Ionic (Saline) Hydrides of the s-Block

Drop a piece of sodium metal into an atmosphere of pure hydrogen gas. Warm it gently past 300 °C. Something conceptually strange happens: hydrogen, the element that forms H⁺ in acids and H⁺(aq) in every general-chemistry course, is forced to accept an electron instead. The product, NaH, is a white crystalline solid. It is built from Na⁺ cations and H⁻ anions. These stack in the same rock-salt lattice as NaCl. This is the defining act of saline (ionic) hydride chemistry. A metal so electropositive can reduce hydrogen all the way to the hydride ion, H⁻. This species exists nowhere in aqueous chemistry, because it instantly rips a proton from water. This matters because the saline hydrides sit at a genuine boundary in the periodic table's bonding character. They sit alongside the Classification of Binary Hydrides that separates ionic, metallic, and covalent hydrogen compounds. Only the most electropositive metals form true saline hydrides with rock-salt-type structures. These are the alkali metals Li–Cs and the heavier alkaline earths Ca, Sr, Ba. Beryllium and magnesium hydrides, by contrast, are covalent/polymeric. This divergence is itself a clean lesson. It shows how electronegativity and ionic radius govern the ionic-covalent transition. Getting this classification right is essential scaffolding for everything downstream. So is understanding why it falls where it does. That includes the extreme reducing power and hydridic reactivity of these compounds. It also includes their industrial roles as reducing agents, drying agents, and hydrogen-storage materials. The big picture is a thermodynamic cycle. Forming M⁺H⁻(s) from M(s) and ½H₂(g) requires atomizing and ionizing the metal. Crucially, it also requires supplying an electron to hydrogen to make H⁻. This process is endothermic, because hydrogen's electron affinity is small. The second electron must overcome real electron-electron repulsion in the compact 1s² shell. The whole enterprise is rescued from thermodynamic disaster only by a large, favorable lattice energy. This is exactly the Lattice Energy and the Born–Haber Cycle machinery. This is why saline hydride stability tracks lattice energy trends. It is largest for the smallest cations. It does not track electron affinity or reduction potential in any simple way. This article works through the Born–Haber energetics that make H⁻ salts possible at all. It covers the NaCl-type (and modified) structures adopted by LiH–CsH and CaH₂–BaH₂. It also covers periodic trends in thermal stability, lattice enthalpy, and reactivity across both groups. Three fully worked numerical examples anchor the qualitative trends in real numbers. These are a foundational Born–Haber cycle for NaH, an applied comparison of thermal stability across the alkali hydride series, and an exam-level lattice-energy/Kapustinskii calculation for CaH₂. You can check these numbers yourself.

Key takeaways
  • Saline hydrides (LiH–CsH, CaH₂–BaH₂) contain discrete H⁻ ions in ionic lattices, formed only by the most electropositive s-block metals.
  • LiH–CsH adopt the NaCl (rock-salt) structure; CaH₂–BaH₂ adopt a distorted PbCl₂-type structure with more complex coordination.
  • Formation is thermodynamically possible only because a large negative lattice enthalpy outweighs the endothermic cost of making H⁻ from H₂.
  • Thermal stability and lattice enthalpy both fall as the cation gets larger (Li→Cs, Ca→Ba), tracking the r₊ + r₋ dependence of Coulombic lattice energy.
  • H⁻ is a powerful Brønsted base and reducing agent: saline hydrides react violently with water/protic solvents and reduce many oxides and halides.
  • Ionic radius of H⁻ (~130–150 pm, compressible) is unusually variable because the anion has no nuclear charge to hold its diffuse electron cloud rigid.
  • Key idea: Saline hydride stability is a lattice-energy story — H⁻ formation is energetically uphill on its own, and only the strong electrostatic attraction in a small-cation rock-salt lattice makes the ionic hydride thermodynamically viable.

Where this lives today

Saline hydrides remain genuinely important reagents rather than historical curiosities. Sodium hydride, NaH, is a workhorse strong, non-nucleophilic base in organic and organometallic synthesis, used to deprotonate alcohols, amides, and active-methylene compounds cleanly, generating H₂ gas as the only byproduct — a direct, everyday consequence of H⁻'s extreme Brønsted basicity discussed above, and a preview of the reducing/hydridic reactivity explored further in Reducing and Hydridic Reactivity of s-Block Hydrides. Calcium hydride, CaH₂, is one of the most common laboratory desiccants for drying organic solvents (particularly hydrocarbons and ethers) because it reacts quantitatively and irreversibly with trace water, releasing measurable, stoichiometric H₂ that also serves as a built-in indicator of solvent dryness. Beyond the bench, saline and related complex hydrides are studied intensively as solid-state hydrogen-storage materials, since the volumetric hydrogen density of a compact ionic hydride lattice can exceed that of compressed or even liquid H₂ gas; LiH and related complex hydrides (LiAlH₄, NaBH₄ — themselves built partly on the hydridic character established here) are central to ongoing work in Industrial Uses of s-Block Hydrides and Metals, including battery and fuel-cell adjacent chemistry. Lithium hydride specifically has found application as a lightweight neutron shielding and moderating material in nuclear technology, exploiting hydrogen's exceptional efficiency at slowing fast neutrons through elastic collisions, while also serving historically as a component in thermonuclear weapons research (lithium-6 deuteride). In metallurgy, calcium hydride is used as a reducing agent to extract certain reactive and refractory metals from their oxides, exploiting H⁻'s powerful reducing character — a theme that generalizes across the entire family of saline hydrides and connects directly to their industrial value as clean, byproduct-friendly (H₂-releasing) reducing and desiccating agents across the chemical industry.

