Atomic Structure

— Chemistry · AS Level (7405) · Crash Course —

★ Exam Tip: Sir's words: this is one of the easiest chapters — mostly memorisation + understanding, hardly any calculation, and low weightage. Questions mostly come as Paper 1 MCQs (P2 examiners "don't like asking such childish things"). But it is very easy to make silly mistakes (1–2 marks lost) — build habits, don't rush!

Sir's method: The periodic table is your best friend. Do NOT memorise extra rules (Aufbau lists, energy orders…) — read everything off the periodic table. Have one open for every question.

What this chapter covers: ① subatomic particles & electric-field deflection ② Bohr → Quantum model ③ electronic configuration (atoms, ions, shorthand, Cr & Cu) ④ orbital shapes ⑤ ionisation energies & all their trends (they are the evidence for the quantum model)

1. Subatomic Particles

Subatomic particles = the particles inside an atom: protons, electrons, neutrons.

	Relative charge	Relative mass
proton	+1	1
neutron	0	1
electron	−1	11840 ≈ 0

Note (board slip): the handwritten table on the board has the rows shifted (electron shown as "0" charge). The correct values are the ones above — neutron 0 charge / mass 1, electron −1 charge / mass 1⁄1840.

★ Relative = relative to a proton, so no units (no g, no amu!). Mass: divide by proton mass.

Electron mass: for all practical purposes (weighing scale) it is 0, but the true value is 1/1840 of a proton — use the real number in calculations (e.g. electron velocity in Physics), otherwise the calculation dies at zero.

Behaviour in an electric field

e⁻ → + plate (bends most) · p⁺ → − plate (bends less) · n straight

Opposite charges attract: electrons → positive plate, protons → negative plate.
Neutrons have no charge ⇒ undeflected (straight).
Electron path is drawn shorter / more curved — it hits the plate first.

Why do electrons deflect MORE? (exam answer)The greater the charge/mass ratio, the more the deflection.
electron: Qm = −11/1840 → huge | proton: +11 = 1 | neutron: 01 = 0

Magnitude of charge is same for e⁻ and p⁺ (1 each) — the difference is only the tiny mass of the electron ("less mass" / larger Q/m). Order: electrons > protons > neutrons (0).

Trick (sir): dividing by a very small number gives a very big number — the calculator shows "Math error" ≈ ∞ (which is only a concept, not a real number). Real value = −1 ÷ (1/1840) = −1840.

Possible exam twist: a mixed beam of e⁻, p⁺, n → "separate them": pass through an electric field (one plate +, one −) and they split into 3 beams by themselves.

Electronic structure at O-level — the Bohr model

Bohr: nucleus 11p 12n; shells 2 · 8 · 1

Sodium (Na): 11 protons, neutrons = 23 − 11 = 12 (from the periodic table: atomic no. 11, mass no. 23). Electrons = 11 → shells fill 2, 8, 1 (draw crosses or dots).

This "shells only" model is called the Bohr model (Niels Bohr — his work reappears in A2 Physics: Quantum Physics). It worked well: ionic bonding (Na loses its 1 valence e⁻), octets, 11+ nucleus with 11 e⁻…

Problem with Bohr: it is a simplistic model. It cannot explain (i) the trends & dips in ionisation energies (later in this chapter) and (ii) atomic emission / absorption spectra (Physics A2). We needed a better understanding ⇒ the Quantum model. The nucleus story (Rutherford/Bohr) stays the same — the quantum model only changes how the electrons are arranged.

Sir's aside: "What you learnt earlier was right for that time — you'd have blown a fuse if we taught everything at once. Humanity progressed the same way: new phenomena appeared, the old model couldn't explain them, so a better one was built."

2. The Quantum Model

Sir: "This slide is for understanding, not rote-learning. Pointers can appear as MCQs, but the periodic-table hack does the real work."

Electrons are arranged in shells, n = 1, 2, 3 … — n = principal quantum number (Bohr said shells too — not wrong, just incomplete).
Shells increase in energy as n increases (a few exceptions — e.g. 4s vs 3d, see later).
Each shell has subshells — the number of subshells = its principal quantum number (n = 1 → 1, n = 2 → 2 …). ← new in the quantum model (Bohr stopped at shells).
Subshells are named s, p, d, f — always in that order (never 1, 2, 3, 4).
Each subshell has one or more orbitals = little "boxes" where the electrons actually sit: s → 1, p → 3, d → 5, f → 7 (odd numbers going up).

shell (number) ⊃ subshell (letter) ⊃ orbital (box) ⊃ electrons

Common mistake (MCQ trap!): students mix up shell ↔ subshell ↔ orbital. Shell = the number. Subshell = the letter. Orbital = the box inside a subshell. e.g. "3p" is a subshell and contains 3 orbitals.

Shell	No. of subshells	Names of subshells	Orbitals present
n = 1	1	1s	1s
n = 2	2	2s, 2p	2s, 2p_x, 2p_y, 2p_z
n = 3	3	3s, 3p, 3d	3s, 3p_x, 3p_y, 3p_z, (5 × 3d)
n = 4	4	4s, 4p, 4d, 4f	4s, 4p_x, 4p_y, 4p_z, (5 × 4d), (7 × 4f)

Only need: s has just "s"; p → p_x, p_y, p_z; d → just say "5 × 3d" (individual d-orbital names are A2). The 1s / 2s "orbital names" are given only for consistency — nobody asks the table.

Electron spin & filling rules

Sir's convention: 1 electron → draw ↑ (spin up). 2 electrons → ↑↓. "First electron is always up" — everyone does it, keeps you consistent.

Electrons have a property called spin: spin up ↑ or spin down ↓ (drawn with a full or half arrowhead).
Electrons fill orbitals in order of increasing energy — work it out from the periodic table.
An orbital holds a maximum of 2 electrons; if 2, they must have opposite spin (↑↓ — "anti-parallel").
Orbitals of equal energy (e.g. the three 2p) → each gets 1 electron first, and only then start pairing up (Hund).

