Solubility and Thermal Stability Trends of Group 2 Salts

Magnesium hydroxide is so insoluble that it forms the antacid milk of magnesia. Barium hydroxide, by contrast, is soluble enough to make a usable titration standard. Yet both salts contain the same anion and come from the same group. Magnesium sulfate dissolves so readily that Epsom salt is sold by the kilogram for garden and bath use. Barium sulfate, however, is so insoluble that patients drink it as an X-ray contrast agent. They do so without absorbing a dangerous dose of a toxic metal. These are not isolated curiosities. Hydroxide solubility rises down Group 2 while sulfate solubility falls. The two trends run in exactly opposite directions for a systematic thermodynamic reason. A second, equally sharp pattern governs how these salts behave in fire rather than in water. Magnesium carbonate decomposes to the oxide below 550 °C. Barium carbonate survives past 1300 °C, and the nitrates show the same widening gap. Chemists have long organized this behavior with a slogan. The slogan is "the smaller cation destabilizes the larger anion." But a slogan is not an explanation. The explanation lives in the competition between lattice energy terms. These terms scale with how tightly two ions of very different sizes can pack. It can be made quantitative with a short thermodynamic cycle for each family of compounds. The big picture is that every one of these trends is a size-mismatch effect. It is viewed from two different thermodynamic vantage points. Solubility trends are read off a Born–Haber-type cycle. This cycle compares lattice energy to hydration energy. Decomposition-temperature trends are read off a different cycle. This one compares the lattice energy of the reactant salt to that of the product oxide. Both cycles hinge on the same physical fact. As the cation radius increases from Mg²⁺ to Ba²⁺, lattice energy falls off steeply when paired with a small, compact anion (O²⁻, OH⁻). It falls off only gently when paired with a large, diffuse anion (SO₄²⁻, CO₃²⁻, NO₃⁻). This article builds that reasoning from the ground up. First comes the qualitative size-mismatch picture with structural diagrams. Then the Kapustinskii-based lattice energy expression and the solvation-cycle equations follow. These equations convert lattice/hydration energy differences into solubility and decomposition-temperature trends. Next come three fully worked numerical examples. These span a beginner comparison, an applied industrial case, and a genuinely quantitative competition-style estimate. Finally, there is a survey of where these trends show up. That includes cement chemistry, medicine, pyrotechnics, and water treatment.

Key takeaways

Group 2 hydroxide solubility increases down the group (Mg(OH)₂ insoluble → Ba(OH)₂ soluble); sulfate solubility decreases down the group (MgSO₄ soluble → BaSO₄ insoluble).
The reversal comes from the size mismatch: lattice energy falls faster with increasing cation radius when the anion is small (OH⁻) than when the anion is large (SO₄²⁻), because lattice energy depends on $1/(r_+ + r_-)$ .
Thermal decomposition of carbonates and nitrates to oxide + gas becomes harder (higher $T_{dec}$ ) down the group, tracking the same size-mismatch logic applied to a lattice-energy-difference cycle rather than a solubility cycle.
Both trends are quantified with the same tool: a thermodynamic cycle combining lattice energy (Kapustinskii/Born–Landé estimate) with hydration energy or with the lattice energy of the oxide product.
Charge density (charge/radius²) of the cation is the single organizing variable: high charge density favors small, tightly bound anions and disfavors large ones, in both solubility and thermal stability.
Key idea: whenever a small, high-charge-density cation reacts with a large, diffuse anion, the difference in lattice energy between competing reference states shrinks as the cation grows, flipping which side of a thermodynamic cycle wins down a group.

The core insight

Down Group 2, increasing cationic radius lowers lattice energy faster for compounds with small anions than for compounds with large anions; this single asymmetry, applied to a dissolution cycle, explains why hydroxide solubility rises and sulfate solubility falls, and applied to a decomposition cycle, explains why carbonates and nitrates become more thermally stable.

Where this lives today

The cement and lime industries depend directly on the carbonate decomposition trend: calcium carbonate is calcined at roughly 900 °C in rotary kilns to produce quicklime (CaO) for cement clinker, a temperature chosen because it sits comfortably above CaCO₃'s decomposition threshold but far below the much higher threshold that would be needed for the more thermally robust strontium or barium carbonates — meaning the "easy" position of calcium in the middle of the group is not an accident of convenience but a direct consequence of the same lattice-energy competition worked through above. Barium carbonate's exceptional thermal robustness, by contrast, is exploited in specialty glass and ceramic glazes where a carbonate additive must survive high processing temperatures without prematurely releasing CO₂. Medicine leans on the solubility trend at both extremes: magnesium hydroxide's near-insolubility (and consequent mild, slow-acting basicity) makes it a gentle antacid and laxative (milk of magnesia), while barium sulfate's extreme insolubility, quantified in Example 2, is precisely what makes barium — an intrinsically toxic heavy metal — safe to swallow in gram quantities as a radiographic contrast agent. Water treatment and geochemistry use the same sulfate trend in reverse: gypsum (CaSO₄·2H₂O) scaling in boilers and pipes, and barite (BaSO₄) as a heavy drilling mud additive in the petroleum industry, both exploit the low but non-negligible solubility of the middle-to-heavy Group 2 sulfates. Pyrotechnics and flame chemistry exploit the decomposition-temperature trend for color and timing control: strontium and barium carbonates and nitrates are chosen as oxidizer/colorant precursors specifically because their higher decomposition temperatures delay oxide/gas release until the flame reaches the desired thermal stage, giving red (strontium) and green (barium) emission colors at controlled points in a firework's burn. Analytical and materials chemistry more broadly use this whole family of trends as a diagnostic and predictive tool: given only ionic radii and charges, a chemist can sketch, before ever opening a solubility table, which Group 2 salts of a newly proposed anion are likely to be soluble and which carbonate or nitrate analogues will be thermally robust — the same reasoning developed further in treatments of Lattice Energy and Thermodynamic Stability of s-Block Salts, and resting throughout on the descriptive foundation laid out in the Descriptive Chemistry of Alkaline Earth Metals.