Ask the tutor: Why does the Kapustinskii equation give a reasonable lattice-enthalpy estimate for CaH₂ even without knowing its true (non-cubic) crystal structure?

The physical picture

Hydrogen is peculiar among elements because it can appear in three fundamentally different bonding environments with metals, and the saline hydrides represent only one extreme. In covalent/molecular hydrides (CH₄, NH₃, B₂H₆) hydrogen shares electrons roughly symmetrically or is bound in electron-deficient multicenter bonds. In metallic/interstitial hydrides (PdH_x, the transition-metal and rare-earth hydrides) hydrogen atoms occupy interstices in a metallic lattice, and the material remains a conductor. In saline hydrides, hydrogen is reduced outright: it accepts an electron from an electropositive metal to become H⁻, a closed-shell 1s² anion isoelectronic with helium, and the compound is best described as a normal ionic solid, M⁺H⁻ or M²⁺(H⁻)₂. The chemical identity of H⁻ is worth dwelling on because it is counter-intuitive. In water, hydrogen is overwhelmingly found as H⁺ (really H₃O⁺), because water's oxygen lone pairs stabilize the proton and because H⁻ would instantaneously protonate from any O–H bond, releasing H₂ gas:

H−+H2O→H2(g)+OH−\text{H}^- + \text{H}_2\text{O} \rightarrow \text{H}_2(g) + \text{OH}^-

This reaction is essentially irreversible and enormously exothermic, which is precisely why saline hydrides must be handled and characterized under strictly anhydrous, often inert-atmosphere conditions, and why H⁻ never survives in aqueous solution. The existence of a stable solid salt of H⁻ is a lattice-energy phenomenon: in the solid state, H⁻ is locked into a low-energy electrostatic environment from which it cannot escape without disrupting the whole crystal, whereas in solution it is a free, wildly basic ion with no such protection. Only metals from Groups 1 and 2 that are sufficiently electropositive form these true ionic hydrides. Down Group 1, all of Li, Na, K, Rb, Cs form MH with the rock-salt (NaCl-type) structure: each M⁺ is octahedrally coordinated by six H⁻ and vice versa, exactly like Na⁺ and Cl⁻ in table salt. Down Group 2, Be and Mg do not form saline hydrides — BeH₂ is a covalent polymeric solid with three-center two-electron Be–H–Be bridges (structurally analogous to (BeH₂)ₙ chains, echoing the electron-deficient bonding seen in boranes), and MgH₂, while ionic in character, adopts the rutile structure with only moderate ionic character and significant covalent contribution. It is only Ca, Sr, and Ba — the larger, more electropositive alkaline earths — that form clearly ionic dihydrides, CaH₂, SrH₂, BaH₂, and these adopt a distorted PbCl₂-type (cotunnite-related) structure rather than a simple fluorite lattice, because H⁻ is not quite the right size/shape for a perfect CaF₂-type arrangement. The size of H⁻ itself deserves comment because it is genuinely unusual among anions. Unlike halide ions, whose radii are set by a nucleus of substantial charge holding a compact electron cloud, H⁻ has only one proton holding onto two electrons. The electron cloud is diffuse, highly polarizable, and its effective radius depends strongly on the surrounding cation — it swells in the field of a weak, large cation and compresses next to a small, polarizing cation. Crystallographic estimates place r(H⁻) anywhere from about 130 pm (in LiH, where the small Li⁺ compresses it) to 150 pm or more (in CsH, where the large, weakly polarizing Cs⁺ lets it expand). This anion "softness" is itself diagnostic of the ionic-but-not-perfectly-ionic character of these solids, and it is the reason simple hard-sphere ionic radius tables give only approximate lattice-energy predictions for hydrides compared to halides.

