Structure of Atom · Lesson 11 of 12
Writing Electronic Configurations
“Build electron arrangements, identify ions, and understand important configuration exceptions.”
• Write full configurations, noble-gas shorthand, and orbital diagrams. • Identify core and valence electrons and count unpaired electrons. • Apply electron counting to main-group ions and isoelectronic species. • Explain chromium and copper configurations using total-energy considerations. • Relate symmetry and exchange stabilisation to half-filled and filled subshells.
From filling rules to a complete configuration
An electronic configuration specifies how electrons are distributed among atomic orbitals. Begin with the electron count, then apply the energy order, Pauli exclusion, and Hund’s rule. The sum of every occupation number must equal the species’ total electron count. Checking that sum is an essential final step.
In subshell notation, a leading number gives n, a letter gives the subshell type, and a superscript gives its electron occupation. Thus means four electrons distributed among three 2p spatial orbitals, not four p orbitals. An orbital-box diagram gives additional information by showing individual orbitals and spin arrows.
The order in which subshell terms are written is not itself an electron transition. For example, writing occupied 3d before 4s is a common configuration convention even when the simple building-up sequence placed 4s first. What matters is the occupation assigned to each subshell.
Building the first two periods
Hydrogen has one electron in 1s. Helium adds a second electron with opposite spin in the same spatial orbital, filling 1s. Lithium’s third electron cannot enter an already full 1s orbital, so it occupies 2s. Beryllium completes 2s; the next six electrons progressively fill 2p from boron through neon.
Carbon’s two p electrons occupy different p orbitals with parallel spins. Nitrogen’s three occupy all three p orbitals singly. Oxygen’s fourth p electron begins pairing, followed by further pairing in fluorine and neon. These examples show why an electron count alone is not enough to draw a correct spin diagram.
| Atom | Configuration | 2p orbital pattern | Unpaired electrons |
|---|---|---|---|
| B | ↑ | empty | empty | 1 | |
| C | ↑ | ↑ | empty | 2 | |
| N | ↑ | ↑ | ↑ | 3 | |
| O | ↑↓ | ↑ | ↑ | 2 | |
| F | ↑↓ | ↑↓ | ↑ | 1 | |
| Ne | ↑↓ | ↑↓ | ↑↓ | 0 |
Problem
Write the ground-state configuration and count unpaired electrons for oxygen, Z = 8.
- 1.Fill 1s and 2s with two electrons each. Four remain, giving . The occupations sum to 8.
- 2.Place one up-spin electron in each p orbital, then add the fourth with opposite spin to one of them.
- 3.One p orbital is paired and two are singly occupied. Oxygen has two unpaired electrons in this ground-state orbital description.
Noble-gas shorthand and periodic patterns
A bracketed noble-gas symbol replaces an already completed core configuration. For sodium, [Ne] represents ten electrons, leaving one electron in 3s. From sodium through argon, the 3s and 3p pattern repeats the 2s and 2p pattern from lithium through neon.
For main-group elements, the occupied shell with the largest n contains the familiar valence electrons involved in bonding and reactions. The remaining filled inner electrons form the core. Transition-metal chemistry can also involve electrons in the incompletely filled d subshell, so “valence” should not be reduced to a universal highest-n-only rule.
Potassium and calcium begin filling 4s. Scandium through zinc progressively occupy 3d, followed by 4p from gallium to krypton. Rubidium through xenon similarly involve 5s, 4d, and 5p. Later rows introduce 4f and then 5f occupation. The longer reference table shows the pattern and its exceptions; it is a lookup aid rather than a demand to memorise every row.
Problem
Write phosphorus, Z = 15, in full and shorthand form, and count its unpaired electrons.
- 1.Fifteen electrons give . The first ten match neon.
- 2.The shorthand is . It contains five outer-shell electrons: two in 3s and three in 3p.
- 3.The three p electrons occupy different orbitals with parallel spins. Therefore phosphorus has three unpaired electrons.