Habit to build: in p (and d) boxes avoid ↑↓ at all costs until there is no empty box left. Rushing and pairing early is the #1 silly mistake in this chapter (e.g. P drawn as ↑↓ ↑ _ instead of ↑ ↑ ↑ is wrong).

3. The Periodic-Table Hack

The periodic table tells you the whole configuration. Split it into blocks named after the subshell being filled:

Rows = periods (1–7) · columns = groups (1–18). H and He both sit in the s-block.

s-block = groups 1 & 2 (also H and He). p-block = groups 13–18 (B → Ne and downwards). d-block = groups 3–12 = transition metals. f-block = the two rows below → ignore.
From now on write the groups as 13, 14, 15, 16, 17, 18 — not 3,4,5,6,7,8 (that's O-level!).
The period number gives the shell number for s & p blocks: e.g. Mg is in period 3 → 3s. Chlorine: period 3, 5th p-box → 3p⁵.

d-block ruled-block takes the number of the PREVIOUS period → in period 4 you fill 3d, period 5 → 4d.
(f-block takes the number 2 periods before.)

Careful: only the d subshell goes "one number back". e.g. As: 4s² 3d¹⁰ 4p³ — the p is 4p (period 4), not 3p. Same for Te: 5s² 4d¹⁰ 5p⁴ — 4d (previous period) but 5s and 5p.

How to use it (walk along the table, counting from H):

Start at H/He (period 1 → 1s), then period 2 (2s across groups 1-2, then 2p), period 3 (3s, 3p), period 4 (4s, then d-block 3d, then 4p) … and stop at your element. The number of boxes you crossed in the last block = electrons in that subshell.
Mg → 1s² 2s² 2p⁶ 3s² · Cl → 1s² 2s² 2p⁶ 3s² 3p⁵ · Ti → …3p⁶ 4s² 3d²

Write 4s before 3d (as the table gives it). Some teachers write 3d before 4s — also accepted, but sir's way reduces mistakes and shows the filling order (no marks lost either way).

4. Electronic Configuration (Z = 1 – 30)

Method: find the element on the table → count back from the start, filling every earlier subshell completely → last one = leftover electrons. For the arrangement (box diagram) draw a box for each orbital (s = 1 box, p = 3 joined, d = 5 joined) and fill: singly first, then pair.

Period 1 & 2 (H → Ne)

ElementElectronic configurationArrangement

1) H₁1s¹1s↑

2) He₂1s²1s↑↓

3) Li₃1s² 2s¹1s↑↓2s↑

4) Be₄1s² 2s²1s↑↓2s↑↓

5) B₅1s² 2s² 2p¹1s↑↓2s↑↓2p↑

6) C₆1s² 2s² 2p²1s↑↓2s↑↓2p↑↑

7) N₇1s² 2s² 2p³1s↑↓2s↑↓2p↑↑↑

8) O₈1s² 2s² 2p⁴1s↑↓2s↑↓2p↑↓↑↑

9) F₉1s² 2s² 2p⁵1s↑↓2s↑↓2p↑↓↑↓↑

10) Ne₁₀1s² 2s² 2p⁶1s↑↓2s↑↓2p↑↓↑↓↑↓

Notes: H → 1s¹ (↑); He → 1s² (↑↓ — opposite spins). B is the first p-electron; C, N put 1 electron in each 2p box; O is the first that must pair (2p⁴ = ↑↓ ↑ ↑); F = ↑↓ ↑↓ ↑; Ne completes 2p⁶.

Period 3 & K, Ca (Na → Ca)

ElementElectronic configurationArrangement

11) Na₁₁1s² 2s² 2p⁶ 3s¹1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑

12) Mg₁₂1s² 2s² 2p⁶ 3s²1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓

13) Al₁₃1s² 2s² 2p⁶ 3s² 3p¹1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑

14) Si₁₄1s² 2s² 2p⁶ 3s² 3p²1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↑

15) P₁₅1s² 2s² 2p⁶ 3s² 3p³1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↑↑

16) S₁₆1s² 2s² 2p⁶ 3s² 3p⁴1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↑

17) Cl₁₇1s² 2s² 2p⁶ 3s² 3p⁵1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑

18) Ar₁₈1s² 2s² 2p⁶ 3s² 3p⁶1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓

19) K₁₉1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑

20) Ca₂₀1s² 2s² 2p⁶ 3s² 3p⁶ 4s²1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓

Watch out — K and Ca: after 3p⁶ the next electrons go into 4s (4s¹ for K, 4s² for Ca), not 3d! (3d exists in shell 3, but 4s fills first — explained below.)

Transition metals (Sc → Zn)

ElementElectronic configurationArrangement

21) Sc₂₁1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑

22) Ti₂₂1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑↑

23) V₂₃1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑↑↑

24) Cr₂₄ ★1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑3d↑↑↑↑↑

25) Mn₂₅1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑↑↑↑↑

26) Fe₂₆1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑↓↑↑↑↑

27) Co₂₇1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑↓↑↓↑↑↑

28) Ni₂₈1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑↓↑↓↑↓↑↑

29) Cu₂₉ ★1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑3d↑↓↑↓↑↓↑↓↑↓

30) Zn₃₀1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↓↑↓↑↓4s↑↓3d↑↓↑↓↑↓↑↓↑↓

★ Cr and Cu are very important! Their real configurations are anomalous: Cr = …4s¹ 3d⁵ and Cu = …4s¹ 3d¹⁰ (details next). All others follow the pattern: 4s² then 3dⁿ.

Board slip: Zn's line is written "4s¹ 3d¹⁰" on the board but the box diagram (and correct answer) is 4s² 3d¹⁰ — Zn's 3d is full and 4s stays full (only Cr, Cu are anomalous).

Sir: d-block boxes — draw 5 joined boxes for d; put 1 electron in each first (Sc → Mn), only then pair (Fe onwards). Fe: 3d⁶ = ↑↓ ↑ ↑ ↑ ↑; Ni: 3d⁸ = ↑↓ ↑↓ ↑↓ ↑ ↑; Zn: five ↑↓ boxes.

Nickel & zinc (sir's quick check): count along the d-block — Ni is the 8th box → 3d⁸, Zn the 10th → 3d¹⁰. Then 4s² is still there in both.