Figure
Rock-salt (NaCl-type) structure of LiH–CsH: alternating M⁺ and H⁻ on a cubic close-packed lattice, each ion octahedrally coordinated by six of the other
NaCl-type lattice: M⁺H⁻ (Li, Na, K, Rb, Cs) M⁺ (Li…Cs), r = 76–170 pm H⁻, r ≈ 130–150 pm (compressible) Each ion: 6-coordinate octahedral; coordination number 6:6
Figure
Born–Haber cycle for NaH(s): why the exothermic overall ΔfH depends on lattice enthalpy overcoming the endothermic electron affinity step
Born–Haber cycle: Na(s) + ½H₂(g) → NaH(s) Na(s) + ½H₂(g) ΔfH° = −56 kJ/mol +108 (Na sublimation) +218 (½ D(H–H)) +496 (IE₁ Na) −73 (EA H, H+e⁻→H⁻) −809 (lattice enthalpy) Na⁺(g) + H⁻(g) → NaH(s) Sum: 108+218+496−73−809 = −60 ≈ −56 kJ/mol (lit. value; small rounding)

The core ideas

The central quantitative relationship governing whether a saline hydride can form is the Born–Haber cycle, which decomposes the overall formation enthalpy into atomization, ionization, electron-affinity, and lattice terms:

ΔfH∘(MH,s)=ΔsubH(M)+12D(H−H)+IE1(M)+ΔEAH(H)+ΔLH(MH)\Delta_f H^\circ(\mathrm{MH,s}) = \Delta_{sub}H(\mathrm{M}) + \tfrac{1}{2}D(\mathrm{H-H}) + IE_1(\mathrm{M}) + \Delta_{EA}H(\mathrm{H}) + \Delta_{L}H(\mathrm{MH})

where:

  • ΔsubH(M)\Delta_{sub}H(\mathrm{M}) = enthalpy to atomize/sublime the solid metal to gaseous atoms (endothermic, +);
  • 12D(H−H)\tfrac12 D(\mathrm{H-H}) = half the H–H bond dissociation enthalpy, to make one mole of H atoms from H₂(g) (endothermic, +);
  • IE1(M)IE_1(\mathrm{M}) = first ionization energy of the metal atom, M(g) → M⁺(g) + e⁻ (endothermic, +);
  • ΔEAH(H)\Delta_{EA}H(\mathrm{H}) = enthalpy change for H(g) + e⁻ → H⁻(g) (this is endothermic, small positive value, unlike halogens);
  • ΔLH(MH)\Delta_L H(\mathrm{MH}) = lattice enthalpy of formation, M⁺(g) + H⁻(g) → MH(s) (strongly exothermic, large negative).

Reading this equation: every step before the last is either endothermic or only mildly exothermic, and critically the electron-affinity step for hydrogen is endothermic — adding a second electron to the compact hydrogen 1s orbital costs energy because of strong electron-electron repulsion in such a small orbital, unlike a halogen atom where the incoming electron fills a larger p-shell more comfortably and net EA is exothermic. The entire viability of saline hydride formation is therefore carried by the last term: the lattice enthalpy must be large and negative enough to make the sum favorable. This is why the reaction only proceeds for the most electropositive metals (low IE₁, or in Group 2 a favorable combination of IE₁+IE₂ and small Δ_sub H) combined with small enough cation radius to make Δ_L H very large. The lattice enthalpy itself is captured, to reasonable accuracy for a rock-salt structure, by the Born–Landé / Born–Mayer type expression:

ΔLH=−NAMz+z−e24πε0(r++r−)(1−1n)\Delta_L H = -\frac{N_A M z_+ z_- e^2}{4\pi\varepsilon_0 (r_+ + r_-)}\left(1 - \frac{1}{n}\right)

where:

  • NAN_A = Avogadro's number;
  • MM = Madelung constant (1.7476 for the NaCl-type structure);
  • z+,z−z_+, z_- = ionic charges (+1 and −1 for MH, or effectively +2 for MH₂ built from M²⁺ and two H⁻);
  • e2/4πε0e^2/4\pi\varepsilon_0 = the Coulomb energy prefactor;
  • r++r−r_+ + r_- = the sum of ionic radii (the interionic distance);
  • nn = the Born exponent, reflecting short-range electron-cloud repulsion (typically 7–10 for these ions).

Reading this equation: lattice enthalpy scales as 1/(r++r−)1/(r_+ + r_-) and as z+z−z_+z_-, so it is most negative (strongest binding) for the smallest cations and for the doubly-charged alkaline-earth hydrides. This single dependence is the quantitative backbone of every trend discussed below: LiH has the largest lattice enthalpy magnitude of the alkali hydrides and CsH the smallest; CaH₂ has a substantially larger lattice enthalpy magnitude (per mole of "MH unit," and certainly per formula unit given the doubled charge) than the alkali hydrides, which correlates with the higher thermal stability of the alkaline-earth dihydrides.

Derivation AWhy hydride formation is only possible for electropositive metals: signs and magnitudes in the Born–Haber sum

Goal. Show, term by term, why the alkali/heavy alkaline-earth metals succeed in making stable ionic hydrides while a hypothetical "hydride" of a less electropositive or more electronegative element would not form, and identify exactly which term is decisive.

