Electron configuration
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| shell/ʃel/ | 的伟大 |
| sub-shell/sʌb ʃel/ | phân lớp |
| orbital/ˈɔːbɪtl/ | orbital |
| ground state/ɡraʊnd steɪt/ | The lowest-energy state available to an atom, particle or other quantum system. |
| principal quantum number/ˈprɪnsɪpl ˈkwɒntəm ˈnʌmbə/ | số lượng tử chính |
| configuration/kənˌfɪɡjəˈreɪʃn/ | cấu hình electron |
| transition metal/trænˈsɪʃn ˈmetl/ | kim loại chuyển tiếp |
Count electrons before choosing their arrangement
- Electrons occupy shells 壳层, sub-shells 亚层 and orbitals 轨道. The ground state 基态 is the lowest-energy arrangement for that atom or ion.
- This syllabus section assesses ground-state species from hydrogen to krypton. First calculate the number of electrons, then fill the available orbitals; do not use the nucleon number as the electron total.
Relate a shell to its orbitals
- The principal quantum number 主量子数 $n$ labels the main shell. Larger $n$ generally means a greater spatial extent, but $n$ alone does not put every sub-shell of a multi-electron atom into energy order.
- Each orbital holds up to two electrons. A shell contains sub-shells; a p sub-shell contains three orbitals, not one six-electron orbital.
| Sub-shell | Number of orbitals | Maximum electrons |
|---|---|---|
| s | 1 | 2 |
| p | 3 | 6 |
| d | 5 | 10 |
Filling the shells
Drag the proton number. Electrons fill the shells from the inside out (2, then 8, then 8) — the basis of the periodic table.
How many electrons can a p sub-shell hold?
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A p sub-shell has 3 orbitals, each holding 2 electrons, so 6 in total (s = 2, d = 10).
Match each sub-shell to how many electrons it holds.
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Each orbital holds 2; s has 1 orbital, p has 3, d has 5 — so 2, 6 and 10 electrons.
Use the filling order with its limits
- This is the usual construction order for the neutral ground-state configurations studied here. Fill lower-energy available states first; chromium and copper have the exceptions shown below.
- Do not use the sequence to claim that 4s is always below occupied 3d levels. For first-row transition-metal cations, remove 4s electrons before 3d electrons.
The diagram supports the filling sequence; apply the separate ion-removal rule when forming transition-metal cations.
In the usual neutral-atom filling sequence used here, which is reached first: 4s or 3d?
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4s is reached before 3d in the usual neutral filling sequence. This does not mean that 4s remains lower in energy in every occupied atom or ion.
Put these sub-shells in the order they fill.
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Fill by energy, not by number: 4s comes before 3d.
Write configurations 电子排布 and check their sum
- Fe has 26 electrons: $1\mathrm{s}^2\,2\mathrm{s}^2\,2\mathrm{p}^6\,3\mathrm{s}^2\,3\mathrm{p}^6\,3\mathrm{d}^6\,4\mathrm{s}^2$, or $[\mathrm{Ar}]\,3\mathrm{d}^6\,4\mathrm{s}^2$. The [Ar] core represents 18 electrons, not one orbital.
- Group terms by $n$ to count shells: Fe has $2,8,14,2$. Its third shell contains 3s, 3p and 3d electrons, even though 4s was reached before 3d in the filling sequence.
The electron configuration of sodium (11 electrons) is 1s² 2s² 2p⁶ ______.
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Sodium has one electron in the 3s sub-shell.
The ground-state configuration of iron (Z = 26) is:
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After argon (18), iron adds 4s² then 3d⁶: [Ar] 3d⁶ 4s².
An Fe atom is [Ar] 3d⁶ 4s². Give the number of electrons in each occupied main shell and explain why the third-shell total is not eight.
The shell totals are 2,8,14,2. The argon core includes 3s² and 3p⁶, then the 3d⁶ electrons also belong to n = 3, giving 2 + 6 + 6 = 14. Filling 4s before 3d does not make 3d part of the fourth shell.
Keep the two neutral-atom exceptions
- Chromium has $[\mathrm{Ar}]\,3\mathrm{d}^5\,4\mathrm{s}^1$ and copper has $[\mathrm{Ar}]\,3\mathrm{d}^{10}\,4\mathrm{s}^1$. These are their ground states, rather than the result of mechanically assigning two 4s electrons in every case.
- A transition metal 过渡金属 ion may have a different occupation from the neutral atom. Start from the correct neutral configuration and remove the stated number of electrons; an ion does not gain electrons because its charge number is positive.
Make the ion, then recount
- Fe²⁺ loses two 4s electrons, giving $[\mathrm{Ar}]\,3\mathrm{d}^6$. Fe³⁺ loses one further 3d electron, giving $[\mathrm{Ar}]\,3\mathrm{d}^5$. The proton number remains 26; the electron totals are 24 and 23.
- For negative ions, add electrons. F⁻ has $9+1=10$ electrons, so it is $1\mathrm{s}^2\,2\mathrm{s}^2\,2\mathrm{p}^6$. Always check both the total and the sub-shell capacities.
For Co²⁺ ($Z=27$), neutral Co is [Ar] 3d⁷ 4s². Remove two 4s electrons: [Ar] 3d⁷, with $18+7=25$ electrons. The charge check is $27-25=+2$.
When iron forms Fe³⁺, which electrons are lost first?
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Transition metals lose 4s before 3d, so Fe³⁺ is [Ar] 3d⁵.
For Mn (Z = 25) and Mn²⁺, write the ground-state shorthand electron configurations and check the electron totals. Explain which sub-shell loses electrons first.
Mn is [Ar] 3d⁵ 4s², with 18 + 5 + 2 = 25 electrons. Mn²⁺ is [Ar] 3d⁵, with 23 electrons: the two 4s electrons are removed first. Its proton number stays 25 and the charge check is 25 − 23 = +2.
Neutral copper (Z = 29) has ground-state configuration [Ar] 3d¹⁰ 4s¹. Deduce the configurations of Cu⁺ and Cu²⁺ and explain why beginning with [Ar] 3d⁹ 4s² would be inappropriate.
Remove the single 4s electron first: Cu⁺ is [Ar] 3d¹⁰, with 28 electrons. Remove one 3d electron next: Cu²⁺ is [Ar] 3d⁹, with 27 electrons. The stated neutral ground state is copper’s exception to the simple filling pattern; [Ar] 3d⁹ 4s² has the right electron total but is not its ground-state configuration.
You've got it
- Shell → sub-shell → orbital; s, p and d capacities are 2, 6 and 10.
- Use the neutral filling sequence, but retain Cr and Cu ground-state exceptions.
- Remove 4s before 3d when forming the first-row transition-metal cations studied here.
- Check electron total, charge, orbital capacity and the shell number of every term.