5. Shorthand Configuration & Anomalies

Shorthand configuration: instead of writing everything, quote the noble gas BEFORE the element in square brackets, then carry on from there.

When can I use it? Only if the examiner asks ("give the shorthand / abbreviated configuration"). If the question just says "give the electronic configuration", write the whole thing even if it takes 10 lines. Call it "shorthand configuration" — that is the name.

a) Mn (25)back → first noble gas = Ar[Ar] 4s² 3d⁵

b) Cl (17)nearest noble gas behind it = Ne (not Ar — Ar is ahead!)[Ne] 3s² 3p⁵

c) Mg (12)back → Ne[Ne] 3s²

d) As (33)back → Ar[Ar] 4s² 3d¹⁰ 4p³

Board slip: the As line on the board ends "…4p³" but is written like "2p³" — the p is 4p³. Sir's extra example: Te (52) = [Kr] 5s² 4d¹⁰ 5p⁴ (4d, not 3d, because d takes the previous period).

Trick: go back from the element until you hit the first noble gas (closest behind it) — Cl: Ne, Mn / As: Ar, Te: Kr. Never the one in front!

★ Anomalous configurations of Cr and Cu

Expected (from the table): Cr = …4s² 3d⁴ and Cu = …4s² 3d⁹. Reality: one electron moves from 4s to 3d.

Expected ✗what the periodic table gives you

Cr₂₄1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴4s↑↓3d↑↑↑↑

Cu₂₉1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹4s↑↓3d↑↓↑↓↑↓↑↓↑

Actual ✓anomalous — remember these two

Cr₂₄ ★1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵4s↑3d↑↑↑↑↑

Cu₂₉ ★1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰4s↑3d↑↓↑↓↑↓↑↓↑↓

Why? A half-filled (3d⁵) or completely filled (3d¹⁰) d subshell is more stable, so one electron is transferred from 4s to 3d. It costs a little energy to move it, but the electron releases more energy once in 3d (all five d orbitals equally occupied) — a net gain in stability. 3d⁴ (one d box empty) or 3d⁹ is less stable.

Sir's advice: don't just memorise the strings — remember "Cr and Cu have a problem". Work out the normal configuration from the table (3d⁴ / 3d⁹), then see it isn't stable → pick one electron from the s and put it into d. The examiner hides these two in a longer list (e.g. a "Sc → Zn" question) — keep eyes open every time they appear!

Also for unpaired-electron counting: normally you only inspect the last subshell — but for Cr and Cu the 4s¹ is also unpaired (Cr: 6 unpaired, Cu: 1 unpaired).

Why is 4s filled before 3d?

4s is LOWER in energy than 3d (when the atom is being built) — it sits closer to the nucleus, so it is easier to put an electron in it ⇒ 4s fills first (K, Ca), then 3d (Sc → Zn).
Bohr's "fill shell by shell" (all of shell 3 before shell 4) would predict 3d first — this is one thing the quantum model explains and Bohr cannot.
Energy order of filling: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p …
Twist: once 4s is filled it goes up above 3d (electrons gained → higher energy). So when transition metals lose electrons, 4s goes first (see ions).

6. Configuration on an Energy Graph

Examiners sometimes ask "show the electrons and relative energies". Students go wrong here because they never practised it. Don't stack the boxes side by side — draw each subshell higher than the previous one (energy on the y-axis), in the same order you wrote the configuration (that already is the energy order, so no extra memorising).

a) Na: 1s² 2s² 2p⁶ 3s¹

b) Ar: 1s² 2s² 2p⁶ 3s² 3p⁶

1s lowest → then 2s → 2p (3 equal-energy boxes side by side) → 3s → 3p …
Equal-energy boxes (the three p's, five d's) are drawn at the same height; fill singly first, then pair.

Si (Z = 14): 3p² = ↑ ↑ _ (Don't pair them!)

Sir's tougher example — yttrium (Z = 39):

Configuration from the table: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹ — stop at 4d¹ (d = previous period). Draw the staircase in the same order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d (one ↑ in the 4d). "Trust the process — you'll never need to remember which is higher in energy."

7. Electronic Configuration of Ions

Rule: write the neutral atom first, then remove (positive ion) or add (negative ion) electrons as shown by the charge. Electrons are removed from the highest-energy (outermost) subshell, added to the next available spot.

a) Na⁺ (lose 1 e⁻)Na = 1s² 2s² 2p⁶ 3s¹Na⁺ = 1s² 2s² 2p⁶

b) O²⁻ (gain 2 e⁻)O = 1s² 2s² 2p⁴O²⁻ = 1s² 2s² 2p⁶

c) Ca²⁺ (lose 2 e⁻)Ca = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²Ca²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶

d) Br⁻ (gain 1 e⁻)Br = …4s² 3d¹⁰ 4p⁵Br⁻ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Board slip: in (b) the board labels the answer "O:" — it should read O²⁻.

Note: Na⁺, O²⁻, Ca²⁺, Br⁻ all end at a noble-gas configuration (Ne, Ne, Ar, Kr). Ca²⁺ loses the whole 4s; Br⁻ fills 4p to 4p⁶.

★ Transition metals (d-block) are different! When they form positive ions, electrons are removed from 4s FIRST, then 3d — even though 3d was written after 4s and 4s was filled first. Reason: once 4s has electrons in it, it rises above 3d in energy ("4s is lower when empty, higher when full"), so 4s electrons are the highest-energy ones and leave first.

Using shorthand configuration:

a) Ti²⁺Ti = [Ar] 4s² 3d²remove 2 from 4s →Ti²⁺ = [Ar] 3d²

b) Fe²⁺Fe = [Ar] 4s² 3d⁶remove 2 from 4s →Fe²⁺ = [Ar] 3d⁶

c) Fe³⁺Fe = [Ar] 4s² 3d⁶4s² gone, 1 more from 3d →Fe³⁺ = [Ar] 3d⁵ (half-filled!)

d) Cu⁺ (start from anomalous Cu!)Cu = [Ar] 4s¹ 3d¹⁰remove the 4s electron →Cu⁺ = [Ar] 3d¹⁰

e) Cu²⁺Cu = [Ar] 4s¹ 3d¹⁰4s¹ gone, 1 more from 3d →Cu²⁺ = [Ar] 3d⁹

Board slip: in part (c) the board labels the answer "Fe²⁺"; it is the Fe³⁺ result, [Ar] 3d⁵.

