Hydrogen typically exhibits an oxidation number of +1 in most compounds, but this value shifts to -1 when hydrogen bonds with metals in metal hydrides. Understanding in which reaction the oxidation number of hydrogen changes requires a solid grasp of redox principles, electronegativity differences, and the specific chemical environments where hydrogen acts as either an oxidizing or reducing agent. This article explores the conditions, reaction types, and practical examples where hydrogen’s oxidation state deviates from its standard +1 assignment, providing a full breakdown for students and chemistry enthusiasts.
The Standard Rule: Hydrogen as +1
Before identifying the exceptions, it is crucial to establish the baseline. Because of that, in the vast majority of covalent compounds—such as water (H₂O), methane (CH₄), hydrogen chloride (HCl), and ammonia (NH₃)—hydrogen carries an oxidation number of +1. Day to day, this assignment stems from hydrogen’s electronegativity (2. 20 on the Pauling scale), which is lower than nonmetals like oxygen (3.44), chlorine (3.Day to day, 16), and nitrogen (3. 04). When bonded to these more electronegative elements, hydrogen loses electron density, effectively "donating" its single electron to the bond, resulting in a +1 oxidation state.
In these scenarios, hydrogen acts as a reducing agent. It is oxidized (loses electrons) or remains in its +1 state during acid-base reactions or precipitation reactions where no redox occurs. Take this case: in the neutralization reaction between HCl and NaOH:
$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$
Hydrogen remains +1 on both the reactant and product sides. No change in oxidation number occurs here; this is a non-redox reaction Worth knowing..
The Major Exception: Metal Hydrides (Oxidation Number -1)
The most significant answer to the question "in which reaction does the oxidation number of hydrogen change" lies in the formation and decomposition of metal hydrides (saline hydrides). These are ionic compounds formed between hydrogen and highly electropositive metals, typically Group 1 (alkali metals) and Group 2 (alkaline earth metals).
Not obvious, but once you see it — you'll see it everywhere.
In compounds like sodium hydride (NaH), calcium hydride (CaH₂), or lithium aluminum hydride (LiAlH₄), the metal is far less electronegative than hydrogen. Sodium has an electronegativity of 0.Because hydrogen is more electronegative than these metals, it attracts the bonding electrons toward itself. 00. 93; calcium is 1.Because of this, hydrogen gains an electron, achieving a -1 oxidation state (hydride ion, H⁻) It's one of those things that adds up. No workaround needed..
Reaction Type 1: Synthesis (Direct Combination) of Metal Hydrides
The primary reaction where hydrogen changes from 0 to -1 is the direct synthesis of metal hydrides from elemental hydrogen gas and the metal.
Example: Formation of Sodium Hydride $2\text{Na}(s) + \text{H}_2(g) \xrightarrow{\text{heat}} 2\text{NaH}(s)$
- Reactants: Na (0), H₂ (0).
- Products: Na (+1), H (-1).
Analysis: Hydrogen is reduced (oxidation number decreases from 0 to -1). It acts as the oxidizing agent. The metal is oxidized. This is a classic redox reaction driven by the high affinity of alkali metals for electrons and the stability of the resulting ionic lattice.
Example: Formation of Calcium Hydride $\text{Ca}(s) + \text{H}_2(g) \xrightarrow{\text{heat}} \text{CaH}_2(s)$
Here, calcium goes from 0 to +2, and hydrogen goes from 0 to -1.
Reaction Type 2: Decomposition of Metal Hydrides
The reverse reaction—thermal decomposition of metal hydrides—also features a change in hydrogen’s oxidation number, this time from -1 back to 0.
Example: Decomposition of Sodium Hydride $2\text{NaH}(s) \xrightarrow{\Delta} 2\text{Na}(l) + \text{H}_2(g)$
- Reactants: H is -1.
- Products: H₂ is 0.
Analysis: Hydrogen is oxidized (oxidation number increases from -1 to 0). It acts as a reducing agent. This reaction requires significant heat because the ionic lattice energy of the hydride is high It's one of those things that adds up. Turns out it matters..
Reaction Type 3: Reaction of Metal Hydrides with Water or Acids (Proton Transfer)
A highly practical and frequently tested reaction involves the reaction of metal hydrides (where H is -1) with water or acids (where H is +1). This is a disproportionation-like redox reaction where hydrogen is both oxidized and reduced simultaneously, producing hydrogen gas (H₂, oxidation state 0) Most people skip this — try not to..
Example: Sodium Hydride reacting with Water $\text{NaH}(s) + \text{H}_2\text{O}(l) \rightarrow \text{NaOH}(aq) + \text{H}_2(g)$
Let’s track the oxidation numbers:
- In NaH: H = -1 (Hydride ion, H⁻).
- In H₂O: H = +1.
- In H₂: H = 0.
Mechanism: The hydride ion (H⁻) is a strong base and a strong reducing agent. It attacks the proton (H⁺) from water.
- The H⁻ (-1) loses 1 electron → becomes H (0) in H₂. (Oxidation)
- The H⁺ (+1) gains 1 electron → becomes H (0) in H₂. (Reduction)
This reaction is a quintessential example of a change in oxidation number for hydrogen. It demonstrates hydrogen’s amphoteric redox nature: it can serve as an electron donor (when -1) or an electron acceptor (when +1) to reach the stable elemental state (0) And it works..
Example: Calcium Hydride as a Drying Agent $\text{CaH}_2(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + 2\text{H}_2(g)$
Calcium hydride is commonly used to dry organic solvents. The reaction produces hydrogen gas, visibly bubbling, confirming the redox change: H⁻ (-1) and H⁺ (+1) $\rightarrow$ H₂ (0) The details matter here..
Reaction Type 4: Reaction with Non-Metal Oxides (Reducing Agent)
Because hydrogen in metal hydrides exists as H⁻ (-1), it is a powerful reducing agent. It can reduce non-metal oxides, where hydrogen is oxidized from -1 to +1 (usually ending up in water or hydroxide).
Example: Reduction of Silicon Dioxide (Sand) using Calcium Hydride $\text{CaH}_2(s) + \text{SiO}_2(s) \xrightarrow{\text{heat}} \text{CaSiO}_3(s) + 2\text{H}_2(g) \quad \text{(Simplified)}$ Actually, at high temps, it often forms silanes or other products, but the oxidation change holds.
A clearer example is the reduction of metal oxides using gaseous hydrogen (where H changes 0 to +1), but for hydrides specifically: $\text{NaH} + \text{CO}_2 \rightarrow \text{NaHCO}_3 \quad \text{(Not redox, acid-base)}$
A true redox example using hydrides as reductants: $3\text{NaH} + \text{BCl}_3 \rightarrow \text{NaBH}_4 + 2\text{NaCl}$ Here, H remains -1 in NaBH₄. No change Most people skip this — try not to..
Let's look at a reaction where hyd
