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Diagonal Relationships: Li-Mg and Be-Al

Diagonal Relationships: Li-Mg and Be-Al | chemca
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Diagonal Relationships

The striking chemical similarities between Li & Mg, and Be & Al.

By chemca Team • Updated Aug 2026

In the periodic table, the first element of a group often exhibits properties that differ significantly from its own group members, but surprisingly resemble the properties of the second element in the next group. This cross-group similarity is known as a Diagonal Relationship.

1. The Cause: Ionic Potential / Polarizing Power

Why does Lithium act like Magnesium instead of Sodium? The answer lies in Fajans' Rules and the concept of Ionic Potential ($\phi$), also known as polarizing power.

$$\text{Ionic Potential } (\phi) = \frac{\text{Charge on Cation}}{\text{Radius of Cation}}$$

The Cancellation Effect:

  • Moving across a period (left to right): Charge increases and radius decreases. $\phi$ increases significantly.
  • Moving down a group (top to bottom): Radius increases significantly. $\phi$ decreases.
  • Moving diagonally (down and right): The increase in charge is almost perfectly canceled out by the increase in radius. As a result, diagonally adjacent elements have nearly identical ionic potentials (polarizing power), electronegativity, and similar atomic/ionic radii.
Because their polarizing power is the same, they distort anion electron clouds to the same extent, giving their compounds very similar covalent character and chemical properties.

2. Diagonal Relationship: Lithium (Li) & Magnesium (Mg)

Lithium (Group 1, Period 2) and Magnesium (Group 2, Period 3) share several striking chemical similarities that are highly tested in competitive exams.

  • Nitride Formation: Unlike other alkali metals, both Li and Mg react directly with nitrogen gas in the air to form stable, ionic nitrides.
    $6Li + N_2 \xrightarrow{\Delta} 2Li_3N$
    $3Mg + N_2 \xrightarrow{\Delta} Mg_3N_2$
  • Thermal Decomposition of Carbonates: Other alkali metal carbonates are thermally stable. Both $Li_2CO_3$ and $MgCO_3$ are relatively unstable and decompose easily on heating to form oxides and $CO_2$.
    $Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2 \uparrow$
    $MgCO_3 \xrightarrow{\Delta} MgO + CO_2 \uparrow$
  • Covalent and Deliquescent Chlorides: Due to their high polarizing power, both $LiCl$ and $MgCl_2$ have significant covalent character. They are soluble in organic solvents (like ethanol) and absorb moisture from the air (deliquescent), crystallizing as hydrates: $\mathbf{LiCl \cdot 2H_2O}$ and $\mathbf{MgCl_2 \cdot 6H_2O}$.
  • Reaction with Water: Both metals react very slowly with cold water, unlike the violent reactions of Sodium, Potassium, Calcium, etc.
  • Solid Bicarbonates: Neither Li nor Mg form solid bicarbonates; their bicarbonates exist only in aqueous solutions.

3. Diagonal Relationship: Beryllium (Be) & Aluminum (Al)

Beryllium (Group 2, Period 2) and Aluminum (Group 13, Period 3) share an even more pronounced diagonal relationship.

  • Amphoteric Nature: While other Group 2 metals are strictly basic, both $Be$ and $Al$ (and their oxides/hydroxides) are amphoteric. They dissolve in both acids and strong alkalis (like $NaOH$) to form complex ions ($[Be(OH)_4]^{2-}$ beryllate, and $[Al(OH)_4]^-$ aluminate) while releasing $H_2$ gas.
  • Passivity with Nitric Acid: Both metals are rendered passive (unreactive) by concentrated Nitric Acid ($HNO_3$) due to the instant formation of a tough, impermeable, protective oxide film on their surface.
  • Covalent, Polymeric Chlorides: Both $BeCl_2$ and $AlCl_3$ are predominantly covalent. In the vapor state, they both exist as chloro-bridged dimers ($(BeCl_2)_2$ and $Al_2Cl_6$). They both act as strong Lewis Acids and are frequently used as catalysts in Friedel-Crafts reactions.
  • Hydrolysis of Carbides: The carbides of both metals ($Be_2C$ and $Al_4C_3$) are classified as methanides. Upon reaction with water (hydrolysis), they both yield Methane ($CH_4$) gas (unlike Calcium carbide, which yields acetylene).
    $Be_2C + 4H_2O \rightarrow 2Be(OH)_2 + CH_4 \uparrow$
    $Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4 \uparrow$

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