Ask the tutor: How would these trends change if we compared Group 2 salts of a mixed-charge anion like phosphate (PO₄³⁻) instead of a 2− anion?

The physical picture

Every Group 2 cation is small and doubly charged, so it carries a high charge density (charge divided by ionic-radius-squared, a useful shorthand for polarizing power and electrostatic binding strength). Descending the group, ionic radius grows steadily — roughly 72 pm for Mg²⁺ up to 135 pm for Ba²⁺ in six-coordination — and charge density falls correspondingly. The consequence for any ionic solid is captured by the lattice-energy relationship, which for a 1:1 stoichiometric charge ratio behaves approximately as

U_L \propto \frac{z_+ z_-}{r_+ + r_-}

The key structural fact is what happens to this ratio when $r_-$ itself is large. If the anion radius $r_-$ is small (as for OH⁻, ~140 pm, or O²⁻, ~140 pm), then $r_+ + r_-$ is dominated by $r_+$ , so increasing $r_+$ from Mg²⁺ to Ba²⁺ produces a large fractional change in the denominator and hence a large fractional drop in $U_L$ . If $r_-$ is large (as for SO₄²⁻, ~230 pm, or CO₃²⁻, ~178 pm, or NO₃⁻, ~179 pm), then $r_+ + r_-$ is dominated by the anion, and the same increase in $r_+$ produces only a small fractional change in the sum — lattice energy barely moves down the group. This asymmetry alone would just say "all lattice energies fall down the group, but sulfate/carbonate/nitrate lattice energies fall more slowly." To get a solubility trend or a decomposition-temperature trend, this lattice-energy term must be compared against a second energetic quantity that has the opposite size-dependence, so that the two together can cross over or diverge systematically. For solubility, that second quantity is hydration energy. Hydration energy is large (very exothermic) for small cations because the electric field at the surface of a small ion is intense, and hydration energy of the anion behaves the same way — but crucially, cation hydration energy also falls steeply as $r_+$ increases, in a way that tracks the cation-dominated lattice-energy term for small anions almost one-for-one. The dissolution free energy is a difference of two large, closely related numbers, so subtle differences in how each falls with $r_+$ decide the sign of $\Delta G_{soln}$ . For thermal decomposition, the second quantity is the lattice energy of the oxide product (MgO, CaO, …), which is enormous and dominated by the very short M²⁺–O²⁻ distance — again a small-anion, cation-radius-sensitive quantity. Comparing a nearly cation-radius-insensitive reactant lattice energy (carbonate, nitrate) against a strongly cation-radius-sensitive product lattice energy (oxide) produces a decomposition energetics that gets less favorable — i.e., a higher decomposition temperature — as the cation grows.

Figure

Ionic radius trend and the size-mismatch idea for Group 2 cations paired with a small vs. large anion

Figure

Two thermodynamic cycles: dissolution vs. carbonate decomposition

The core ideas

The lattice energy of an ionic solid, from the Kapustinskii approximation (a generalized, structure-independent form of the Born–Landé result), is

U_L = \frac{1.202\times10^{5}\, \nu\, z_+ z_- }{r_+ + r_-}\left(1-\frac{d}{r_+ + r_-}\right) \ \text{kJ mol}^{-1}

where: $\nu$ is the number of ions per formula unit, $z_+, z_-$ are the (unsigned) ionic charges, $r_+ + r_-$ is in picometers, and $d\approx 34.5$ pm is an empirical softness constant. The important qualitative content, already visible without the correction term, is that $U_L$ is a decreasing function of $r_+ + r_-$ with a curvature that makes fractional changes in the sum matter more when the sum is small (small anion) than when it is large (big anion) — precisely the asymmetry sketched above. The solubility criterion follows from a Born–Haber-style solvation cycle:

\Delta G_{soln} = -U_L + \Delta H_{hyd}(M^{2+}) + \Delta H_{hyd}(X^{n-}) - T\Delta S_{soln}

where: $U_L$ is the (positive, by convention) lattice energy of the solid, $\Delta H_{hyd}$ terms are the (negative) hydration enthalpies of the separated gas-phase ions, and $T\Delta S_{soln}$ is a generally small, often slightly unfavorable entropy term for ordering water around ions. A salt is soluble when $\Delta G_{soln} < 0$ , i.e. when hydration energy release outpaces the lattice energy cost of separating the ions. Reading: because both $U_L$ and $\Delta H_{hyd}(M^{2+})$ depend on $r_+$ in almost the same functional way when the anion is small, their difference is nearly constant down the group for hydroxides, and the modest residual trend is set by how $\Delta H_{hyd}(M^{2+})$ (falling steeply, since hydration always scales as $1/r$ regardless of anion) outpaces $U_L$ (also falling steeply, but bounded below by the anion's own size contribution) — net effect: solubility rises down the group for hydroxides. For sulfates, $U_L$ barely falls at all down the group (anion-dominated denominator) while $\Delta H_{hyd}(M^{2+})$ still falls sharply (hydration cares only about the cation's own radius, not what anion it will pair with) — so the lattice term wins the trend, $\Delta G_{soln}$ becomes less negative (or more positive), and solubility falls down the group. The decomposition criterion follows from comparing lattice energies of reactant and product directly, since the archetypal decomposition is

\mathrm{MCO_3(s)} \rightarrow \mathrm{MO(s)} + \mathrm{CO_2(g)}

with

\Delta H_{dec} = \big[U_L(\mathrm{MO}) - U_L(\mathrm{MCO_3})\big] + \Delta H_f^{\circ}(\mathrm{CO_2}) - (\text{lattice-independent offset})

where: $U_L(\mathrm{MO})$ is the (very large, short-bond) lattice energy of the oxide product, $U_L(\mathrm{MCO_3})$ is the (smaller in magnitude, weaker cation-radius dependence) lattice energy of the carbonate, and the offset collects the CO₃²⁻ → O²⁻ + CO₂ intrinsic bond-breaking energy, which is roughly constant across the whole series because it does not involve the metal at all. Because $U_L(\mathrm{MO})$ falls steeply with growing $r_+$ (small O²⁻ anion) while $U_L(\mathrm{MCO_3})$ falls only gently (large CO₃²⁻ anion), the bracketed difference shrinks (becomes less favorable, i.e. $\Delta H_{dec}$ rises) down the group, so more thermal energy — a higher decomposition temperature — is required to drive the reaction down the group.

Derivation AWhy hydroxide solubility rises but sulfate solubility falls down Group 2

Goal. Show, from the solvation cycle, why the sign of the solubility trend flips when the anion is switched from small (OH⁻) to large (SO₄²⁻), using only the radius-dependence of $U_L$ and $\Delta H_{hyd}$ .

Step 1 — Write both energetic terms as explicit functions of cation radius. Model lattice energy with the Born-type form $U_L(r_+) = \dfrac{A z_+z_-}{r_+ + r_-}$ , where $A$ collects Madelung and Born-exponent constants and is treated as roughly anion-independent for a fixed structure family. Model cation hydration enthalpy with the classical Born solvation form $\Delta H_{hyd}(M^{2+}) = -\dfrac{B z_+^2}{r_+}$ , where $B$ is a positive constant absorbing the dielectric constant of water and self-energy bookkeeping; note this term depends on $r_+$ alone, never on the anion, because hydration happens after the ions are already separated in the gas phase.

Step 2 — Assemble the solubility free energy difference between two group members. Take a "small" cation with radius $r_1$ (e.g. Mg²⁺) and a "large" cation with radius $r_2 > r_1$ (e.g. Ba²⁺), same anion throughout. Ignoring the anion's own hydration term (identical for both salts since the anion is unchanged) and the small entropy term, the change in $\Delta G_{soln}$ on going from cation 1 to cation 2 is

\Delta\Delta G = \left[-U_L(r_2)+U_L(r_1)\right] + \left[\Delta H_{hyd}(r_2)-\Delta H_{hyd}(r_1)\right]

\Delta\Delta G = A z_+z_-\left(\frac{1}{r_1+r_-}-\frac{1}{r_2+r_-}\right) - B z_+^2\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

Step 3 — Evaluate the sign for small $r_-$ (hydroxide-like). When $r_-$ is small, $r_1+r_-$ and $r_2+r_-$ are close to $r_1$ and $r_2$ themselves, so the first bracket is numerically close to $\left(\frac1{r_1}-\frac1{r_2}\right)$ — nearly matching the magnitude of the second bracket's structure, but the two brackets carry opposite overall signs in $\Delta\Delta G$ (one is $-U_L$ shrinking with size, i.e. becoming less costly to break; one is $\Delta H_{hyd}$ shrinking with size, i.e. becoming less rewarding). Because $A z_+ z_-$ and $B z_+^2$ are both large and the two $1/r$ -type brackets nearly cancel in magnitude when $r_-\to 0$ , the residual $\Delta\Delta G$ is small in magnitude but is empirically negative (favoring dissolution) because hydration free energy — which also carries an entropic water-reorganization penalty that grows for small, structure-making ions like Mg²⁺ — is disproportionately suppressed for small cations relative to lattice energy release upon dissolving. The practical upshot, confirmed by tabulated enthalpies, is that $\Delta H_{hyd}$ falls (becomes less exothermic) slightly faster than $U_L$ falls as $r_+$ grows for small-anion salts, so $\Delta\Delta G$ becomes more negative going down the group: solubility increases down the group for hydroxides.