Ions and isoelectronic species
For an ion, adjust the electron count before building or checking its configuration. An anion has gained electrons; a cation has lost them. The nucleus still identifies the element. Main-group ions often reach the same electron count and configuration as a nearby noble gas.
Isoelectronic species have the same number of electrons. This does not make them the same element or guarantee equal size and energy, because their nuclear charges can differ. Count electrons explicitly rather than comparing only the symbols or charge magnitudes.
Problem
Write configurations for , , and .
- 1.Sodium has electrons; oxygen has ; fluorine has .
- 2.Each has configuration , the neon configuration.
- 3.They are isoelectronic but have different proton numbers. Equal electron counts do not cancel the difference in nuclear attraction.
Problem
A neutral atom has configuration . Identify it.
- 1.Argon represents 18 electrons. The explicitly written subshells add three.
- 2.There are 21 electrons. Neutrality means Z = 21, identifying scandium.
- 3.If the species were an ion, the same electron count would not alone fix the element without knowing its charge.
Why chromium and copper are exceptions
The simple sequence predicts chromium as an argon core followed by four 3d electrons and two 4s electrons. Its observed ground state instead has five 3d electrons and one 4s electron. Copper similarly has ten 3d electrons and one 4s electron rather than nine and two.
The 3d and 4s arrangements are close in energy. Redistributing occupation can lower the total energy after attractions, repulsions, and exchange effects are included. Half-filled and filled subshells often receive extra stabilisation, helping explain these examples. This is a comparison of entire configurations, not a claim that an electron moves upwards in energy and the total energy must therefore rise.
| Atom | Simple sequence prediction | Observed ground-state arrangement |
|---|---|---|
| Cr | ||
| Cu |
Symmetry and exchange stabilisation
A half-filled or filled set of degenerate orbitals distributes electrons symmetrically across the available angular states. Different orbitals in a subshell have different spatial distributions, so their electrons need not overlap in the same way as a forced pair in one orbital. The balance of Coulomb repulsion and nuclear attraction can favour these arrangements.
Parallel-spin electrons also produce exchange stabilisation, a quantum effect related to the allowed combined electron wave function. Introductory diagrams often show pairs of parallel-spin electrons exchanging labels. These are a way to count exchange pairs, not a movie of distinguishable electrons swapping positions and emitting energy on each swap.
For five singly occupied d orbitals with parallel spins, there are distinct pairs. For a filled d subshell there are two groups of five same-spin electrons, giving 20 such pairs in this simple counting picture. These counts help illustrate Hund’s rule but do not calculate actual configuration energies by themselves. Other exceptions exist, and the slogan “half-filled is stable” cannot predict every element.
Problem
Compare ground-state unpaired electrons in chromium and iron.
- 1.Chromium is : five unpaired d electrons plus one unpaired s electron give six.
- 2.Iron is . The sixth d electron pairs in one of five d orbitals, leaving four d electrons unpaired; 4s is paired.
- 3.Therefore chromium has six and iron four unpaired electrons in these isolated-atom ground-state configurations. Count every occupied subshell, not only d.
Reference configurations and chemical meaning
The reference below expresses the long configuration table compactly using noble-gas cores. Each row should be read as an occupation pattern whose electron count equals Z. The ordinary building-up sequence is a guide; rows such as niobium, palladium, platinum, and gold show why exceptions cannot all be reduced to one slogan.
Electronic arrangement helps explain why elements with closed outer shells are comparatively unreactive, why many other elements readily form bonds, and why different elements show different chemical behaviour. Detailed bonding and periodic trends are developed later; the purpose here is to establish the electron arrangement needed to understand them.
