Worked examples — Diagonal relationship — Li - Mg, Be - Al, B - Si

This page is the practice court for the parent note Diagonal relationship. There, we derived the idea. Here we use it — on every kind of question the topic can throw at you.

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The scenario matrix

Before solving anything, let us list every kind of case this topic contains. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
A	Compute & rank $\varphi$ (positive charges, normal ions)	just arithmetic, but read the matched input	Ex 1
B	"Which is the diagonal partner?" (position logic)	down-and-right, not straight down	Ex 2
C	Solubility prediction (carbonate/fluoride vs chloride)	the anion decides — a sign flip in behaviour	Ex 3
D	Amphoterism / acid + base (Be–Al)	both directions must be shown	Ex 4
E	Degenerate input: B³⁺ "does not exist"	$\varphi =z/r$ breaks — use covalent radius instead	Ex 5
F	Limiting case: what if trends cancelled exactly?	tests the "no exact cancellation" idea	Ex 6
G	Real-world word problem (why Li in batteries, MgO in furnaces)	translate chemistry to application	Ex 7
H	Exam twist: odd-one-out / "spot the false claim"	a trap built from a common mistake	Ex 8
I	Zero case: neutral atom vs ion ($z=0$)	$\varphi =0/r=0$ — no polarising power at all	Ex 9

The figure below is a map of this matrix so you can see the nine cells at a glance. Read it as a 3×3 grid: each pastel tile is one case class labelled with its cell letter (A–I) and the example that fills it. The lavender column on the left (A, D, G) is "pure calculation / core chemistry"; the coral column in the middle (B, E, H) is "logic and traps"; the mint column on the right (C, F, I) is "edges and limits" — solubility flips, exact-cancellation limits, and the zero case. Every tile gets solved once below, so when you finish, the whole grid is coloured in.

Figure s01 — The scenario matrix as a 3×3 map: nine case classes (cells A–I), each tagged with its worked example. Left lavender column = calculation/core; middle coral column = logic/traps; right mint column = edges/limits.

We now walk A → I in order. Every numeric result is machine-checked in this note's verify block.

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Example 1 — Cell A: compute and rank $\varphi$

Forecast: Guess before reading on — is Li⁺ closer in size to Na⁺ (its group-mate) or to Mg²⁺ (its diagonal partner)?

1. Compute each $\varphi$. $\varphi ({\text{Li}}^{+})=\frac{1}{76}=0.01316,\varphi ({\text{Na}}^{+})=\frac{1}{102}=0.00980,\varphi ({\text{Mg}}^{2+})=\frac{2}{72}=0.02778$ Why this step? $\varphi$ is the "field strength at the ion's surface" we defined above. Compute it to rank grabbiness.

2. Compare radii, not just $\varphi$. $|r_{\text{Li}}-r_{\text{Na}}|=|76-102|=26$ pm, but $|r_{\text{Li}}-r_{\text{Mg}}|=|76-72|=4$ pm. Why this step? For the Li/Mg pair the parent note warned that $\varphi$ does not match — the radius does. So the honest comparison is size. See Periodic Trends — Atomic and Ionic Radii.

3. State the ranking. Most polarising: ${\text{Mg}}^{2+}>{\text{Li}}^{+}>{\text{Na}}^{+}$. Closest in size: Li⁺ ↔ Mg²⁺ (4 pm apart). Why this step? The question asked two separate things — grabbiness (rank by $\varphi$) and size match (rank by $r$). We state both explicitly so the reader sees that the "diagonal match" for Li/Mg lives in the radius column, not the $\varphi$ column.

Verify: $\varphi ({\text{Mg}}^{2+})/\varphi ({\text{Li}}^{+})=0.02778/0.01316\approx 2.11$ — Mg²⁺ is ~twice as polarising, exactly the "φ does not cancel" fact from the parent. And Li's radius is 6× closer to Mg's than to Na's. Answer consistent. ✓

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Example 2 — Cell B: find the diagonal partner by position

Forecast: Straight down from Be is Mg then Ca. Diagonal means one step down AND one step right — so where do you land?

1. Move down one row, right one column from Be. Be (Grp 2, Per 2) → down to Period 3, right to Group 13 → aluminium (Al). Why this step? The definition in the parent note is literally "one place to the right and one place down." Ca is straight down — a group relationship, not diagonal.