Common mistake: starting Cu or Cr ions from the expected configuration (4s² 3d⁹). Always begin from the anomalous Cu = 4s¹ 3d¹⁰ / Cr = 4s¹ 3d⁵ — "copper, chromium, copper, chromium — when you see them, be alert!" e.g. Ni²⁺ = [Ar] 3d⁸ (4s² removed; not 3d⁶ 4s²).

p-block ions: remove from the outermost, highest-energy subshell first (the last one written — p before s). Ge → Ge⁴⁺: first two e⁻ from 4p, then the next two from 4s, (3d stays because 3d is now lower than 4s) → order 4p, 4p, 4s, 4s (see practice Q11).

8. Shapes of s and p Orbitals

s orbital: a 3-D sphere centred on the nucleus (like a ball extending along x, y and z). 2s and 3s are also spheres — the higher the energy, the bigger the sphere (further from the nucleus).

1s < 2s < 3s in energy ⇒ each drawn a larger sphere (all still spheres)

p orbital: a dumbbell / "infinity sign" (figure of eight) — two lobes, centred on an axis that its subscript names: 2p_x on the x-axis, 2p_y on the y-axis, 2p_z on the z-axis. All three 2p are identical in shape/size ⇒ equal energy.

2p_x = 2p_y = 2p_z in energy; 3p_z = same shape as 2p_z but larger ⇒ higher energy

Summary (sir): s = sphere; want higher energy → just draw it bigger. p = dumbbell along the axis in its subscript; higher energy → bigger. Axes can be drawn any orientation — just centre the lobes on the right one. d-orbital shapes are A2, not needed now.

Exam link: the "infinity sign" shown in a question = a p orbital (see practice Q1(b)(iii)). "State the number of electrons in all orbitals of this type in F" → F is 1s² 2s² 2p⁵ ⇒ 5 electrons in p orbitals.

9. Past-Paper Practice — Structure & Configuration

Sir: past-paper questions are the real revision — keep your solved topicals and re-read them while revising (the notes cannot hold everything).

Q1 (a)structured · beams in an electric field

Diagram: separate beams of electrons and protons from one source between two vertical plates. (i) Give the charge on each plate. (ii) Draw the path of a beam of neutrons.

electrons bend to the + plate, protons to the − plate

(i) electrons (−) are attracted to the left plate ⇒ positive; protons (+) to the right plate ⇒ negative [1]

(ii) neutrons: straight, undeflected line from the source (no charge) [1]

Q1 (b)(i)structured · number of occupied orbitals in Kr (Z = 36)

Kr = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶s: 1s, 2s, 3s, 4s → 4 (1 orbital each)p: 2p, 3p, 4p → 3 subshells × 3 orbitals = 9d: 3d → 1 subshell × 5 orbitals = 5

Trap: count orbitals, not subshells! A p subshell = 3 orbitals, d = 5.

type of orbital	s	p	d
number of orbitals	4	9	5

Q1 (b)(ii)structured · diagram for Si (Z = 14) [2]

Complete the energy-level diagram (2 marks): Si = 1s² 2s² 2p⁶ 3s² 3p². Draw boxes in rising order, 4s box left empty, and the two 3p electrons unpaired (↑ ↑ _).

1s↑↓2s↑↓2p↑↓↑↓↑↓3s↑↓3p↑↑4s

Common mistake: pairing the 3p electrons (↑↓ _ _) — equal-energy boxes fill singly first.

Q1 (b)(iii)structured · electrons in "infinity-sign" orbitals of F

The orbital drawn is a figure-of-eight = p orbital. F (Z = 9) = 1s² 2s² 2p⁵ ⇒ total electrons in p orbitals = 5 electrons.

Q1 (c)structured · hydron ions ¹H⁺, ²H⁺, ³H⁺ [1]

State, in terms of subatomic particles in the nucleus, what is the same and what differs.

same: proton number | different: neutron number (mass number)

Sir: all three have no electrons (H⁺ = a bare proton) — true, but the question says "in the nucleus", so don't mention electrons or you'd be wrong for that question.

Q1 (iv)structured · 3 marks · first IE of Sc → Cu very similar

The first ionisation energies of Sc (21) to Cu (29) are very similar. In all of them a 4s electron is lost in the first ionisation. Suggest why the values are very similar.

Use the three factors: nuclear charge, shielding, atomic radius. Reason by elimination (sir's working): nuclear charge increases (21 → 29). If shielding decreased or stayed constant, the pull on the 4s electron (and so the IE) would rise a lot — but the IEs are "very similar" ⇒ shielding must also increase, cancelling the extra nuclear charge.

Why does shielding increase? The 4s² is already filled; each new electron goes into the 3d, which is lower in energy / inner ⇒ the 3d subshell shields the 4s electrons (inner-subshell shielding, not an extra inner shell).

Model answerNuclear charge increases and shielding also increases as more electrons are added to the 3d subshell, which shields the 4s electrons. These cancel each other out, so the atomic radius remains roughly constant and the first ionisation energies are very similar.

Q11 (MCQ)order of electrons lost forming Ge⁴⁺

Ge = [Ar] 3d¹⁰ 4s² 4p². Which order are the electrons lost?

A ✓ 4p, 4p, 4s, 4sB ✗ 4p, 4p, 3d, 3d — 3d is lower than 4s here, so 4s goes firstC ✗ 4s, 4s, 4p, 4p — 4p is highest energy (written last) ⇒ it leaves firstD ✗ 3d first — never

Write it in sir's order: [Ar] 4s² 3d¹⁰ 4p² → remove from the end (4p) first, then 4s (never 3d before 4s). Answer: A

Q3 (MCQ)atomic orbitals occupied in phosphorus

P = 1s² 2s² 2p⁶ 3s² 3p³. Options: A 1p 2s 2p ✗ (1p doesn't exist) · B 2s 2p 2d ✗ (no 2d) · C 2s 2p 3s ✓ · D 2p 3s 3d ✗ (3d exists but is empty in P).