Step 1 — Tally the endothermic cost of getting to M⁺(g) + H(g). For any metal, three steps must be paid up front: sublimation of the solid metal to atoms, dissociation of H₂ to H atoms (halved, since one H is needed per M⁺), and ionization of the metal atom. For an alkali metal these sum to roughly 300–650 kJ mol⁻¹ depending on the metal (dominated by IE₁, which ranges from 520 kJ mol⁻¹ for Li down to 376 kJ mol⁻¹ for Cs). For an alkaline-earth metal, one must pay both IE₁ and IE₂ (since M²⁺ is required), a substantially larger cost — for Ca this is 590 + 1145 = 1735 kJ mol⁻¹ just for ionization, far more than any alkali metal.

Step 2 — Add the endothermic hydrogen electron-affinity step. Unlike a halogen X, for which X(g) + e⁻ → X⁻(g) is strongly exothermic (Cl: −349 kJ mol⁻¹), hydrogen's analogous step is endothermic by a small positive amount, of order +30 to +80 kJ mol⁻¹ depending on the source and sign convention used (values vary because H⁻'s electron affinity from H is small and some estimates place the true EA of hydrogen — the reverse process, H⁻(g) → H(g)+e⁻ — near +73 kJ mol⁻¹, meaning forming H⁻ costs ≈+73 kJ mol⁻¹). This term never helps; it only ever adds a modest energetic penalty, unlike halide formation which contributes a large favorable term.

Step 3 — Recognize that only the lattice enthalpy term can rescue the cycle. Since every earlier step is unfavorable (endothermic) or at best a wash, the sum of the first four terms is positive — often several hundred kJ mol⁻¹. For the overall ΔfH∘\Delta_f H^\circ to be negative (spontaneous, stable compound), the lattice enthalpy magnitude must exceed this positive sum. Because lattice enthalpy scales as z+z−/(r++r−)z_+z_-/(r_++r_-), it is maximized when the cation is small and highly charged and the anion is small. This is exactly satisfied by Li⁺, Na⁺, K⁺, and by Ca²⁺, Sr²⁺, Ba²⁺ combined with the small, compressible H⁻ ion, but is not well satisfied by large, singly-charged, weakly-charge-dense cations of low electropositivity, nor by any attempt to force a genuinely electronegative element to accept an electron into an already electron-rich orbital pattern.

Step 4 — Conclude the periodicity of saline hydride formation. The pattern that only Group 1 and the heavier Group 2 metals form true ionic hydrides is therefore not accidental: it requires simultaneously (i) low enough ionization energy that steps 1–2 are not prohibitive, and (ii) small enough cationic radius that the lattice term is large enough to compensate. Be and Mg fail on both counts in combination — although Mg²⁺ is reasonably small, the very high IE₁+IE₂ of Be and Mg and the covalent-directional character of their small, highly polarizing cations push MgH₂ and BeH₂ toward covalent/polymeric bonding rather than simple ionic lattices, exactly as predicted by Fajans' rules for small, highly charged cations.

Go deeper · Derivation B · 2 minTrend in thermal stability down each group: why LiH is the most thermally robust alkali hydride and BaH₂ decomposes more readily than CaH₂
Derivation BTrend in thermal stability down each group: why LiH is the most thermally robust alkali hydride and BaH₂ decomposes more readily than CaH₂

Goal. Derive, from the same lattice-enthalpy dependence, why thermal stability (resistance to the reverse decomposition MH(s) → M(s) + ½H₂(g)) decreases down each group, in direct correlation with decreasing lattice enthalpy magnitude.

Step 1 — Write the decomposition as the reverse of formation. Thermal decomposition is simply the reverse reaction, so its enthalpy is −ΔfH∘(MH)-\Delta_f H^\circ(\mathrm{MH}). If ΔfH∘\Delta_f H^\circ becomes less negative (formation less favorable) down a group, decomposition becomes easier (less endothermic), and the temperature at which the equilibrium H₂ pressure reaches 1 atm — a practical index of thermal stability — falls.

Step 2 — Track how each Born–Haber term changes down Group 1. Moving Li → Na → K → Rb → Cs: the sublimation enthalpy decreases modestly, and first ionization energy decreases substantially (520 → 496 → 419 → 403 → 376 kJ mol⁻¹), both of which by themselves would make formation more favorable for heavier metals. The hydrogen electron-affinity term is constant (same H⁻ each time). The decisive term is the lattice enthalpy, which becomes markedly less negative down the group because cation radius increases sharply (Li⁺ 76 pm → Cs⁺ 170 pm), and ΔLH∝−1/(r++r−)\Delta_L H \propto -1/(r_++r_-).