2. Sanity of the chemistry. Ca²⁺ is large ($r\approx 100$ pm) and low-$\varphi$ → ionic, basic oxide (CaO). Al is high-$\varphi$ → covalent chlorides, amphoteric oxide — matching Be, not Ca. Why this step? Position gives the candidate; $\varphi$ confirms the chemistry.

Verify: Be's diagonal partner = Al. Both amphoteric (Cell D below), both covalent chlorides — checks against the parent's Be/Al evidence. ✓

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Example 3 — Cell C: solubility, where the anion flips the answer

Forecast: All four contain a small high-$\varphi$ cation. Do they all behave the same? (Trap!)

1. Recall the energetics. Solubility is a tug-of-war: lattice energy (holds solid together) vs hydration energy (water pulls ions apart). See Lattice Energy vs Hydration Energy. Why this step? Solubility is not about the cation alone — it is a balance of two energies.

2. Big cation + big anion (carbonate) → high lattice energy. CO₃²⁻ is large and doubly charged; with small Li⁺/Mg²⁺ the lattice is very stable → Li₂CO₃, MgCO₃ sparingly soluble. Why this step? When both ions are small-ish and highly charged, lattice energy wins.

3. Same cation + small singly-charged anion (Cl⁻) → hydration wins. LiCl, MgCl₂ have modest lattice energy but strong hydration of the small cation → soluble. Why this step? This is the parent note's explicit warning: not all Mg salts are insoluble — only carbonate, phosphate, fluoride.

4. Table the result:

Salt	Behaviour
Li₂CO₃	sparingly soluble
MgCO₃	sparingly soluble
LiCl	soluble
MgCl₂	soluble

Why this step? Laying the four salts side by side makes the anion-driven flip visible in one glance: hold the cation fixed and the carbonate/chloride switch alone decides the answer — which is the whole point of Cell C.

Verify: The diagonal similarity holds anion-by-anion: Li and Mg agree for carbonate (both insoluble) and agree for chloride (both soluble). Contrast Na₂CO₃ — soluble — which is where Li breaks its group. ✓

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Example 4 — Cell D: amphoterism (both acid AND base)

Forecast: A "normal" basic oxide (like Na₂O) reacts only with acid. Does BeO stop there?

1. Reaction with acid (oxide acts as a base). $\text{BeO}+2{\text{H}}^{+}\to {\text{Be}}^{2+}+{\text{H}}_2\text{O}$ Why this step? Every metal oxide with cation charge can neutralise acid; this is the "easy" half.

2. Reaction with base (oxide acts as an acid). $\text{BeO}+2{\text{OH}}^{-}+{\text{H}}_2\text{O}\to [{\text{Be(OH)}}_4{]}^{2-}$ Why this step? This second reaction is the signature of amphoterism — see Amphoterism. Only high-$\varphi$ (small, strongly polarising) cations do this.

3. Mirror it with Al₂O₃ (from the parent note): ${\text{Al}}_2{\text{O}}_3+6{\text{H}}^{+}\to 2{\text{Al}}^{3+}+3{\text{H}}_2\text{O},{\text{Al}}_2{\text{O}}_3+2{\text{OH}}^{-}+3{\text{H}}_2\text{O}\to 2[{\text{Al(OH)}}_4{]}^{-}$ Why this step? Identical dual behaviour = diagonal match.

The figure below is the picture of this dual behaviour. Read it left-to-right: the lavender box in the centre is the oxide (BeO, and equally Al₂O₃). Follow the coral arrow to the left — that is the oxide acting as a base, gobbling $2{\text{H}}^{+}$ from an acid to give the cation plus water. Follow the mint arrow to the right — that is the same oxide now acting as an acid, taking $2{\text{OH}}^{-}$ from a base to give the $[{\text{Be(OH)}}_4{]}^{2-}$ ion. One box, two arrows in opposite directions: that visual two-way split is what "amphoteric" means, and it is the fingerprint the Be/Al pair share.

Figure s02 — Amphoterism of BeO (and Al₂O₃): the central lavender oxide box reacts to the left with acid (coral arrow, acting as a base → Be²⁺ + H₂O) and to the right with base (mint arrow, acting as an acid → [Be(OH)₄]²⁻). Two arrows in opposite directions from one oxide = amphoteric.