Step 4 — Evaluate the sign for large $r_-$ (sulfate-like). When $r_-$ is large (comparable to or larger than $r_+$ itself), $\dfrac1{r_1+r_-}-\dfrac1{r_2+r_-}$ becomes small because both denominators are anion-dominated and nearly equal; the first ( $-U_L$ ) bracket in $\Delta\Delta G$ shrinks toward zero. The second bracket, $-Bz_+^2\left(\frac1{r_1}-\frac1{r_2}\right)$ , is completely unaffected by the anion's size and remains the same large negative-favoring-dissolution number it was for hydroxide. But now it stands alone against a near-zero lattice contribution, meaning the actual reference point for $\Delta G_{soln}$ itself (not just its group-to-group change) is now dominated by the fact that $U_L$ for the large-anion salt was never very sensitive to $r_+$ in the first place — so the absolute lattice energy stays large and roughly constant while hydration energy keeps dropping down the group. Since $U_L$ was already the dominant, harder-to-overcome term for a big, highly charged anion like SO₄²�¯ (high lattice energy that hydration must outcompete), and hydration keeps weakening down the group while lattice energy does not correspondingly weaken, $\Delta G_{soln}$ becomes progressively less negative (more positive): solubility decreases down the group for sulfates.

Step 5 — State the crossover condition explicitly. The two behaviors are reconciled by one inequality: solubility rises down the group whenever

\left|\frac{d\,U_L}{dr_+}\right| < \left|\frac{d\,\Delta H_{hyd}}{dr_+}\right|

which holds when $r_-$ is small enough that $U_L$ is cation-radius-sensitive but not so much more sensitive than hydration that it dominates; it fails (reverses) once $r_-$ is large enough that $U_L$ becomes essentially cation-radius-insensitive, at which point hydration's steep $1/r_+$ fall is the only thing changing and the salt's solubility must fall down the group. The anion radius at which the inequality flips depends on the specific constants $A$ and $B$ , but the qualitative crossover — small anion → rising solubility, large anion → falling solubility — is robust across the whole family of anions from OH⁻/F⁻ to SO₄²⁻/CO₃²⁻.

Go deeper · Derivation B · 2 minWhy carbonate and nitrate decomposition temperatures rise down Group 2

Derivation BWhy carbonate and nitrate decomposition temperatures rise down Group 2

Goal. Use the same lattice-energy asymmetry to show that $\Delta H_{dec}$ for $\mathrm{MCO_3 \rightarrow MO + CO_2}$ becomes more endothermic (and hence $T_{dec}$ rises) as the cation grows.

Step 1 — Set up the lattice-energy difference that drives the reaction. Write the decomposition enthalpy as a sum of a metal-independent "intrinsic" carbonate-breaking term $\delta$ (the energy to convert an isolated gas-phase CO₃²⁻ into O²⁻ + CO₂, a fixed quantity because it involves no metal atom at all) and a metal-dependent lattice-energy difference:

\Delta H_{dec}(r_+) = \delta + \big[U_L^{MCO_3}(r_+) - U_L^{MO}(r_+)\big]

Here $U_L^{MO}(r_+)$ is written with a minus sign because forming the stable oxide lattice releases energy that offsets the cost of breaking up the carbonate lattice; decomposition becomes easier (lower $\Delta H_{dec}$ ) the more negative this bracket is, i.e. the more the oxide's lattice energy exceeds the carbonate's.

Step 2 — Insert the Kapustinskii-type radius dependence for each lattice energy. Using $U_L \propto \dfrac{1}{r_++r_-}$ (charges and $\nu$ fixed within each compound class, with the caveat that MO is 1:1 while MCO₃ is also effectively 1:1 in Madelung constant terms for these rock-salt-related structures), the bracket becomes

U_L^{MCO_3}(r_+) - U_L^{MO}(r_+) = A\left(\frac{1}{r_++r_{CO_3}} - \frac{1}{r_++r_{O}}\right)

Since $r_{CO_3} \approx 178\ \text{pm} \gg r_O \approx 140\ \text{pm}$ , the second term (oxide) is always the larger, more negative-favoring one — the oxide lattice is always the "prize" that pulls decomposition forward. The question is only how the size of this favorable gap changes with $r_+$ .

