Diagonal relationship — Li - Mg, Be - Al, B - Si: Visual walkthrough

We assume you know only this: atoms are tiny balls with a positive centre (the nucleus) and a fuzzy cloud of negative electrons around them. Everything else we define as we go.

Step 1 — Two atoms, one tug-of-war

WHAT. Picture a small positive ball (a cation — an atom that has lost electrons, so it is positive) sitting next to a bigger, softer negative ball (an anion — an atom that gained electrons, so it is negative). The positive ball pulls on the outer electrons of the negative ball.

WHY. All of diagonal-relationship chemistry comes down to how hard that pull is. A gentle pull → the two ions stay separate, ionic and salt-like. A fierce pull → the positive ball drags the negative cloud partly onto itself, sharing electrons — that is covalent behaviour. So "how hard is the pull?" is the question we must quantify.

PICTURE. The burnt-orange arrows are the cation tugging the plum electron cloud out of shape.

Figure — Diagonal relationship — Li - Mg, Be - Al, B - Si

Step 2 — Turn the pull into a number: the ionic potential

WHAT. We build a single number that captures the pull. Two things obviously matter:

the charge of the cation, written $z$ (a $+ 2$ ball pulls harder than a $+ 1$ ball),
the size of the cation, its radius $r$ (a smaller ball lets the anion get closer, so the pull at the surface is fiercer).

Combine them:

WHY this exact form and not another? The electric pull between charges follows Coulomb's idea: force grows with charge and weakens with distance. Charge belongs on top, distance on the bottom — that is literally what " $\frac{z}{r}$ " says. We use a ratio (not $z - r$ or $z \times r$ ) because doubling the charge should double the pull while halving the size should double it too — division captures both cleanly.

PICTURE. Left: big cation, small $z$ → weak field (few thin field lines). Right: tiny cation, big $z$ → dense crowded field lines. The number $ϕ$ is "how crowded the lines are at the surface."

Step 3 — Lay the elements on a map and walk RIGHT

WHAT. Draw the periodic table as a grid. Walk one step across a period (left → right), e.g. from Li to Be to B.

WHY this move. Moving right, the nucleus gains protons but electrons pile into the same shell. More protons pulling the same shell means the cation ends up smaller ( $r ↓$ ) and typically more charged ( $z ↑$ ). See Effective Nuclear Charge (Zeff) and Periodic Trends — Atomic and Ionic Radii.

Feed that into our fraction: $ϕ = \frac{z ↑}{r ↓} ⟹ ϕ rises steeply.$ Here $z ↑$ (bigger top) and $r ↓$ (smaller bottom) both push $ϕ$ up — a double whammy.

PICTURE. The burnt-orange right-arrow; $ϕ$ shown climbing as a rising bar.

Step 4 — Now walk DOWN

WHAT. From the same start, walk one step down a group instead, e.g. Li → Na, or Be → Mg.

WHY this move. Going down, a whole new shell is added. The cation gets bigger ( $r ↑$ ); the charge $z$ usually stays the same.

$ϕ = \frac{z}{r ↑} ⟹ ϕ falls.$ The top ( $z$ ) is unchanged, the bottom ( $r$ ) grows, so the fraction shrinks.

PICTURE. The deep-teal down-arrow; the $ϕ$ bar dropping.

Step 5 — Combine the two arrows: the diagonal step

WHAT. Do both moves at once: one right and one down. That is the diagonal step ↘, e.g. Li ↘ Mg.

WHY it matters. Right pushes $ϕ$ up; down pushes $ϕ$ down. They point in opposite directions, so they partly cancel. You do not end up "hotter" (like a pure right step) nor "cooler" (like a pure down step) — you land back near where you started in polarising behaviour.

$ϕ ↑ right + ϕ ↓ down = ϕ roughly preserved diagonal$

PICTURE. The orange right-arrow and teal down-arrow add tip-to-tail into a plum diagonal arrow; the two vertical $ϕ$ nudges (up then down) net out small.

Step 6 — The honest numbers: what exactly matches?

WHAT. Plug in real ionic radii (in picometres, pm — trillionths of a metre) and compute $ϕ = z / r$ .

Ion	$z$	$r$ (pm)	$ϕ = z / r$
Li⁺	1	76	0.013
Na⁺	1	102	0.010
Mg²⁺	2	72	0.028
Be²⁺	2	45	0.044
Al³⁺	3	53	0.057

WHY read it carefully. Look at Li → Mg: $ϕ$ goes $0.013 \to 0.028$ — it roughly doubles, it does not match! But the radius matches beautifully: 76 vs 72 pm. So for Li/Mg the shared personality comes from near-equal size, which governs how salts pack (lattice energy) versus how they dissolve (hydration energy) — see Lattice Energy vs Hydration Energy.

Look at Be → Al: now $ϕ$ itself is close (0.044 vs 0.057) and both are huge → both polarise ferociously → both bond covalently.

PICTURE. Two panels: (left) Li⁺ and Mg²⁺ drawn near-equal in size but different $ϕ$ bars; (right) Be²⁺ and Al³⁺ with closely matched tall $ϕ$ bars.

Step 7 — The degenerate case: B/Si, where no ion exists

WHAT. For B and Si, a cation like B³⁺ would need such fierce $ϕ$ that it never forms — bonding is covalent from the start. So $z / r$ has no ion to plug in.

WHY it still works. We switch the matched quantity from ionic radius to covalent radius and electronegativity ( $χ$ = how greedily an atom hogs shared electrons). Compare: B ( $r \approx 84$ pm, $χ \approx 2.0$ ) vs Si ( $r \approx 111$ pm, $χ \approx 1.9$ ). Both small, both comparably electron-greedy ⇒ both metalloids that form acidic oxides and volatile, water-sensitive hydrides. This is the same "matched-input" logic, just with the covalent inputs.

PICTURE. B and Si shown as covalent networks (no lone cation), with matched size + $χ$ tags side by side.

The one-picture summary

The whole story on one map: start anywhere in period 2; right heats $ϕ$ , down cools $ϕ$ , the diagonal nets them out — so the period-3 neighbour below-right shares the surviving matched input (size for Li/Mg, high $ϕ$ for Be/Al, size+ $χ$ for B/Si) and therefore the chemistry.

Recall Feynman retelling — say it back in plain words

A little positive ball tugs on a big soft negative cloud. How hard? Two things: how charged the ball is (goes on top) and how small it is (goes on the bottom). That fraction, $z / r$ , is the pull. Now walk the periodic map. Step right and the ball shrinks and charges up — it pulls harder. Step down and the ball fattens — it pulls softer. Step diagonally — do both — and the two effects fight, so you end up pulling about the same as where you began. That is why an element resembles its diagonal neighbour: something important stayed the same. For Li and Mg it's the size; for Be and Al it's the pull itself (and it's huge, so both go covalent); for B and Si there's no ball at all, so we match size and electron-greed instead. Same idea every time: a diagonal step leaves one key ingredient unchanged, and chemistry follows that ingredient.

Recall Quick self-test

Why does the diagonal step preserve chemistry? ::: Moving right raises $ϕ$ and moving down lowers it — opposing trends — so a matched input survives (size or $ϕ$ itself) For Li→Mg, what matches — $ϕ$ or radius? ::: The radius (76 vs 72 pm); $ϕ$ roughly doubles Why can't we use $z / r$ for B/Si? ::: No real B³⁺ cation exists; compare covalent radius and electronegativity instead

Related: Anomalous Behaviour of First Element · Amphoterism · Effective Nuclear Charge (Zeff)