Verify (atom balance, BeO + 2OH⁻ + H₂O → [Be(OH)₄]²⁻): Be: $1=1$ ✓. O: $(1\text{ from BeO})+(2\text{ from }2{\text{OH}}^{-})+(1{\text{ from H}}_2\text{O})=4$, right has 4 O in $[{\text{Be(OH)}}_4{]}^{2-}$ ✓. H: $(2\text{ from }2{\text{OH}}^{-})+(2{\text{ from H}}_2\text{O})=4$ = 4 in four OH ✓. Charge: left $=2(-1)=-2$ = right $-2$ ✓. Balanced. ✓

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Example 5 — Cell E: the degenerate case, B³⁺ "does not exist"

Forecast: Can we plug B into $\varphi =z/r$ the way we did for Li and Be?

1. Spot the degenerate input. Removing 3 electrons from boron costs enormous energy — a bare B³⁺ never forms; boron bonds covalently. So there is no real ionic $r$ to put in $\varphi =z/r$. Why this step? $\varphi$ assumes an ionic cation. Feed it a non-ion and the formula is meaningless — a genuine degenerate case, like dividing by a quantity that does not exist.

2. Switch tools: use covalent radius + electronegativity. B (covalent $r\approx 84$ pm, $\chi \approx 2.0$) vs Si ($r\approx 111$ pm, $\chi \approx 1.9$). Why this step? When the ionic model breaks, compare the properties that do apply to covalent atoms — size and electron-pulling power ($\chi$).

3. Read the match. $|{\chi }_{\text{B}}-{\chi }_{\text{Si}}|=|2.0-1.9|=0.1$ — nearly identical → both small, comparably electronegative metalloids → acidic oxides (B₂O₃, SiO₂), hydrolysable volatile hydrides. Why this step? The similarity is now a qualitative size/electronegativity analogy, exactly as the parent note insisted.

Verify: Electronegativity gap $=0.1$ (tiny); covalent-radius gap $=27$ pm. The $z/r$ route gives nonsense (no B³⁺), the $\chi$/size route gives a real match. ✓

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Example 6 — Cell F: limiting case, "what if $\varphi$ cancelled exactly?"

Forecast: If $z$ and $r$ both double, does the ratio $z/r$ change?

1. Compute both $\varphi$. $\varphi ({\text{X}}^{+})=\frac{1}{50}=0.02,\varphi ({\text{Y}}^{2+})=\frac{2}{100}=0.02$ Identical — exact cancellation, by construction. Why this step? It shows the "trends cancel" slogan could be true only if the two inputs scaled in perfect lockstep.

2. Compare to reality (Li→Mg). Real: $z$ doubled (1→2) but $r$ barely changed (76→72 pm, a drop, not a doubling). So $\varphi$ rose from 0.0132 to 0.0278 — it did not cancel. Why this step? This is the parent note's key honesty point: opposing trends, partial offset, not exact cancellation. The matched input for Li/Mg is radius, not $\varphi$.

Verify: Hypothetical ratio $0.02=0.02$ (exact). Real ratio $0.0278/0.0132\approx 2.11\ne 1$. The idealised cancellation and reality disagree — proving the slogan is only a cartoon. ✓

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Example 7 — Cell G: real-world word problem

Forecast: Why pick the small high-$\varphi$ member each time?

1. MgO melting point via lattice energy. Mg²⁺ is small and doubly charged (high $\varphi$); O²⁻ is small and doubly charged. Lattice energy $\propto \frac{z_{+}{z}_{-}}{r_{+}+r_{-}}$ — big numerator, small denominator → huge lattice energy → very high melting point (~2850 °C). Why this step? The same high-$\varphi$ that links Mg to Li also makes MgO refractory — the diagonal cation is engineered-in.

2. Contrast Na₂O. Na⁺ is large, singly charged (low $\varphi$) → smaller lattice energy → lower melting, and Na₂O is only basic (not tough). See Effective Nuclear Charge (Zeff) for why Na is bigger. Why this step? Shows why the group-mate fails the job the diagonal-type small ion does.