Step 3 — Show the lattice term dominates over the ionization-energy term. Numerically, IE₁ falls by only about 144 kJ mol⁻¹ from Li to Cs, while the magnitude of the lattice enthalpy falls by considerably more (roughly 809 kJ mol⁻¹ for LiH down to about 649 kJ mol⁻¹ for CsH, a drop of ~160 kJ mol⁻¹, but combined with the smaller radius-driven changes in the other terms, e.g. sublimation enthalpy, the net effect is that ΔfH∘\Delta_f H^\circ becomes less negative down the group). Because the favorable trend in IE₁ is outpaced by the unfavorable trend in lattice enthalpy, the net formation enthalpy decreases in magnitude (becomes less exothermic) from LiH to CsH, so LiH is thermodynamically the most stable and CsH the least, and correspondingly LiH survives to the highest temperature before measurable decomposition.

Step 4 — Apply the identical logic to Group 2, and note the doubled-charge amplification. Moving Ca → Sr → Ba, cation radius again increases (Ca²⁺ 100 pm → Ba²⁺ 135 pm), and because lattice enthalpy for a 2:1 salt scales with z+z−=2z_+z_- = 2 rather than 1, the radius-driven fall in lattice enthalpy is numerically larger per formula unit than for the alkali hydrides, reinforcing the same qualitative trend: CaH₂ is more thermally robust than BaH₂. Practically, CaH₂ can be heated to several hundred degrees Celsius without significant decomposition and is used industrially as a drying agent and hydrogen source precisely because of this robustness, whereas BaH₂ decomposes at correspondingly lower temperatures.

Worked examples

Example 1 (Foundational): Complete Born–Haber cycle for NaH and confirm the sign logic. Given data (all in kJ mol⁻¹, standard literature values): ΔsubH(Na)=+108\Delta_{sub}H(\mathrm{Na}) = +108; D(H−H)=+436D(\mathrm{H-H}) = +436 so 12D=+218\tfrac12 D = +218; IE1(Na)=+496IE_1(\mathrm{Na}) = +496; the "electron affinity" step for hydrogen, H(g) + e⁻ → H⁻(g), taken as ΔEAH=−73\Delta_{EA}H = -73 kJ mol⁻¹ using the convention that this is slightly exothermic overall in some compilations (values in the literature range from mildly exothermic, ≈ −73 kJ mol⁻¹, to mildly endothermic depending on source and definition; here we use the commonly tabulated Born–Haber value of −73 kJ mol⁻¹ for internal consistency with the experimental ΔfH); lattice enthalpy ΔLH(NaH)=−809\Delta_L H(\mathrm{NaH}) = -809 kJ mol⁻�1 (calculated/experimental composite for the rock-salt lattice). Find ΔfH∘(NaH)\Delta_f H^\circ(\mathrm{NaH}) and compare to the experimental value of −56 kJ mol⁻¹. Sum the cycle:

ΔfH∘=(+108)+(+218)+(+496)+(−73)+(−809)\Delta_f H^\circ = (+108) + (+218) + (+496) + (-73) + (-809)
ΔfH∘=108+218+496−73−809=−60 kJ mol−1\Delta_f H^\circ = 108 + 218 + 496 - 73 - 809 = -60 \ \text{kJ mol}^{-1}

This is close to the experimental value of −56 kJ mol⁻¹ (the 4 kJ mol⁻¹ discrepancy reflects the approximate nature of the tabulated lattice enthalpy and the sensitivity of the H⁻ electron-affinity value to the source used). The important qualitative check: without the lattice term, the sum of the first four terms is

108+218+496−73=+749 kJ mol−1108 + 218 + 496 - 73 = +749 \ \text{kJ mol}^{-1}

strongly endothermic — confirming that it is the lattice enthalpy alone that converts an otherwise wildly unfavorable process into a stable, isolable ionic solid. As a limiting-case sanity check: if the lattice enthalpy magnitude were only, say, −400 kJ mol⁻¹ (roughly what a much larger, singly-charged cation with r₊ ≈ 300 pm might give), ΔfH∘\Delta_f H^\circ would be +349 kJ mol⁻¹ and NaH-like compound would simply not exist as a stable solid — precisely the situation for metals too large or too weakly charged to generate sufficient lattice stabilization.

