Before you can read the parent note, you must own every symbol it throws at you. Below, each idea is built from nothing: plain words → the picture → why the topic needs it. Read top to bottom; each rung stands on the one below.
Figure s01 — A boron atom in cross-section: a black nucleus at the centre, a full inner shell of 2 black electrons, and an outer shell holding exactly 3 red valence electrons. The red count "3" is the seed of the whole chapter.
Look at the red electrons in the outer ring of the figure — those are the valence electrons. Boron has exactly 3 of them, drawn in red. That number "3" is the seed of the entire chapter.
The label 2pz (which the parent uses constantly) reads: "the dumbbell pocket pointing along z, in the 2nd shell."
Figure s02 — Boron's ground-state valence orbitals as an energy diagram: the round 2s holds a pair of black electrons; the three 2p dumbbells sit one energy level up, with one red electron in 2px and 2py, 2pzempty (by Hund and Aufbau). Five electrons total: 1s22s22p1.
Recall What is boron's ground-state configuration?
Five electrons filled by Aufbau then Hund. ::: 1s22s22p1 — one lone electron in one 2p, the other two 2p empty.
Figure s05 — Left: the pure atomic set (2s + 2px + 2py mixed, 2pz untouched). Right: three black sp2 hybrid arms at 120° in a plane (each ready to bond, e.g. to F in BF3), with the pure 2pz drawn as a red dumbbell sticking straight up, empty.
Recall After
sp2 hybridisation, which orbital is empty?
Three sp2 arms bond; one orbital is untouched. ::: The pure 2pz, perpendicular to the plane, is empty.
Why the topic needs it: this single limitation ("shell 2 has no d") is what makes boron differ from Al and underlies its whole anomaly in the parent note.
This is why the parent writes :NH3 — those two dots are nitrogen's lone pair, ready to donate. See Lewis Acids and Bases for the full donor/acceptor language.
The parent's number ΔiH1+ΔiH2+ΔiH3≈6887 kJ/mol just says: "stripping 3 electrons off boron costs a fortune." That is why no free B3+ exists — see Ionisation Enthalpy Trends.
Figure s03 — A 2×3 slice of the periodic table (B, C, N on top; Al, Si, P below). A red arrow runs diagonally from B down to Si. Black arrows show the two competing trends: electronegativity rises to the right and falls going down — they cancel along the red diagonal.
Figure s04 — Boron (left) shows its empty 2pz as a red outline dumbbell; a halogen X (right) shows a filled black lone-pair dumbbell. A curved red arrow carries the halogen's lone pair sideways into boron's empty pocket — this is back-bonding.
The red arrow shows a halogen's lone pair donating back into boron. Why the topic needs this: it is the only reason the acidity order BF3<BCl3<BBr3<BI3 runs "backwards." Good sideways overlap needs matching orbital sizes — F's small 2p matches B's small 2p, so back-bonding is strong and B feels satisfied. Full treatment in Back-bonding (pπ–pπ) and Boron Trihalides.
Why the topic needs it: it names boron's defining behaviour. Everything in §3 of the parent is just "empty pocket meets lone pair." Details in Lewis Acids and Bases.
Figure s06 — Two borons with an empty hybrid arm each pointing inward, and a bridging hydrogen's round 1s orbital between and slightly below them. The three orbitals merge into one red curved (banana) cloud holding one electron pair — a 3-centre 2-electron B−H−B bond.
Why the topic needs it: borazine has a six-membered ring where each nitrogen's lone pair (2 electrons × 3 N = 6π electrons) pours into boron's empty 2pz, hitting the same 4n+2=6 count as benzene — so it mimics aromatic benzene ("inorganic benzene"). But because N is more electronegative, the cloud is lopsided (Nδ−/Bδ+), making it more reactive. Compare with Benzene Aromaticity.
Read it downward: the single fact "boron is small, and after bonding has an empty pocket" (top) fans out into every headline result (bottom row) of the parent note.