Answer: C

Q31 (MCQ)common features of ³³₁₅P³⁻ and ³⁷₁₉K⁺

P³⁻: protons 15, electrons = 15 + 3 = 18, neutrons = 33 − 15 = 18K⁺: protons 19, electrons = 19 − 1 = 18, neutrons = 37 − 19 = 18

1 ✓ same electrons · 2 ✓ same neutrons · 3 ✗ protons differ (15 vs 19). Answer: 1 and 2 only.

Q2 (MCQ)valence-shell and unpaired electrons in S (1s² 2s² 2p⁶ 3s² 3p⁴)

Valence shell = shell 3 → 3s² + 3p⁴ = 6. 3p⁴ = ↑↓ ↑ ↑ ⇒ 2 unpaired (1s², 2s², 2p⁶ are full).

A ✗ 2, 1B ✗ 4, 2C ✗ 6, 0D ✓ 6, 2

Only Cr and Cu need you to look at 4s as well. For everything else, just inspect the last subshell.

Q1 (MCQ)particle with equal protons & neutrons and structure 1s² 2s² 2p⁶ 3s² 3p⁶ (18 e⁻)

³⁹₁₈Ar: p = 18, n = 39 − 18 = 21 → not equal ✗⁴⁰₂₀Ca²⁺: p = 20, n = 20 ✓; e⁻ = 20 − 2 = 18 ✓¹⁶₈O²⁻: p = 8, n = 8 ✓ but e⁻ = 10 = 1s² 2s² 2p⁶ ✗³²₁₆S: p = n = 16 ✓ but it's a neutral atom = 3p⁴, not 3p⁶ ✗

Answer: B (Ca²⁺)

Sir's "katam-pooti method": when unsure, cross out wrong options — 25% chance becomes 33% after one elimination, 50% after two.

Q8 (MCQ)electronic configuration of Ni²⁺

Ni = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸lose 2 e⁻ from 4s firstNi²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸

A ✗ …3d⁶ 4s²B ✗ …3d⁷ 4s¹C ✗ …3d¹⁰ 4s²D ✓ …3d⁸ (4s emptied)

Board discrepancy: the circle on the printed board sits on A, but the working (and sir's spoken answer) gives D — A has the wrong number of d electrons anyway.

Q2 (MCQ)pair where BOTH species have only ONE unpaired p electron

A: Ar⁺ = 3p⁵ (↑↓ ↑↓ ↑ → 1 unpaired) ✓ but C⁻ = 2p³ (↑ ↑ ↑ → 3) ✗B: B = 2p¹ (1) ✓ but Ti⁺ = 4s¹ 3d² — no unpaired p electrons at all ✗C: F = 2p⁵ (1) ✓ and Ga = 4p¹ (1) ✓ → both ✓D: Se⁻ = 4p⁵ (1) ✓ but Si⁻ = 3p³ (3) ✗

Answer: C (F and Ga)

Tricky one: Ti⁺ — a transition-metal ion loses from 4s first (leaving 4s¹ 3d²), so it has no p electrons.

10. Ionisation Energies

Now the quantum model gets linked to ionisation energies (IE): learn the trends and be able to explain each one — they are the evidence that Bohr's model was not enough. IEs can be 1st, 2nd, 3rd … as many as you like.

First ionisation energy — definition (learn word-for-word)The minimum amount of energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of 1+ charged gaseous ions.

Equations (examples — first IE, always one electron off a gaseous atom):

Na(g) → Na⁺(g) + e⁻Cl(g) → Cl⁺(g) + e⁻

O(g) → O⁺(g) + e⁻Ar(g) → Ar⁺(g) + e⁻

★ Marks are lost for: ① missing state symbols (g) — everything gaseous (Na is a solid normally, but for IE it's Na(g)) ② using molecules — write Cl(g), not Cl₂ (a single gaseous atom; ignore how the element normally exists) ③ writing the ion Cl⁻ — IE always makes a positive ion (you remove an electron), even for chlorine that "normally" forms Cl⁻ ④ "1 mole" missing (moles of electrons, moles of atoms, moles of ions). (Electrons have no state symbol.)

Second ionisation energy: the minimum energy to remove 1 mole of electrons from 1 mole of 1+ gaseous ions to form 1 mole of 2+ gaseous ions. (Equation shows the definition — don't cram it.)

Na⁺(g) → Na²⁺(g) + e⁻Cl⁺(g) → Cl²⁺(g) + e⁻

O⁺(g) → O²⁺(g) + e⁻Ar⁺(g) → Ar²⁺(g) + e⁻

…and so on for other ionisation energies. e.g. 4th IE of Ar: Ar³⁺(g) → Ar⁴⁺(g) + e⁻ · 6th IE of Mg: Mg⁵⁺(g) → Mg⁶⁺(g) + e⁻ (the nth IE starts from the (n−1)+ ion.)

Why is IE always endothermic (energy required)? The positive nucleus attracts the negative electrons — that's why they stay. To pull one away (like separating two magnets) you must put energy in. It's an unusual, forced process (nobody normally ionises atoms like this) and the energies are huge.

11. Successive Ionisation Energies of One Element

Remove electrons one after another from the same element (1st IE, 2nd IE, 3rd, …). Example: sodium (Bohr structure 2 · 8 · 1) — plot lg IE against number of electrons removed:

Na: 1 | 8 | 2 pattern = shells 3, 2, 1 ⇒ evidence for Bohr's shells

Huge jump ⇒ change of shell

After the 1st e⁻ (3rd shell) is gone, you remove electrons from an inner shell, closer to the nucleus ⇒ attracted much more strongly ⇒ far more energy. Another huge jump when you reach the 1st shell (electron 10). So the jump tells you: electron being removed from a shell closer to the nucleus.