Step 3 — Differentiate qualitatively with respect to $r_+$ . Both terms decrease in magnitude as $r_+$ grows, but the fractional decrease is larger for the small-anion (oxide) term, because $r_+$ is a larger fraction of $r_+ + r_O$ than of $r_+ + r_{CO_3}$ . Concretely, going from $r_+=72$ pm (Mg²⁺) to $r_+=135$ pm (Ba²⁺):

\frac{1}{72+140} - \frac{1}{135+140} = \frac{1}{212} - \frac{1}{275} = 4.717\times10^{-3} - 3.636\times10^{-3} = 1.081\times10^{-3}\ \text{pm}^{-1}

for the oxide-like term, versus

\frac{1}{72+178} - \frac{1}{135+178} = \frac{1}{250} - \frac{1}{313} = 4.000\times10^{-3} - 3.195\times10^{-3} = 8.05\times10^{-4}\ \text{pm}^{-1}

for the carbonate-like term. The oxide term changes by more ( $1.081\times10^{-3}$ vs $8.05\times10^{-4}$ pm⁻¹), confirming that the favorable oxide-formation contribution to $\Delta H_{dec}$ shrinks faster down the group than the carbonate-breaking cost shrinks.

Step 4 — Combine into the net trend. Because the oxide lattice energy (the term that makes decomposition favorable) drops faster than the carbonate lattice energy (the term that resists decomposition) as $r_+$ grows, the net driving force for decomposition weakens down the group: the bracket $U_L^{MCO_3} - U_L^{MO}$ becomes less negative, so $\Delta H_{dec}$ increases (decomposition becomes more endothermic). A higher activation-adjacent enthalpy translates operationally into a higher temperature at which the equilibrium CO₂ partial pressure reaches one atmosphere (or the reaction becomes spontaneous under furnace conditions), so decomposition temperature rises down the group: MgCO₃ < CaCO₃ < SrCO₃ < BaCO₃.

Step 5 — Note that nitrate decomposition follows the identical logic with a different intrinsic step. For nitrates, the relevant reaction is $2\mathrm{M(NO_3)_2 \rightarrow 2MO + 4NO_2 + O_2}$ (Group 2 nitrates, unlike alkali nitrates, decompose all the way to oxide rather than stopping at nitrite because the M²⁺ charge density is high enough to destabilize the nitrite intermediate as well). The same $U_L^{MO}$ vs. $U_L^{M(NO_3)_2}$ competition applies, with $r_{NO_3^-}\approx 179$ pm playing the same "large, radius-insensitive" role that CO₃²⁻ played above, so nitrate decomposition temperature rises down the group by the same mechanism, in the same order Mg < Ca < Sr < Ba.

Worked examples

Example 1 (foundational) — Ranking hydroxide vs. sulfate solubility using tabulated data. Given approximate literature values (all in kJ mol⁻¹, hydration enthalpies for the 2+ cation only, lattice energies for the full salt):

Cation	$r_+$ (pm)	$\Delta H_{hyd}(M^{2+})$	$U_L$ , MSO₄	$U_L$ , M(OH)₂
Mg²⁺	72	−1921	2848	2993
Ca²⁺	100	−1577	2653	2657
Ba²⁺	135	−1305	2423	2313

For hydroxides, track $-U_L + \Delta H_{hyd}(M^{2+})$ (a proxy for the sign of $\Delta G_{soln}$ , ignoring the constant OH⁻ hydration and entropy terms, which shift the whole series by roughly the same amount and so don't affect the trend):

\text{Mg(OH)}_2:\ -2993+(-1921) = -4914\ \text{kJ mol}^{-1}

\text{Ba(OH)}_2:\ -2313+(-1305) = -3618\ \text{kJ mol}^{-1}

The Mg(OH)₂ combination is more negative, meaning the lattice-energy penalty dominates and net dissolution is comparatively less favorable — consistent with Mg(OH)₂ being the least soluble hydroxide and Ba(OH)₂ the most soluble, since the proxy becomes progressively less negative (more favorable to dissolution) down the group. For sulfates, do the same subtraction:

\text{MgSO}_4:\ -2848+(-1921) = -4769\ \text{kJ mol}^{-1}

\text{BaSO}_4:\ -2423+(-1305) = -3728\ \text{kJ mol}^{-1}

Here the numbers also become less negative down the group by this crude proxy — the real discriminator is not visible in this simplified two-term subtraction alone, because both terms include the full anion-independent hydration collapse. The genuine reversal requires comparing the rate of change relative to the anion's own hydration enthalpy (roughly constant at about −1330 kJ mol⁻¹ for SO₄²⁻ across the series and about −460 kJ mol⁻¹ for OH⁻ per ion, doubled for two ions): adding those constants, Mg(OH)₂ gives $-4914 - 920 = -5834$ kJ mol⁻¹, Ba(OH)₂ gives $-3618-920=-4538$ kJ mol⁻¹ (difference of 1296 kJ mol⁻¹ across the group), while MgSO₄ gives $-4769-1330=-6099$ kJ mol⁻¹ and BaSO₄ gives $-3728-1330=-5058$ kJ mol⁻¹ (difference of 1041 kJ mol⁻¹). The sulfate proxy changes less across the group in absolute terms relative to its starting depth, and once the (small, but real) lattice-energy entropy and second-hydration-shell corrections are included, the accepted experimental result is unambiguous: solubility of MSO₄ falls from freely soluble (Mg, Ca partially) to essentially insoluble (Sr, Ba), the reverse of the hydroxide trend, exactly as the qualitative mechanism predicted.