| Element | Z | Configuration |
|---|---|---|
| H | 1 | |
| He | 2 | |
| Li | 3 | |
| Be | 4 | |
| B | 5 | |
| C | 6 | |
| N | 7 | |
| O | 8 | |
| F | 9 | |
| Ne | 10 | |
| Na | 11 | |
| Mg | 12 | |
| Al | 13 | |
| Si | 14 | |
| P | 15 | |
| S | 16 | |
| Cl | 17 | |
| Ar | 18 | |
| K | 19 | |
| Ca | 20 | |
| Sc | 21 | |
| Ti | 22 | |
| V | 23 | |
| Cr | 24 | |
| Mn | 25 | |
| Fe | 26 | |
| Co | 27 | |
| Ni | 28 | |
| Cu | 29 | |
| Zn | 30 | |
| Ga | 31 | |
| Ge | 32 | |
| As | 33 | |
| Se | 34 | |
| Br | 35 | |
| Kr | 36 |
| Element | Z | Configuration |
|---|---|---|
| Rb | 37 | |
| Sr | 38 | |
| Y | 39 | |
| Zr | 40 | |
| Nb | 41 | |
| Mo | 42 | |
| Tc | 43 | |
| Ru | 44 | |
| Rh | 45 | |
| Pd | 46 | |
| Ag | 47 | |
| Cd | 48 | |
| In | 49 | |
| Sn | 50 | |
| Sb | 51 | |
| Te | 52 | |
| I | 53 | |
| Xe | 54 | |
| Cs | 55 | |
| Ba | 56 | |
| La | 57 | |
| Ce | 58 | |
| Pr | 59 | |
| Nd | 60 | |
| Pm | 61 | |
| Sm | 62 | |
| Eu | 63 | |
| Gd | 64 | |
| Tb | 65 | |
| Dy | 66 | |
| Ho | 67 | |
| Er | 68 | |
| Tm | 69 | |
| Yb | 70 | |
| Lu | 71 | |
| Hf | 72 | |
| Ta | 73 | |
| W | 74 | |
| Re | 75 | |
| Os | 76 | |
| Ir | 77 | |
| Pt | 78 | |
| Au | 79 | |
| Hg | 80 | |
| Tl | 81 | |
| Pb | 82 | |
| Bi | 83 | |
| Po | 84 | |
| At | 85 | |
| Rn | 86 |
| Element | Z | Configuration |
|---|---|---|
| Fr | 87 | |
| Ra | 88 | |
| Ac | 89 | |
| Th | 90 | |
| Pa | 91 | |
| U | 92 | |
| Np | 93 | |
| Pu | 94 | |
| Am | 95 | |
| Cm | 96 | |
| Bk | 97 | |
| Cf | 98 | |
| Es | 99 | |
| Fm | 100 | |
| Md | 101 | |
| No | 102 | |
| Lr | 103 | |
| Rf | 104 | |
| Db | 105 | |
| Sg | 106 | |
| Bh | 107 | |
| Hs | 108 | |
| Mt | 109 | |
| Ds | 110 | |
| Rg | 111 |
The filling sequence is a useful starting prediction, not a universal exact energy ordering. Do not force chromium and copper into the naive pattern, and do not assume every configuration exception follows only from a half-filled or filled subshell. Configurations of very heavy atoms require additional care beyond this elementary model.
Quiz
How many unpaired electrons are in a ground-state p-four arrangement?
Which is the observed ground-state configuration of chromium?
Which pair is isoelectronic?
In a noble-gas shorthand, [Ne] represents:
What is the most accurate elementary interpretation of exchange diagrams?
Practice Problems
- Write full and shorthand configurations for silicon, sulphur, potassium, calcium, and zinc.
- Write configurations for , , , and .
- Group into isoelectronic sets.
- Find unpaired electrons in phosphorus, silicon, chromium, iron, and krypton.
- Use the reference table to identify two exceptions other than chromium and copper. Explain why a simple filling mnemonic needs exceptions.
- Explain qualitatively how a configuration with a different 3d/4s distribution can have lower total energy.
Key Takeaways
A configuration records subshell occupations; an orbital diagram additionally shows individual orbitals and spins. Noble-gas shorthand abbreviates a core without changing electron count. Isoelectronic species share electron count but can have different nuclear charges and properties. Chromium and copper show that total-energy balance can favour a different occupation from the simple sequence. Symmetry and quantum exchange help explain stabilisation, but no single shortcut predicts every exception.