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

The pairs are Li/Mg, Be/Al, B/Si. The rule: take a period-2 element, step one place to the right and one place down into period 3. That element is its diagonal partner. In symbols: $\text{Li}\searrow \text{Mg}$, $\text{Be}\searrow \text{Al}$, $\text{B}\searrow \text{Si}$.

Read off $\varphi =z/r$ (charge/pm):

Order (least → most polarising): Na⁺ < Li⁺ < Mg²⁺ < Al³⁺. Notice Li⁺ sits above Na⁺ despite same charge — because it is smaller. That anomalously high $\varphi$ is exactly why Li breaks the alkali-metal pattern.

$\varphi ({\text{Li}}^{+})=\frac{1}{76}=0.0132,\varphi ({\text{Mg}}^{2+})=\frac{2}{72}=\mathrm{0.0278.}$ These are not equal — Mg²⁺ is about twice as polarising. What is strikingly matched is the radius: 76 pm vs 72 pm (a difference of only ~5%). Conclusion: for the Li/Mg pair the driver of similar chemistry is near-equal ionic size, which controls the balance of Lattice Energy vs Hydration Energy and hence solubility — not an exact $\varphi$ cancellation.

$\varphi ({\text{Be}}^{2+})=\frac{2}{45}=0.0444,\varphi ({\text{Al}}^{3+})=\frac{3}{53}=\mathrm{0.0566.}$ Here the $\varphi$ values themselves are close and both very high. High $\varphi$ ⇒ strong polarisation of any anion ⇒ covalent bonding for both. So:

• Li/Mg similarity ⇐ matched radius.
• Be/Al similarity ⇐ matched (and high) $\varphi$ itself. Same phenomenon, but the matched input of $\varphi =z/r$ is different for each pair.

The decomposition reaction: ${\text{Li}}_2{\text{CO}}_3\to {\text{Li}}_2\text{O}+{\text{CO}}_2$ (same shape as ${\text{MgCO}}_3\to \text{MgO}+{\text{CO}}_2$). The mechanism in words: the small, high-$\varphi$ cation sits right up against the big, floppy carbonate ion ${\text{CO}}_3^{2-}$ and polarises it — pulling electron density away from one C–O bond, weakening it, so the ion falls apart into oxide + CO₂ on gentle heating.

• Li⁺ ($\varphi =0.013$) and Mg²⁺ ($\varphi =0.028$) are both small and polarising → both destabilise the carbonate.
• Na⁺ ($\varphi =0.010$) is bigger and gentler → it does not polarise the carbonate enough → Na₂CO₃ survives strong heating. That is why Li tracks the diagonal partner Mg, not its group-mate Na.

Fajans' Rules say a bond becomes more covalent when the cation is small and highly charged (high $\varphi$) and the anion is large and easily deformed (Cl⁻ is fairly large).

• Be²⁺: $\varphi =0.044$. Al³⁺: $\varphi =0.057$. Both are high and close.
• Such strong tugging distorts the Cl⁻ electron cloud so heavily that electron density piles up between the nuclei — the definition of a covalent (shared) bond. Result: BeCl₂ forms polymeric chains and AlCl₃ dimerises to Al₂Cl₆; both fume in air and act as Lewis acids. Compare NaCl (Na⁺, low $\varphi$) which stays a clean ionic lattice.

Why $\varphi$ fails for B/Si: $\varphi =z/r$ needs a real ion. But a bare B³⁺ cation never actually exists — boron chemistry is entirely covalent, so there is no ionic radius to plug in. The same is true of Si. What we use instead: compare covalent radius and electronegativity (the pull an atom has on shared electrons):

• B: covalent $r\approx 84$ pm, $\chi \approx 2.0$.
• Si: covalent $r\approx 111$ pm, $\chi \approx 1.9$. Both are small, comparably electronegative — so both favour strong covalent, multi-centre bonding (Anomalous Behaviour of First Element explains why B refuses to be C-like). Two confirming facts (any two):

1. Both are metalloids forming hard, refractory covalent-network solids.
2. Both form acidic oxides (B₂O₃, SiO₂).
3. Both have volatile, flammable, water-hydrolysed hydrides (boranes ~ silanes), unlike the stable, inert alkanes of carbon.

With acid (behaves as a base): ${\text{Al}}_2{\text{O}}_3+6{\text{H}}^{+}\to 2{\text{Al}}^{3+}+3{\text{H}}_2\text{O}$ With base (behaves as an acid): ${\text{Al}}_2{\text{O}}_3+2{\text{OH}}^{-}+3{\text{H}}_2\text{O}\to 2[{\text{Al(OH)}}_4{]}^{-}$ Why amphoteric? Because Be²⁺/Al³⁺ have high $\varphi$: they polarise the O–H bonds enough that the oxide can release H⁺-attracting oxide (basic side) yet also accept OH⁻ to form hydroxo-anions (acidic side). A low-$\varphi$ metal (Na⁺, K⁺) gives a purely basic oxide. See Amphoterism.

Refutation. The claim over-generalises. Only carbonate, phosphate, and fluoride of Mg mirror Li's insolubility:

• Insoluble/sparingly soluble (matched): Li₂CO₃ ↔ MgCO₃, Li₃PO₄ ↔ Mg₃(PO₄)₂, LiF ↔ MgF₂.
• Soluble in both (no special link needed): LiCl, LiNO₃, MgCl₂, Mg(NO₃)₂, MgSO₄.

The $\varphi$-based reason (why anion matters): Solubility is a tug-of-war between Lattice Energy vs Hydration Energy.

• For small, high-charge anions (CO₃²⁻, PO₄³⁻, F⁻), the small high-$\varphi$ cation packs tightly → very large lattice energy that hydration cannot overcome → insoluble.
• For large, singly-charged anions (Cl⁻, NO₃⁻), lattice energy is modest and hydration wins → soluble. So the diagonal match shows up only where lattice energy dominates — i.e. with small hard anions. That is why the pattern is anion-specific, not universal.

1. Across a period (→): charge $z$ ↑ and radius $r$ ↓ (higher Effective Nuclear Charge (Zeff) pulls electrons in — see Periodic Trends — Atomic and Ionic Radii), so $\varphi =z/r$ rises strongly.
2. Down a group (↓): charge $z$ is unchanged but radius $r$ ↑ (new shell), so $\varphi$ falls.
3. Diagonally (↘) = both at once: the rise from going right is partly offset by the fall from going down — the two effects push in opposite directions.
4. They do not cancel exactly, but a diagonal step leaves one input of $\varphi$ nearly matched.
5. For Li/Mg the matched input is the radius (76 ≈ 72 pm) → similar lattice-vs-hydration behaviour → matching solubilities, decompositions, nitride formation.
6. For Be/Al the matched input is $\varphi$ itself (0.044 ≈ 0.057, both high) → both strongly covalent → covalent chlorides, amphoteric oxides.
7. Therefore each period-2 element resembles its down-right partner more than its group neighbour — a cross-trend effect, never a same-electron-count (group) effect.