3. Li in batteries. Li⁺ (76 pm) is the smallest, lightest metal cation → highest charge density → high cell voltage and light weight — the same small-size property that gives it Mg-like chemistry. Why this step? The application is a direct payoff of the anomalous small size (from Anomalous Behaviour of First Element).

Verify (lattice-energy proxy $z_{+}{z}_{-}/(r_{+}+r_{-})$, radii in pm): MgO: $r({\text{O}}^{2-})=140$, so $\frac{2\cdot 2}{72+140}=\frac{4}{212}=0.01887$. Na₂O: $\frac{1\cdot 2}{102+140}=\frac{2}{242}=0.00826$. MgO proxy is $0.01887/0.00826\approx 2.28\times$ larger → higher melting. Consistent with MgO being the furnace choice. ✓

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Example 8 — Cell H: exam twist, spot the false claim

Forecast: Three are straight from the evidence list — one is a classic trap. Which?

1. Test (a). $6\text{Li}+{\text{N}}_2\to 2{\text{Li}}_3\text{N}$; Mg → Mg₃N₂. True (Na does neither). Why this step? Nitride formation is a listed Li/Mg similarity; check it against the known reactions before moving on — no trap here.

2. Test (b). High matched ionic potential polarises Cl⁻ (Fajans) → covalent. True. Why this step? This is the Be/Al covalent-chloride evidence; the reason given ($\varphi$ matched) is exactly right, so the statement stands.

3. Test (c). The parent's mistake box: only carbonate, phosphate, fluoride of Mg are insoluble; MgCl₂, Mg(NO₃)₂, MgSO₄ dissolve. The word all makes it FALSE. ← answer. Why this step? Watch the quantifier — a single over-general word ("all") turns a true pattern into a false claim. This is the built-in trap.

4. Test (d). ${\text{Li}}_2{\text{CO}}_3\to {\text{Li}}_2\text{O}+{\text{CO}}_2$; MgCO₃ likewise. True (Na₂CO₃ is stable). Why this step? Thermal decomposition of carbonate is a listed Li/Mg match; confirm it is stated without over-generalising, so it is true.

Verify: Exactly one statement is false — (c) — and it fails on the word "all," the classic solubility trap. The other three (a, b, d) each map to a confirmed entry in the parent note's evidence list, so they are true. ✓

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Example 9 — Cell I: the zero case, $z=0$

Forecast: Plug $z=0$ into $\varphi =z/r$ — what comes out, and is that sensible?

1. Compute. Using the neutral atomic radius $r\approx 160$ pm and charge $z=0$: $\varphi =\frac{z}{r}=\frac{0}{160}=0.$ Why this step? We are testing the edge of the model. A neutral atom has no net charge, so the numerator of $\varphi$ is zero — and any zero-over-nonzero is exactly $0$. The value of $r$ never even matters once $z=0$.

2. Interpret physically. Polarisation (the Fajans-style pull that distorts an anion into covalent character) needs a charged cation to create a field. With $z=0$ there is no field, so a neutral atom exerts zero polarising power — it cannot bend Cl⁻'s electron cloud at all. Why this step? It shows why the whole diagonal argument is framed around ions, never neutral atoms — and it explains why B (which stays effectively neutral/covalent, Ex 5) also fell outside the $z/r$ model.

3. Conclusion. $\varphi (\text{neutral Mg})=0$. The formula behaves correctly at this boundary: no charge ⇒ no polarising power ⇒ the diagonal similarity story simply does not apply to uncharged species.

Verify: $0/160=0$ exactly, and the same $0$ results for any $r>0$ — confirming the numerator, not the radius, controls this limit. Together with the non-ionic B³⁺ case (Ex 5), this marks the two edges where $\varphi =z/r$ stops being the right tool. ✓

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Recall Quick self-test

Which cell is "MgCl₂ is soluble but MgCO₃ is not"? ::: Cell C — the anion flips the lattice-vs-hydration balance Why can't we compute $\varphi$ for B in B–Si? ::: Cell E — B³⁺ never forms; use covalent radius and electronegativity instead Does $\varphi$ cancel exactly along a real diagonal? ::: No (Cell F) — for Li/Mg it roughly doubles; the radius matches What is $\varphi$ for a neutral atom? ::: Zero (Cell I) — no charge, no polarising field