3rd shell electron goes first (easy) → then 8 in shell 2 → then 2 in shell 1

Q. Why a slight increase within the same shell?

① Charge: as electrons are removed the ion becomes more positive: there are more protons than electrons (e.g. Na⁺ 11p / 10e⁻; Na²⁺ 11p / 9e⁻), so the attraction on each remaining electron is stronger. (Write it as "more protons per electron".)
② Repulsion: fewer electrons ⇒ less inter-electron repulsion ⇒ the electrons move closer to the nucleus (radius/shell shrinks) ⇒ the attraction is stronger.
Hence slightly more energy is needed to remove the next electron.

Note: successive IE evidence supports Bohr's shells (2 · 8 · 1), not the quantum model's subshells — later trends (the dips) are what prove the quantum model.

Finding the group from successive IEs

A huge increase after the nth electron means the first n electrons were valence electrons ⇒ the element has n valence electrons ⇒ it is in group n (for n = 1, 2) or group 10 + n (n = 3 – 8: groups 13 – 18). Use A-level group numbers (3 valence e⁻ = group 13, not 3!).

Say it fully: write "third and fourth ionisation energy" — do not abbreviate (3rd/4th, IE, "e⁻"…) — and state the conclusion "meaning A has 3 valence electrons".

Qstructured · group from IEs

(a) The first five IEs of element Y (kJ mol⁻¹): 578 · 1817 · 2745 · 3538 · 11 578. To which group does Y belong?

Y: jump between 4th and 5th

Element A (p-block): jump 3rd → 4th

Y: huge increase between the 4th and 5th IE ⇒ 4 valence electrons ⇒ Group 14. (The board note says "between 3rd and 4th" — a slip: the big jump 3538 → 11 578 is 4th → 5th.)

(b) Element A is in the p block; the graph of the first ten successive IEs jumps after the 3rd electron ⇒ group 13. Explanation: there is a huge increase in ionisation energy from the third electron to the fourth electron, meaning A has 3 valence electrons. [2]

12. Trends in First Ionisation Energy

First IE against atomic number (H → Kr; approximate values) — know this general shape

★ Always discuss THREE things in any explanation about IE / radius / (later: electronegativity, period 3 …): ① nuclear charge ② shielding ③ atomic radius. Miss one ⇒ you lose marks. Sir will use these three "again and again for the whole course".

Meaning: Nuclear charge = number of protons (charge of nucleus) · Atomic radius = radius of the atom · Shielding = see below.

Effective charge = protons − inner-shell electrons (only the inner shells shield; valence electrons don't shield each other).

Shielding: the electrons in the inner shells shield the valence electrons from experiencing the full nuclear charge (they repel the outer electron). Each inner-shell electron approximately cancels the effect of 1 proton. (It's not that charge disappears — the outer electron just feels less than the full nucleus.)

Li: 3p, 2 inner e⁻ → valence e⁻ feels 3 − 2 = +1

Be: 4p, 2 inner e⁻ → valence e⁻ feel 4 − 2 = +2

Na: 11 − 10 inner e⁻ = +1 (like Li, far away)

① Across a period: first IE generally increases

Explanation (write in this order)1) The nuclear charge increases across the period (more protons).
2) Shielding by inner-shell electrons remains (approximately) constant — the inner shell is the same, valence electrons can't shield each other.
3) So the valence electrons experience a stronger effective charge ⇒ the atomic radius decreases, the force of attraction between nucleus and valence electrons increases ⇒ more energy is required to remove an electron.

Effective charge Li → Ne: +1, +2, +3, +4, +5 … (3−2, 4−2, 5−2 …) — inner shell always 2 e⁻, so shielding can't cancel the growth. All Period-2 elements have the same 2 inner electrons; only nuclear charge rises.

Sir's "why not…": Bilal asked, "Won't radius be constant?" — no: radius decreases gradually, because the growing effective charge pulls valence electrons closer.

② The two dips inside every period — anomalies

These need the quantum model (subshells, paired electrons) — Bohr can't explain them, which is exactly why the quantum model was developed. Examiners expect you to know them — memorise the reasons.

Dip 1 — group 2 → group 13Be → B, Mg → Al (also Ca → Ga, Sr → In, Ba → Tl …)

Be1s² 2s²1s↑↓2s↑↓

B1s² 2s² 2p¹1s↑↓2s↑↓2p↑

In Be the electron is removed from the 2s, in B from the 2p. 2s is a lower-energy orbital than 2p (closer to the nucleus, held more tightly), so removing an electron from it is harder ⇒ B has the lower IE.

Mg1s² 2s² 2p⁶ 3s²

Al1s² 2s² 2p⁶ 3s² 3p¹

Same argument: 3s is lower in energy than 3p ⇒ harder to remove from Mg's 3s than Al's 3p.

Rule: every group 2 → group 13 step dips (group 2 ends s², group 13 starts p¹ — an outer p electron is higher in energy / further from nucleus). Don't memorise the answer — change 2s/2p to 3s/3p etc. for the element asked.

Dip 2 — group 15 → group 16N → O, P → S

N1s² 2s² 2p³1s↑↓2s↑↓2p↑↑↑

O1s² 2s² 2p⁴1s↑↓2s↑↓2p↑↓↑↑

N (2p³) has 3 unpaired electrons — none are repelled by another in the same box. O (2p⁴) has one paired electron: two electrons in the same small box already repel each other, so it is easier to remove a paired electron than an unpaired one ⇒ O has lower IE.

P 3p³ (3 unpaired) → S 3p⁴ (2 unpaired + 1 pair): same explanation. Group 15 always ends p³ (half-filled), group 16 always p⁴.

Sir's analogy: two people in one small room already dislike sharing it — one leaves as soon as you give him a little push (energy). Alone in their own rooms, everyone is happy and you must pay more to move one out.

Exam wording: Mg → Al: "the electron in magnesium is removed from a 3s orbital, whereas in aluminium it is removed from a 3p orbital. 3p is higher in energy than 3s so less energy is required (it is further from the nucleus)." P → S: "in phosphorus the electron removed is unpaired; in sulfur it is paired, and paired electrons are easier to remove because of electron–electron repulsion."

