Si, As, Te to At separates them; elements next to it (Si, Ge, As, Sb, Te) are metalloids, showing properties of both.
Key formulas and definitions
- Period = highest n in the configuration
- s-block: ns¹⁻² ; group = ns electrons
- p-block: ns² np¹⁻⁶ ; group = 10 + ns + np electrons
- d-block: (n−1)d¹⁻¹⁰ ns⁰⁻² ; group = ns + (n−1)d electrons
- f-block: (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns²
- Elements per period = 2, 8, 8, 18, 18, 32, 32
Worked examples
1. Find the period, group and block of Mg (Z = 12).
Mg: 1s² 2s² 2p⁶ 3s² = [Ne] 3s². Highest n = 3 → period 3. Last electron in s → s-block. Group = ns electrons = 2. Answer: period 3, group 2, s-block.
2. Find the position of phosphorus (Z = 15).
[Ne] 3s² 3p³. n = 3 → period 3. Last electron in p → p-block. Group = 10 + 2 + 3 = 15. Answer: period 3, group 15, p-block.
3. Find the position of iron (Z = 26).
[Ar] 3d⁶ 4s². Highest n = 4 → period 4. Last electron enters 3d → d-block. Group = 2 + 6 = 8. Answer: period 4, group 8, d-block.
4. Find the position of bromine (Z = 35).
[Ar] 3d¹⁰ 4s² 4p⁵. n = 4 → period 4. Last electron in 4p → p-block. Group = 10 + 2 + 5 = 17. Answer: period 4, group 17, p-block (a halogen).
5. Write the configuration of Cr (Z = 24) and find its group.
Expected [Ar] 3d⁴ 4s², but the real one is [Ar] 3d⁵ 4s¹ (half-filled d is more stable). Period 4, d-block, group = 1 + 5 = 6.
6. An element has outer configuration 5s² 5p⁴. Find Z, period, group and block.
n = 5 → period 5. p-block. Group = 10 + 2 + 4 = 16. Full configuration: [Kr] 4d¹⁰ 5s² 5p⁴ → Z = 36 + 10 + 2 + 4 = 52 (tellurium).
7. An element is in period 4, group 11. Write its configuration and Z.
Group 11 is d-block: ns + (n−1)d = 11. Normal filling gives 4s² 3d⁹, but the stable form is 3d¹⁰ 4s¹. So [Ar] 3d¹⁰ 4s¹, Z = 18 + 11 = 29 (copper).
Common mistakes
- Taking the period from the last subshell written. For Fe the last filled is 3d but the period is 4 (from 4s).
- Using "group = number of outer electrons" for the p-block. Add 10: chlorine has 7 outer electrons but is in group 17.
- Writing Cr as 3d⁴ 4s² and Cu as 3d⁹ 4s². The stable forms are 3d⁵ 4s¹ and 3d¹⁰ 4s¹.
- Putting helium in the s-block column of the table. It is 1s² but sits in group 18 with the noble gases.