Example 2 (Applied): Rank the thermal stability of LiH, NaH, and KH using lattice enthalpy and ionic radius data, and predict relative decomposition temperatures. Using cation radii (6-coordinate, Shannon): Li⁺ = 76 pm, Na⁺ = 102 pm, K⁺ = 138 pm, and a fixed effective H⁻ radius of ≈ 137 pm (an intermediate compromise value, acknowledging H⁻'s compressibility), estimate the relative lattice enthalpies using the simple Coulombic scaling ΔLH∝−1/(r++r−)\Delta_L H \propto -1/(r_++r_-), normalized so LiH's calculated value matches its accepted lattice enthalpy of −909 kJ mol⁻¹ (a slightly different literature value from Example 1's rock-salt composite, since different compilations use different H⁻ radii; we use −909 kJ mol⁻¹ here for LiH specifically to illustrate the scaling method). Interionic distances:

r++r−(LiH)=76+137=213 pmr_+ + r_- (\mathrm{LiH}) = 76 + 137 = 213\ \text{pm}
r++r−(NaH)=102+137=239 pmr_+ + r_- (\mathrm{NaH}) = 102 + 137 = 239\ \text{pm}
r++r−(KH)=138+137=275 pmr_+ + r_- (\mathrm{KH}) = 138 + 137 = 275\ \text{pm}

Scaling lattice enthalpy inversely with distance (holding Madelung constant and Born exponent fixed, a reasonable first approximation since all three share the NaCl structure type):

ΔLH(NaH)≈−909×213239=−810 kJ mol−1\Delta_L H(\mathrm{NaH}) \approx -909 \times \frac{213}{239} = -810\ \text{kJ mol}^{-1}
ΔLH(KH)≈−909×213275=−704 kJ mol−1\Delta_L H(\mathrm{KH}) \approx -909 \times \frac{213}{275} = -704\ \text{kJ mol}^{-1}

These estimates (−909, −810, −704 kJ mol⁻¹ for Li, Na, K respectively) track the experimentally observed order of thermal stability: LiH is stable to the highest temperature (decomposing only above ~720 °C, close to its melting point), NaH decomposes noticeably around 425 °C, and KH is thermally the least robust of the three, consistent with the monotonic fall in lattice enthalpy magnitude. As a check on the scaling assumption, note that the ratio of lattice enthalpies (810/909 = 0.891) is close to the inverse ratio of distances (213/239 = 0.891) by construction — confirming the arithmetic — while the physically meaningful result is that this ordering matches the experimentally measured decomposition temperature order, validating the simple 1/(r++r−)1/(r_++r_-) model as a good first approximation even though it neglects small differences in Born exponent and Madelung geometry distortion.

Example 3 (Exam-level): Estimate the lattice enthalpy of CaH₂ using the Kapustinskii equation and use it to rationalize CaH₂'s superior thermal stability relative to NaH. The Kapustinskii equation estimates lattice enthalpy without needing a known crystal structure, useful for MH₂ where the true structure (distorted PbCl₂-type, coordination number 7–9) is more complex than simple rock-salt:

ΔLH≈−1.2025×105 ν z+z−r++r−(1−34.5r++r−) kJ mol−1\Delta_L H \approx -\frac{1.2025\times10^{5}\, \nu\, z_+ z_- }{r_+ + r_-}\left(1 - \frac{34.5}{r_+ + r_-}\right)\ \text{kJ mol}^{-1}

where ν\nu is the number of ions per formula unit (3 for CaH₂: one Ca²⁺ + two H⁻), z+z−=(+2)(−1)=−2z_+z_- = (+2)(-1) = -2 (we take ∣z+z−∣=2|z_+ z_-| = 2 for the magnitude calculation), and r++r−r_+ + r_- is in picometers, with r+(Ca2+)=100r_+(\mathrm{Ca}^{2+}) = 100 pm (6-coordinate Shannon value, used as an approximation despite CaH₂'s higher coordination number) and r−(H−)=130r_-(\mathrm{H}^-) = 130 pm (compressed value appropriate to a small, doubly-charged, polarizing cation). Compute the interionic distance:

r++r−=100+130=230 pmr_+ + r_- = 100 + 130 = 230\ \text{pm}

Compute the bracket correction term:

1−34.5230=1−0.150=0.8501 - \frac{34.5}{230} = 1 - 0.150 = 0.850

Assemble the full expression:

ΔLH≈−1.2025×105×3×2230×0.850\Delta_L H \approx -\frac{1.2025\times10^{5} \times 3 \times 2}{230}\times 0.850
ΔLH≈−7.215×105230×0.850=−3137×0.850=−2666 kJ mol−1\Delta_L H \approx -\frac{7.215\times10^{5}}{230}\times 0.850 = -3137 \times 0.850 = -2666\ \text{kJ mol}^{-1}

This value (≈ −2666 kJ mol⁻¹) is roughly 3.3 times larger in magnitude than NaH's lattice enthalpy (−809 kJ mol⁻¹ from Example 1), driven by the z+z−=2z_+z_- = 2 factor (doubling the numerator relative to a hypothetical 1:1 charge) compounded by the smaller cation radius and the extra ion per formula unit (ν=3\nu=3 rather than 2). Physically: even though Ca must pay a much larger ionization cost (IE₁+IE₂ ≈ 1735 kJ mol⁻¹ versus Na's 496 kJ mol⁻¹), the roughly 1850 kJ mol⁻¹ extra lattice stabilization more than compensates, explaining why CaH₂ is a stable, thermally robust solid used industrially as a drying agent and portable hydrogen source, decomposing only well above 600 °C, in clear contrast to the more modest thermal ceiling of the alkali hydrides. As a limiting-case check: if instead z+z−z_+z_- were only 1 (hypothetically, as if Ca⁺H⁻ existed with singly-ionized calcium), the numerator would halve to give roughly −1333 kJ mol⁻¹, comparable in magnitude to the alkali hydrides — confirming that it is specifically the doubled charge, not merely the smaller radius, that drives CaH₂'s markedly larger lattice enthalpy.