③ Down a group: first IE decreases

e.g. He > Ne > Ar > Kr > Xe (group 18); Li > Na > K > Rb (group 1). No anomalies — a smooth downward trend.

Explanation (three points)1) Nuclear charge increases (more protons down the group) — and 2) shielding also increases (more inner shells / more inner electrons) ⇒ these two cancel each other out (effective charge constant, e.g. Li +1, Na +1 [11 − 10]).
3) Atomic radius increases (more shells) ⇒ valence electron further from nucleus ⇒ pulled less strongly ⇒ easier to remove ⇒ first IE decreases.

Effective charge check: Li 3 − 2 = +1; Na 11 − 10 = +1 — same. So the deciding factor is atomic radius (3rd shell vs 2nd shell).

④ Noble gas → next group 1: big drop

e.g. He → Li, Ne → Na: 1) nuclear charge increases (2+ → 3+, 10+ → 11+) · 2) shielding also increases (a new shell: neon shell 2 → sodium shell 3 ⇒ more inner-shell electrons) — they cancel out · 3) atomic radius increases (more shells) ⇒ the valence electron is pulled less strongly ⇒ easier to remove ⇒ large dip.

★ Conclusion (write this): the variation of ionisation energies across the periodic table provides evidence for the quantum model (subshells and orbital pairing explain the dips). Successive IEs (huge jumps) supported the Bohr model of shells; the group 2/13 and 15/16 dips need subshells ⇒ the quantum model.

Sketching the period-3 graph (sir wants the "general shape")

General shape of first IE across period 3 (approximate)

Ne (top) → huge dip to Na → ↑ Mg → dip at Al (group 13; still above Na!) → ↑ Si → ↑ P (group 15) → dip at S (group 16; not below Si / group 14) → ↑ Cl → Ar (highest).
Your sketch needn't be perfect, but the relative positions matter (Al > Na, S > Si, Ar highest). Exam questions on this ask about period 3.

Atomic radius across period 3 decreases — draw a reciprocal curve, never a straight line

Why not a straight line? A straight line can be extended until radius = 0 at some atomic number — impossible. The curve levels off horizontally on the right and (like a reciprocal) never reaches zero.

13. Past-Paper Practice — Ionisation Energy

Sir: "The periodic table is your best friend, and you must have the general trend in your head — many MCQs are solved from that alone."

MCQRn, Fr, Ra (consecutive proton numbers): order of first IE, least → most endothermic

Rn (noble gas, top) → Fr (group 1, huge dip) → Ra (group 2, up)

Trap: "consecutive proton numbers" ≠ "across a period"! Check the table: Rn is group 18 (end of period 6) → Fr (group 1, period 7) → Ra (group 2).

Least endothermic = smallest IE = Fr; then Ra; largest = Rn (noble gas).

Answer: A (Fr, Ra, Rn)

MCQwhich equation is the second ionisation energy of X?

A ✗ X(g) → X²⁺(g) + 2e⁻ — two electrons from the atom (2 steps in one)B ✓ X⁺(g) → X²⁺(g) + e⁻ — 1 electron, from a 1+ gaseous ionC ✗ X(g) + 2e⁻ → X²⁻(g) — gaining electronsD ✗ X⁻(g) + e⁻ → X²⁻(g) — gaining

Check: 1 mole of electrons removed, from a 1+ ion, all in gaseous state.

MCQfrom which particle is removal of an electron the MOST difficult? Cl⁻(g), F⁻(g), K⁺(g), Na⁺(g)

Negative species already repel their outer electron ⇒ easy to remove (Cl⁻ ✗ and F⁻ ✗). A positive ion holds its electrons tightly ⇒ hardest: K⁺ vs Na⁺ (both 1+). Na⁺ and K⁺ have the same charge, so compare radius: Na⁺ (2 filled shells) is smaller than K⁺ (3 filled shells) ⇒ attraction to the nucleus is stronger ⇒ hardest.

(Shielding and nuclear charge both increase down the group and cancel — so use atomic radius as the deciding factor.) Answer: D (Na⁺)

MCQ"Three successive elements" — pattern of first IE in the diagram: what could the first element be?

Pattern: 1st element mid-height, 2nd element highest, 3rd element lowest of all.

A big drop after the highest point ⇒ 2nd element = noble gas (group 18), 3rd = group 1, so the 1st = group 17. Group 17 among the options C, N, F, Na ⇒ F (C).

Other places a dip occurs are ruled out: group 2 → 13 dip (can't fall below group 1) and group 15 → 16 dip (can't fall below group 14) don't match this "highest then lowest" picture.

Q34 (MCQ)graph of first IE of successive elements A … T — which statements are correct?

Peaks = noble gases (huge dip follows each). The peaks are B, J, R.

1 ✓ B, J, R are in Group 0 / 18 (old papers say group 0 — either is fine).2 ✓ D and L: two electrons in the outer shell — the element right after a noble gas is group 1 (C, K), the next (D, L) is group 2.3 ✓ G and O: count back from the peak J (18): I = 17, H = 16, G = 15; count on from K (group 1): L = 2, M = 13, N = 14, O = 15. Group 15 ends in p³ = half-filled p orbitals ✓.

Answer: 1, 2 and 3 are all correct.

Q13 (MCQ)which marked element is an alkali metal? (P, Q, R, S)

Alkali metal = group 1 = the huge dip right after a noble-gas peak. R is the peak (noble gas) ⇒ S (next, huge drop) is group 1. P, Q are before the peak (rising towards group 18), so they are not group 1.

Answer: D (element S)

Fig 1.2 / 1.3structured · six successive elements A–F (12 marks)

A = noble gas, B = group 1, then C (G2), D (G13), E, F ↑

(i) Define first IE [2] → "The minimum amount of energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of 1+ charged gaseous ions." (gaseous atoms, not Cl₂.)

(ii) Why is B's first IE much lower than A's? [3] — A = noble gas (end of a period), B = group 1 (start of next):

The electron in B is removed from a higher energy level (new shell). From A to B the nuclear charge increases and so does the shielding, so they cancel each other out. The atomic radius increases (extra shell), so the nucleus pulls the valence electron less strongly ⇒ less energy is needed to remove an electron.

