Ionic Hydrides and Their Reactions

Introduction

In a salt-like hydride such as NaH, hydrogen is commonly modelled as H⁻. That is unusual compared with H at +1 in water and many acids. The hydride ion strongly accepts a proton, and the resulting two hydrogen atoms form H₂. This reaction can be described as proton transfer and also analysed through formal oxidation-state changes. Its products are predictable only when the reacting hydride and proton donor are specified.

Core explanation

Sodium hydride is represented as Na⁺H⁻. Sodium is +1 and hydride hydrogen −1. With water, the balanced molecular equation is NaH + H₂O → NaOH + H₂. The Na⁺ associates with hydroxide, while one H from hydride and one H from water form H₂ in an atom-tracking description. Left side has Na 1, O 1 and H 3; right side also has Na 1, O 1 and H 3. This is not an equation for Na metal reacting with water, although both processes can yield NaOH and H₂; their stoichiometric coefficients differ.

The net proton-transfer expression is H⁻ + H₂O → H₂ + OH⁻. On the left charge is −1, and on the right OH⁻ gives −1. Hydride accepts a proton from water, so it acts as a strong Brønsted base. Water becomes hydroxide after losing that proton. In formal redox terms, the H that began in hydride goes −1 → 0, while the H that began in water goes +1 → 0. One is oxidised and the other reduced. The same net event can be viewed through acid–base and redox accounting without contradiction.

With a stated acid such as HCl, NaH + HCl → NaCl + H₂. The hydride H⁻ accepts a proton supplied by the acid; the counter-ion joins sodium in the salt. An ionic summary is H⁻ + H⁺ → H₂. In aqueous acid, H⁺ is shorthand for solvated proton species. The simple formula equation shows the stoichiometry but does not describe every solvent interaction.

Calcium hydride contains two hydride hydrogens per CaH₂ unit. Its water reaction is CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. Count Ca 1, O 2 and H 6 on each side. One mole of CaH₂ can theoretically yield two moles of H₂ from this stoichiometric equation, assuming sufficient water and complete reaction. Those two H₂ molecules contain four H atoms in total: two came from hydride and two from water in the simple origin count. It is wrong to say all hydrogen gas atoms came from the hydride solid.

Hydride is also a reducing agent in many reactions because H at −1 can rise in oxidation state. But “basic” and “reducing” are distinct descriptions. Basicity concerns accepting a proton; reducing power concerns causing another species' reduction while hydride is oxidised. In the water reaction both descriptions are useful because a proton is transferred and formal H oxidation states converge to zero. In other reaction contexts the details can differ, so name the actual species and products.

Ionic hydrides of very active metals are not interchangeable with covalent molecules such as CH₄ or NH₃. Methane does not contain a lattice of free H⁻ ions that instantly gives H₂ on contact with water. The bonding model explains their contrasting behaviour. Even among metal hydrides, degree of ionic character can vary, so the H⁻ model is strongest for classic salt-like examples such as group 1 hydrides and CaH₂.

For calculations, separate chemical stoichiometry from gas volume. The equation gives H₂ moles; converting to a gas volume requires specified temperature and pressure or a valid molar-volume convention. “One mole of CaH₂ gives two moles H₂” is a theoretical mole result, not automatically a fixed number of litres under all conditions.

Step-by-step reasoning

1. Identify the metal hydride formula and count its hydride H centres. 2. Identify a proton donor such as water or a stated acid. 3. Pair each hydride H with one donated proton to form H₂. 4. Balance the metal-containing hydroxide or salt and then all atoms and charge. 5. Use coefficients for theoretical moles and specify conditions for any gas-volume conversion.

Visual explanation

Draw H⁻ from NaH at the left and H⁺ from water at the right, with arrows converging on H—H. Leave OH⁻ after the water proton departs and pair it with Na⁺. Under the diagram write H(−1) → H(0) and H(+1) → H(0). This picture connects the base reaction and redox formalism without confusing the origins of the two gas atoms.

Real-world analogy

Two complementary puzzle pieces join to make one pair. One piece comes from the hydride and the other from the proton donor. Counting only the hydride piece would undercount the final hydrogen gas atoms. The analogy is about atom tracking; actual reaction involves electron redistribution and solvent chemistry.

Real-world example

CaH₂ can serve as a chemical source of hydrogen gas when it reacts with water. Its formula and the balanced equation determine the maximum H₂ amount. The source is not simply “stored hydrogen gas” in the solid: water provides half the H atoms in the generated H₂ under the simple reaction equation.

Why?

Why is H⁻ a strong base? It can combine with a proton to form stable H₂. In water it removes a proton from H₂O, leaving OH⁻. The product equation captures both the proton-transfer tendency and the change in formal oxidation state.

Common misconception

“Every H atom in the evolved gas originally belonged to NaH.” NaH contributes one H atom per H₂ molecule; the other comes from water or acid. Atom tracing through the balanced equation exposes the mistake.

Worked example

Calculate the theoretical H₂ amount from 0.050 mol CaH₂ with excess water. First write CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. The ratio is two moles H₂ per mole CaH₂, so n(H₂) = 2 × 0.050 = 0.100 mol. The equation also requires 0.100 mol water at minimum and produces 0.050 mol Ca(OH)₂. A gas volume cannot be reported without temperature and pressure information.

Quick check

1. In NaH + H₂O → NaOH + H₂, where does the second hydrogen atom in each H₂ molecule come from? Answer: From water; the first comes from hydride H in NaH.

Exam focus

Balance the entire hydride reaction rather than assuming the metal's water reaction coefficients. Use H⁻ as a formal bonding model for classic ionic hydrides, identify the proton donor, and trace H atoms on both sides. Convert to gas volume only with stated conditions.

Advanced insight

The reaction H⁻ + H⁺ → H₂ is both acid–base combination and formal redox convergence: H −1 and +1 meet at 0. Oxidation-state labels help describe the electron redistribution, but a detailed molecular mechanism in solution can involve associated ions and solvent structures beyond this simple net equation.

Summary

Ionic hydrides such as NaH and CaH₂ contain formally H⁻ centres. They accept protons from water or acid, producing H₂ and hydroxide or a salt. The hydrogen gas includes atoms from both the hydride and proton donor, and its theoretical amount follows the balanced coefficients.

Practice questions

1. Balance sodium hydride with water. Answer: NaH + H₂O → NaOH + H₂. 2. How many H₂ moles can 0.020 mol CaH₂ theoretically yield with sufficient water? Answer: 0.040 mol H₂, from the 1:2 CaH₂:H₂ ratio. 3. Why is the hydride–water reaction also describable with oxidation states? Answer: Hydride H rises from −1 to 0 while water H falls from +1 to 0 in H₂.