Limiting cases, scope & common traps

SituationWhat actually happensWhy
Li, Na, K, Rb, Cs + H₂Form MH, NaCl-type structure, 6:6 coordinationSmall/moderate cation, low IE₁, strong lattice enthalpy dominates cycle
Ca, Sr, Ba + H₂Form MH₂, distorted PbCl₂-type structurez₊=2 gives very large lattice enthalpy, compensating high IE₁+IE₂
Be, Mg + H₂Covalent/polymeric hydrides (BeH₂ chains, rutile-type MgH₂ with covalent character)Cations too small/polarizing (Fajans' rules); electrostatic-only model breaks down
MH⁻ in waterInstant, violent reaction: H⁻ + H₂O → H₂ + OH⁻H⁻ is one of the strongest Brønsted bases known; no aqueous H⁻ chemistry exists
Heating MH near/above decomposition TMH(s) → M(s) + ½H₂(g), equilibrium H₂ pressure rises with TReverse of exothermic formation; entropy-driven at high T (ΔS>0 favors gas release)
r(H⁻) "constant" assumptionH⁻ radius actually varies ~130–150 pm across compoundsH⁻ has no core to resist deformation; strongly cation-dependent, unlike halides
Applying simple Born–Landé model to MH₂Works qualitatively but coordination geometry (7–9, not 6) reduces quantitative accuracyPbCl₂-type structure is a distorted, lower-symmetry packing, not ideal NaCl/fluorite

The chemist behind it

The conceptual foundation for classifying hydrides by bonding type — ionic/saline, covalent/molecular, and metallic/interstitial — is generally credited to the systematic structural and thermodynamic work carried out in the early-to-mid twentieth century, building on the broader ionic-crystal framework established by Max Born and Alfred Landé, whose lattice-energy theory (the Born–Landé equation) provided the first quantitative tool for testing whether a proposed ionic formula for a solid was energetically self-consistent. Applying that machinery specifically to metal hydrides — recognizing that H⁻ could be treated as a genuine anion with a crystallographic radius, and that its "abnormal," strongly cation-dependent size reflected weak nuclear control over a two-electron cloud — was a natural and important extension of ionic-crystal theory into a new and initially counter-intuitive chemical territory.

Origins · 1 minMax Born and how this was discovered
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Max Born
1882–1970 · Ionic crystal theory and the lattice-energy framework underlying saline hydride thermodynamics

Max Born, working alongside Alfred Landé and later Fritz Haber and Max Born (the Born–Haber cycle bears both names), developed the quantitative theory of ionic crystal lattice energies in the 1910s, treating a crystal as an array of point charges held together by Coulombic attraction and stabilized against collapse by short-range repulsive forces between overlapping electron clouds. This framework — refined into the Born–Landé and later Born–Mayer equations — remains the standard quantitative tool for testing whether any proposed ionic formulation, including the seemingly implausible idea of a stable solid salt of H⁻, is thermodynamically self-consistent. Born's broader legacy includes the probabilistic interpretation of the wavefunction in quantum mechanics, for which he received the Nobel Prize in Physics in 1954, but his crystal lattice theory, developed decades earlier, is the direct intellectual ancestor of every Born–Haber cycle used to rationalize hydride, halide, and oxide stability today.

History
- 1916–1919 — Born and Landé develop the quantitative theory of ionic lattice energies (Born–Landé equation).
- 1919 — Fritz Haber and Max Born formalize the thermodynamic cycle connecting atomization, ionization, electron affinity, and lattice energy (Born–Haber cycle).
- Early 1900s–1930s — Systematic preparation and structural characterization of alkali metal hydrides (LiH, NaH, KH) establishes the NaCl-type structural motif.
- Mid-20th century — X-ray and neutron diffraction studies confirm the rock-salt structure of alkali hydrides and the distorted PbCl₂-type structure of the heavier alkaline-earth dihydrides.
- 1950s onward — Systematic classification of binary hydrides into ionic/saline, covalent/molecular, and metallic/interstitial categories becomes standard in inorganic chemistry pedagogy.
Why is H⁻ described as "unusual" compared to halide anions of similar formal charge?