Limiting-case check: if the anion were artificially shrunk to the point $r_- \to 0$ (a bare point charge), every salt in the family would behave like the hydroxide extreme — solubility should rise monotonically and steeply down the group, which is indeed the observed behavior for fluorides (small, hard anion) more so even than hydroxides.

Example 2 (applied) — Why BaSO₄ is safe as an X-ray contrast agent but MgSO₄ is not. Radiographic barium meals rely on BaSO₄'s solubility product being vanishingly small ( $K_{sp}\approx 1.1\times10^{-10}$ at 25 °C), so that even though Ba²⁺ is acutely toxic (it blocks potassium channels and can cause fatal cardiac arrhythmia), essentially none of it enters solution while passing through the gastrointestinal tract. Using $K_{sp}=[\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]$ and assuming a saturated solution with only the salt itself present (so $[\mathrm{Ba^{2+}}]=[\mathrm{SO_4^{2-}}]=s$ ):

s = \sqrt{K_{sp}} = \sqrt{1.1\times10^{-10}} = 1.05\times10^{-5}\ \text{mol L}^{-1}

Converting to a mass concentration (using $M(\mathrm{BaSO_4}) = 233.4\ \mathrm{g\,mol^{-1}}$ ):

1.05\times10^{-5}\ \text{mol L}^{-1} \times 233.4\ \text{g mol}^{-1} = 2.45\times10^{-3}\ \text{g L}^{-1}

This is roughly 2.5 mg per liter, far below any toxic dose even for repeated imaging. Contrast this with MgSO₄, whose solubility exceeds 350 g per liter at room temperature — a difference of more than five orders of magnitude in mass solubility for chemically analogous salts, entirely explained by the lattice-energy/hydration-energy crossover: Mg²⁺'s intense hydration enthalpy (−1921 kJ mol⁻¹) easily outcompetes the SO₄²⁻ lattice penalty for such a small cation, while Ba²⁺'s much weaker hydration enthalpy (−1305 kJ mol⁻¹) cannot overcome a lattice energy (2423 kJ mol⁻¹) that has barely fallen from Mg's value (2848 kJ mol⁻¹) despite the huge radius increase — exactly the "anion-dominated, radius-insensitive" lattice term predicted by the size-mismatch mechanism.

Limiting-case check: as the anion is made progressively larger still (e.g., comparing to the much bigger, softer perchlorate or perrhenate anions), lattice energy differences across Group 2 shrink toward zero for all these salts, and essentially every Group 2 perchlorate becomes freely soluble — consistent with lattice energy ceasing to be cation-radius-sensitive at all once the anion is large enough to dominate $r_++r_-$ entirely.

Example 3 (exam/competition level) — Estimating the relative decomposition temperature shift between MgCO₃ and BaCO₃ from a lattice-energy argument. Given (illustrative Kapustinskii-consistent values, kJ mol⁻¹): $U_L(\mathrm{MgO}) = 3795$ , $U_L(\mathrm{BaO}) = 3054$ , $U_L(\mathrm{MgCO_3}) = 3181$ , $U_L(\mathrm{BaCO_3}) = 2611$ , and take the metal-independent intrinsic carbonate-breaking term as $\delta = +1400$ kJ mol⁻¹ (representing $\mathrm{CO_3^{2-}(g) \rightarrow O^{2-}(g) + CO_2(g)}$ , endothermic because it breaks C–O bonding character and destroys ionic stabilization of the extra charge). Compute $\Delta H_{dec}$ for each metal using

\Delta H_{dec} = \delta + \big[U_L(\mathrm{MCO_3}) - U_L(\mathrm{MO})\big]

For magnesium:

\Delta H_{dec}(\mathrm{Mg}) = 1400 + (3181 - 3795) = 1400 - 614 = 786\ \text{kJ mol}^{-1}

For barium:

\Delta H_{dec}(\mathrm{Ba}) = 1400 + (2611-3054) = 1400 - 443 = 957\ \text{kJ mol}^{-1}

The barium pathway is 171 kJ mol⁻¹ more endothermic — decomposition genuinely requires more energy input, consistent with BaCO₃ needing a furnace temperature above 1300 °C versus MgCO₃ decomposing below 550 °C. To convert this into an approximate decomposition temperature shift, use the crude assumption that decomposition proceeds once $\Delta G_{dec} = \Delta H_{dec} - T\Delta S_{dec} = 0$ , with $\Delta S_{dec}\approx +175\ \mathrm{J\,K^{-1}mol^{-1}}$ (dominated by the entropy of releasing one mole of CO₂ gas, roughly constant across the series since it is essentially the translational/rotational entropy of the gas itself):

T_{dec} = \frac{\Delta H_{dec}}{\Delta S_{dec}}

For magnesium carbonate:

T_{dec}(\mathrm{Mg}) = \frac{786\times10^3\ \text{J mol}^{-1}}{175\ \text{J K}^{-1}\text{mol}^{-1}} = 4491\ \text{K, i.e. } 4491\ \text{K (unphysically high — see below)}

This absurdly high number flags an important pedagogical point: $\delta$ , $U_L$ values, and $\Delta S_{dec}$ used here are illustrative order-of-magnitude constructs, not a self-consistent thermochemical dataset, so the absolute $T_{dec}$ from this toy calculation should not be trusted numerically — only the difference mechanism (larger $\Delta H_{dec}$ for Ba than Mg, translating into a higher required temperature) is the reliable, examinable conclusion. Using real tabulated thermodynamic values (standard enthalpies and entropies of formation), the accepted experimental decomposition onset temperatures are approximately 350 °C (MgCO₃), 840 °C (CaCO₃, well-known industrial lime-kiln value), roughly 1100–1200 °C (SrCO₃), and above 1300 °C (BaCO₃) — a monotonic rise consistent with the sign of the lattice-energy-difference argument even though the toy numbers above should not be quoted as literal predictions.

Limiting-case check: setting $\delta \to 0$ (imagining an anion that requires no intrinsic reorganization energy to release its gas) would make $\Delta H_{dec}$ equal purely to the lattice-energy difference, and the same monotonic-rise trend down the group would persist unchanged, because $\delta$ is a constant offset that shifts every member of the series equally and cannot by itself create or destroy a trend.

Limiting cases, scope & common traps

Situation	What happens	Why
Anion is very small (F⁻, O²⁻, OH⁻)	Solubility rises steeply down Group 2	Lattice energy is strongly cation-radius-sensitive; hydration collapse outpaces it
Anion is very large (SO₄²⁻, CrO₄²⁻, ClO₄⁻)	Solubility falls (or stays uniformly high for very soft anions) down Group 2	Lattice energy nearly cation-radius-insensitive; hydration collapse dominates and is unopposed
Anion of intermediate size (CO₃²⁻, oxalate)	Solubility is uniformly low across the whole group	Lattice energy sensitivity and hydration sensitivity are comparably matched; small differences do not overturn a generally unfavorable $\Delta G_{soln}$
Carbonate/nitrate thermal decomposition	Decomposition temperature rises steadily Mg → Ba	Oxide product lattice energy (small anion) drops faster with $r_+$ than reactant salt lattice energy (large anion)
Anion made artificially enormous (e.g. very large complex anions)	All Group 2 salts become similarly soluble/similarly thermally fragile	Lattice energy becomes almost entirely anion-controlled, erasing cation-dependent trends
Be salts (not strictly Group 2 in the periodic trend sense)	Frequently anomalous, often covalent character dominates	Be²⁺ is so small and polarizing that simple ionic lattice/hydration models break down; partial covalency intervenes

The chemist behind it

The quantitative backbone of this article — the idea that lattice energy can be estimated from ionic charges and radii alone, without needing a full crystal structure determination — comes from the Russian crystallographer and thermodynamicist Anatoli Fedorovich Kapustinskii, whose 1933 to 1956 body of work generalized the Born–Landé lattice-energy expression into a form usable for any ionic solid given only its formula unit and approximate ionic radii. Kapustinskii noticed empirically that the ratio of the true Madelung constant to the number of ions per formula unit is nearly the same across many different crystal structure types, which let him replace a structure-specific Madelung constant with a simple counting rule — a dramatic simplification that made lattice-energy estimation accessible for compounds whose crystal structures were unknown or ambiguous, exactly the situation faced by chemists trying to rationalize solubility trends across an entire periodic group before detailed diffraction data existed for every salt.

Origins · 1 minAnatoli Fedorovich Kapustinskii and how this was discovered

Anatoli Fedorovich Kapustinskii

1906-1960 · Soviet physical chemistry and crystal thermodynamics, Moscow

Kapustinskii worked within the Soviet school of physical chemistry associated with the Karpov Institute, applying electrostatic lattice models to problems of practical industrial relevance — including the thermal stability and processing behavior of carbonate and sulfate ores, the very compound classes this article is built around. His equation remains the standard back-of-envelope tool taught alongside the Born–Landé and Born–Mayer equations wherever lattice energy is introduced.

History

- 1918 — Max Born and Alfred Landé formulate the Born–Landé equation, expressing lattice energy from electrostatic and repulsive terms.

- 1933 — Anatoli Kapustinskii proposes his generalized, structure-independent lattice-energy formula.

- 1938–1950s — Systematic tabulation of Group 2 hydroxide and sulfate solubility products clarifies the opposite-direction trends discussed here.

- 1950s — Industrial lime and cement chemistry drives detailed study of carbonate decomposition temperatures across Group 2 and beyond.

- 1970s onward — Modern thermochemical cycles (Born–Haber, solvation cycles) formalize the hydration-vs-lattice-energy competition using calorimetric hydration enthalpy data.

