What NEET Asks
- Conceptual questions on types of hydrides formed with different elements.
- Reaction identification and balancing, especially involving hydrogen as a reducing agent.
- Conditions required for specific reactions (temperature, pressure, catalyst).
Key Points
- Hydrogen reacts with highly electropositive metals (Group 1 & 2) to form ionic/saline hydrides (e.g., NaH, CaH₂).
- It reacts with non-metals to form covalent hydrides (e.g., H₂O, NH₃, HX).
- With oxygen, H₂ burns to form water:
2H₂(g) + O₂(g) → 2H₂O(l)(highly exothermic). - With halogens (X₂), it forms hydrogen halides (HX), with reactivity decreasing from F₂ to I₂.
- Hydrogen acts as a reducing agent, reducing oxides of less active metals (e.g., CuO, Fe₂O₃) to their respective metals.
- Conditions are crucial: high temperature often required, catalysts for specific reactions (e.g., Haber's process for NH₃).
Must-Know Formula / Reaction
Reduction of metal oxides: MO(s) + H₂(g) → M(s) + H₂O(l/g)
- MO: Oxide of a less active metal (e.g., CuO, Fe₂O₃)
- H₂: Hydrogen gas, acts as a reducing agent
- M: Pure metal
- H₂O: Water
Common Mistakes
- Students often confuse ionic and covalent hydrides; remember ionic hydrides are formed with highly electropositive metals.
- Don't forget reaction conditions; for instance, the reaction with nitrogen (Haber's process) requires specific temperature, pressure, and catalyst.
- Misinterpreting hydrogen's role: it's primarily a reducing agent in reactions with metal oxides, not an oxidizing one.
Rapid Revision
Hydrogen forms ionic hydrides with active metals and covalent hydrides with non-metals. It reduces metal oxides to metals and forms water with oxygen. Reactivity with halogens decreases down the group. Always check specific reaction conditions.