(c)(i) Sketch atomic radius across period 3 [1] → decreasing reciprocal curve (see previous page; not a straight line).

(ii) Explain [3] — the nuclear charge increases while the shielding by inner electrons remains constant across the period. Hence the valence electrons experience a greater effective charge and the atomic radius decreases.

Sir's writing tip: in the real paper keep the answer on the lines (one extra line at most, 3–4 words); write close to the line because scans cut off the margins. If you must cross something out, ask for an extra sheet or neatly cross it out and write above it.

Period 3 graph questionstructured · Na → Ar

(a) Define first IE — as above.

(b) Equation with state symbols for the second IE of sodium:

Na⁺(g) → Na²⁺(g) + e⁻

Common error: putting the electron on the wrong side or forgetting the 1+ ion; e⁻ has no state symbol.

(c) Graph: mark crosses for Na, P, S on the given grid; join. Na lower than Mg (group 1 → 2 up); Al dips but not below Na; P higher than Si (group 14 → 15 up); S dips below P but not below Si; then Cl, Ar. Examiner wants approximate positions only.

(ii) & (iii)explanations · 3 marks each · Na → Ar and Al vs Mg

(ii) Why first IE generally increases from Na to Ar

Nuclear charge increases but shielding by inner-shell electrons remains constant. Therefore the valence electrons feel a stronger pull, the atomic radius decreases and more energy is required to remove an electron.

(iii) Why is first IE of Al less than Mg? (Mg: 1s² 2s² 2p⁶ 3s²; Al: 1s² 2s² 2p⁶ 3s² 3p¹)

In Mg the electron is removed from a 3s orbital, which is lower in energy and closer to the nucleus than the 3p orbital from which the electron of Al is removed. So more energy is required to remove an electron from Mg.

(d)order S⁺, S, S⁻ by increasing first IE (lowest first)

S⁻ (3p⁵): radius ↑ , has a paired electron — easy (already repelled)S (3p⁴ = ↑↓ ↑ ↑): paired electron — easy-ishS⁺ (3p³ = ↑ ↑ ↑): radius ↓, all unpaired — hard

Lowest → highest: S⁻ < S < S⁺

Reasoning: positive species pull electrons harder (fewer electrons, smaller radius, more protons per electron); negative species repel. S⁺ also has no paired electrons (harder), S⁻ has the biggest radius and repulsion (easiest).

14. Exam Technique & Study Advice

✓ Habits that save marks (from sir)

Have the periodic table open — every config, ion, IE question starts there.
Write full words ("third and fourth ionisation energy", "electrons") — no abbreviations/short forms.
Use A-level group numbers (13 not 3; 14 not 4; 18 or 0).
IE / equation questions: state symbols, gaseous atoms (Cl not Cl₂), one electron, always positive ion.
Explaining a trend? Nuclear charge → shielding → atomic radius → conclusion (all three, in that order).
Config: write like the table (4s before 3d), fill p/d boxes singly first, watch Cr & Cu, remember d takes previous period, and TM ions lose 4s first.
MCQ: cross out wrong options one by one — every elimination raises your odds.
Keep answers within the lines; write close to the ruled line (scan cuts margin); cross out neatly, write above if no extra sheet.
Past papers are the key to A levels: keep solved topicals, re-read them when revising, then attempt fresh ones — you learn what the examiner expects.

15. Full Chapter Cheat-Sheet

✓ Quick Recap — Structure & configuration

p⁺ (+1, 1) · n (0, 1) · e⁻ (−1, 1/1840). Field: e⁻ → + plate (most), p⁺ → − plate, n straight; more charge/mass ratio ⇒ more deflection.
Bohr = shells only (2, 8, 1…). Quantum = shells (n) ⊃ subshells (n of them: s p d f) ⊃ orbitals (s 1, p 3, d 5, f 7) ⊃ ≤ 2 e⁻ (opposite spins).
Fill lowest energy first; equal-energy orbitals singly first (↑ ↑ ↑ then pair). Periodic-table blocks: s (1, 2 + H, He), p (13–18), d (3–12, uses n − 1).
Cr = [Ar] 4s¹ 3d⁵ · Cu = [Ar] 4s¹ 3d¹⁰ (half/full d is more stable). 4s is lower than 3d when filling.
Shorthand = nearest noble gas behind the element in [ ], only when asked.
Ions: remove from the outermost/highest-energy (p, then s); transition metals: 4s first, then 3d. Negative ions: add into next spot. Ge⁴⁺: 4p, 4p, 4s, 4s.
s orbital = sphere (bigger = higher energy); p = dumbbell on x, y or z axis (2p_x = 2p_y = 2p_z energy; 3p larger).

✓ Quick Recap — Ionisation energy

1st IE: min. energy to remove 1 mol e⁻ from 1 mol gaseous atoms → 1 mol 1+ gaseous ions. Equation: X(g) → X⁺(g) + e⁻. 2nd IE: X⁺(g) → X²⁺(g) + e⁻. Always endothermic.
Successive IEs: huge jump = the next electron comes from a shell closer to the nucleus → number of electrons removed before the jump = valence electrons = group. Slight rise inside a shell: more p per e⁻ + less repulsion → smaller radius.
Across a period ↑: nuclear charge ↑, shielding constant, radius ↓, stronger pull. Down a group ↓: nuclear charge ↑ & shielding ↑ cancel, radius ↑, weaker pull. Noble gas → group 1: big drop (new shell).
Dips: group 2 → 13 (s → p: p higher in energy, easier) · group 15 → 16 (paired electron repulsion: paired e⁻ easier to remove).
Shape of period-3 graph: Na low, Mg ↑, Al ↓ (above Na), Si ↑, P ↑, S ↓ (above Si), Cl ↑, Ar highest. Atomic radius across period: decreasing reciprocal curve.
Negative species easier to ionise (repulsion, larger radius); positive harder. Same charge: smaller radius = harder.
IE variation across the table = evidence for the quantum model (Bohr can't explain the dips).