Why does the same "smaller cation, higher charge density" logic give opposite solubility trends for hydroxides and sulfates?

Because solubility is governed by a difference between two competing energies — lattice energy (favoring the solid) and hydration energy (favoring the dissolved ions) — and those two energies do not respond to cation size in the same way for every anion. Both terms fall as the cation grows, but the lattice-energy term's fall depends strongly on how much the anion's own radius already dominates the interionic distance. For a small anion like OH⁻, lattice energy is highly cation-radius-sensitive and falls fast, letting hydration's steep fall win the trend (solubility rises). For a large anion like SO₄²⁻, lattice energy is nearly cation-radius-insensitive, so it barely falls at all, while hydration still falls sharply — leaving lattice energy to dominate and solubility to fall. The anion, not the cation alone, decides which trend you see.

Does this size-mismatch reasoning apply to alkali metal (Group 1) salts as well?

Only weakly, because Group 1 cations carry a single charge rather than a double charge, so their lattice and hydration energies are both much smaller in magnitude and much less sensitive to radius changes relative to a fixed anion size. The same qualitative direction of argument survives — for instance, lithium salts of small anions can show anomalously low solubility (e.g. LiF) compared to heavier alkali fluorides — but the effect is far less dramatic than in Group 2, precisely because $z_+z_-$ in the lattice-energy expression is smaller (1 instead of 2), making the whole lattice-energy scale, and hence its sensitivity to radius mismatch, correspondingly weaker.

Why do Group 2 carbonates decompose to the oxide, while Group 1 carbonates (except lithium) barely decompose at all below very high temperatures?

Group 1 cations are singly charged and generally larger than their Group 2 neighbors, giving them much lower charge density and far less polarizing power. A carbonate ion sitting next to a large, weakly polarizing Group 1 cation experiences little electrostatic distortion, so its C–O bonding framework remains stable to very high temperatures. Lithium carbonate is the partial exception because Li⁺ is small enough (76 pm, similar to Mg²⁺) to exert comparable polarizing power despite its single charge, which is why Li₂CO₃ decomposes at a noticeably lower temperature than the other alkali carbonates — reinforcing that charge density, not charge alone, is the operative variable.

How reliable is the Kapustinskii equation for predicting these trends quantitatively, rather than just qualitatively?

The Kapustinskii equation typically reproduces experimental lattice energies (from Born–Haber cycles) to within about 5% for simple ionic salts with well-behaved, hard, spherical ions, which is good enough to reproduce the correct sign and rough magnitude of the trends discussed here. It becomes less reliable for anions with significant covalent character in their internal bonding (like carbonate and nitrate, which are themselves covalently bonded polyatomic ions with delocalized π systems) or for cations with significant polarizing power that induces covalency (notably Be²⁺). For these article's purposes — ranking trends across a well-behaved group of ions — the equation's qualitative reliability is excellent even where its absolute numerical accuracy is only approximate.

Why is BeSO₄ or Be(OH)₂ often excluded or treated as an outlier in these trend discussions?

Beryllium's exceptionally small ionic radius (about 27–45 pm depending on coordination) and very high charge density push it toward genuinely covalent, rather than purely ionic, bonding in many of its compounds, so the electrostatic point-charge assumptions behind lattice energy and hydration energy models start to break down. Be²⁺ in water is more accurately described as forming a strongly covalent, kinetically inert aqua complex $[\mathrm{Be(H_2O)_4}]^{2+}$ rather than a simple hydrated point charge, and its hydroxide and other salts show anomalies (e.g., pronounced amphoterism) that don't fit smoothly onto the ionic-radius trend line used for Mg through Ba.

Is the entropy term (

T\Delta S

) ever large enough to overturn these trends?

For dissolution, the entropy term is usually a modest correction (tens of kJ mol⁻¹ at most) compared to lattice and hydration enthalpies that run into the thousands of kJ mol⁻¹, so it shifts absolute solubility values but rarely reverses the qualitative direction of a trend across a whole group. For thermal decomposition, entropy is far more important in an absolute sense — releasing one mole of gas contributes a large, roughly constant positive $\Delta S$ that is why decomposition is thermodynamically possible at all at finite temperature — but because that entropy contribution is nearly identical for every member of the Group 2 carbonate or nitrate series, it does not change the relative ordering of decomposition temperatures, only the absolute temperature scale at which each one occurs.

What laboratory observation would most directly test whether the size-mismatch mechanism, rather than some other explanation, is really responsible for these trends?

A clean test is to compare a series of anions of systematically increasing size against the same cation series and look for the predicted crossover: as the anion grows from something F⁻/OH⁻-sized to something CrO₄²⁻/SO₄²⁻-sized, the solubility trend down Group 2 should flip from increasing to decreasing at some intermediate anion size, tracking the point where lattice energy's cation-radius sensitivity falls below hydration energy's. This crossover is observed experimentally (fluorides and hydroxides rise in solubility down the group; sulfates and chromates fall), which is strong support that the mechanism — rather than some anion-specific chemistry unrelated to size — is the operative cause.