H⁻ has only a single proton holding onto two electrons, so its electron cloud is diffuse, highly polarizable, and poorly defended against distortion by the surrounding cation. Halide ions (F⁻, Cl⁻, etc.) have substantial nuclear charge and inner electron shells that keep their radius comparatively fixed across different compounds. H⁻'s effective ionic radius instead varies considerably, roughly 130 pm in LiH up to 150 pm or more in CsH, because a small, strongly polarizing cation like Li⁺ compresses the anion's electron cloud while a large, weakly polarizing cation like Cs⁺ allows it to expand. This variability is a direct fingerprint of H⁻'s minimal nuclear "grip" on its own electrons.

Why don't saline hydrides exist for transition metals or p-block metals?

Saline hydride formation requires the metal to be electropositive enough that losing an electron (or two) to reduce hydrogen is not prohibitively costly, and simultaneously requires the resulting lattice to be strongly stabilized electrostatically. Transition metals typically have higher ionization energies and, more importantly, tend to form metallic/interstitial hydrides instead, where hydrogen occupies lattice interstices without full electron transfer, because the d-band structure of the metal accommodates hydrogen's electron density differently. p-Block metals are generally not electropositive enough, and where hydrides do form (e.g., AlH₃, GaH₃) they are covalent or polymeric rather than ionic, consistent with Fajans' rules for higher-charge, smaller, more polarizing cations.

How does the PbCl₂-type structure of CaH₂ differ from the rock-salt structure of NaH?

In the NaCl-type structure adopted by the alkali hydrides, both cation and anion sit on equivalent, highly symmetric cubic close-packed sublattices with clean 6:6 octahedral coordination. The heavier alkaline-earth dihydrides (CaH₂, SrH₂, BaH₂) instead adopt a lower-symmetry, distorted PbCl₂-type (cotunnite-related) structure in which the two crystallographically distinct hydride sites have different, non-octahedral coordination environments (typically 7-coordinate and 9-coordinate around the cation, or equivalently different H⁻ environments), reflecting the need to pack two H⁻ ions per M²⁺ efficiently given the specific radius ratio and directional packing constraints — a geometry the simple cubic rock-salt arrangement cannot achieve for a 1:2 stoichiometry.

Is it correct to say saline hydrides are "ionic" in exactly the same sense as NaCl?

Largely yes for the alkali hydrides and the heavier alkaline-earth dihydrides, but with an important caveat: because H⁻'s electron cloud is so diffuse and polarizable, there is somewhat more covalent/polarization character mixed into M–H bonding than in the corresponding M–Cl bonding, especially for smaller, more polarizing cations like Li⁺. The classification as "ionic/saline" is a good first-order description validated by the rock-salt structure, high electrical conductivity of the melt (a classic ionic-solid test), and the clean Born–Haber energetics, but chemists understand it as the dominant bonding character rather than a claim of perfectly non-polarized point charges.

Why does CaH₂ survive heating to much higher temperatures than NaH before decomposing?

Thermal decomposition is the reverse of the exothermic formation reaction, so a more exothermic (more stable) hydride resists decomposition to higher temperatures. CaH₂'s formation is far more exothermic overall than NaH's despite Ca requiring a much larger ionization energy input (needing both IE₁ and IE₂), because the resulting lattice enthalpy is dramatically larger — driven by the doubled ionic charge (z₊z₋ = 2 instead of 1) and a comparably small cation radius. This large lattice-enthalpy "surplus" more than compensates for the extra ionization cost, giving CaH₂ a substantially more negative net formation enthalpy and correspondingly higher thermal stability.

What happens if you try to make a saline hydride from a very large, singly-charged cation, hypothetically?

The Born–Haber cycle predicts failure. As cation radius increases, lattice enthalpy magnitude falls (it scales as 1/(r++r−)1/(r_++r_-)), while the endothermic cost of forming H⁻ from H₂ remains essentially fixed. If the cation is also not especially electropositive (higher ionization energy), the sum of endothermic terms can exceed even a moderately large lattice enthalpy, making ΔfH∘\Delta_f H^\circ positive — the compound would not be thermodynamically stable as an ionic solid at ordinary conditions. This is effectively why the ionic hydride family is restricted to the most electropositive, small-to-moderate-radius metals of Groups 1 and 2, and does not extend to large, weakly electropositive metals.

Why is CaH₂ used as a drying agent rather than, say, NaH, in many labs?

Both react irreversibly and stoichiometrically with water to release H₂, so both can dry solvents in principle. CaH₂ is often preferred for bulk solvent drying (especially hydrocarbons and some ethers) because it is a relatively inexpensive, easy-to-handle solid dihydride that reacts controllably, is less pyrophoric/reactive in bulk handling than very finely divided NaH dispersions, and its reaction produces easily monitored H₂ evolution as a visual/quantitative dryness indicator. NaH is more commonly reserved for its role as a strong non-nucleophilic base in synthesis (often as an oil dispersion for safe handling) rather than as the default bulk drying agent, though both exploit the same underlying hydridic reactivity of H⁻.

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