Group 15 (Nitrogen Family)

The ns²np³ configuration means Group 15 elements can form three covalent bonds using the three half-filled p orbitals, giving a −3 state (e.g., NH₃, PH₃). They can also lose or share electrons with more electronegative atoms. Nitrogen achieves +1 (N₂O), +2 (NO), +3 (HNO₂), +4 (NO₂), and +5 (HNO₃). Phosphorus commonly forms PCl₃ (+3) and PCl₅ (+5) by using its empty 3d orbitals to expand its octet — something nitrogen cannot do, because the n=2 shell has no d orbitals.

Nitrogen behaves differently from its heavier congeners in several important ways. (i) Maximum covalency of 4: Nitrogen lacks d orbitals in its valence shell, so it cannot expand beyond an octet. It forms at most four bonds (e.g., NH₄⁺). Phosphorus, by contrast, forms PCl₅ using 3d orbitals. (ii) Ability to form pπ–pπ multiple bonds: Nitrogen's small atomic radius and high charge density allow effective sideways overlap of p orbitals, producing strong π bonds in N₂ (bond enthalpy ≈ 945 kJ mol⁻¹) and NO₂. Heavier elements prefer single bonds because their larger atomic radii reduce p-orbital overlap. (iii) Lone pair basicity: The lone pair on N in NH₃ is highly available for donation, making ammonia a far better Lewis and Brønsted base than PH₃, AsH₃, or SbH₃, whose lone pairs are in more diffuse, less accessible orbitals. (iv) Anomalously low boiling point: Unlike phosphorus (white), which is a molecular solid, nitrogen is a gas at room temperature because N₂ is held only by very weak London dispersion forces, while P₄ molecules are larger and more polarisable.

Ammonia is manufactured by the direct combination of nitrogen and hydrogen over an iron catalyst promoted with K₂O and Al₂O₃: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH° = −92 kJ mol⁻¹. The reaction is exothermic and decreases moles of gas (4 → 2), so by Le Chatelier's principle, high pressure (150–300 atm) and low temperature favour ammonia formation at equilibrium. However, low temperature slows the rate. A compromise temperature of 400–500 °C is used, giving a yield of about 15–25% per pass; unreacted gases are recycled. The iron catalyst lowers the activation energy by adsorbing N₂ and weakening the N≡N triple bond, enabling dissociative chemisorption.

The Ostwald process converts ammonia to nitric acid in three catalytic steps. Step 1 — Catalytic oxidation: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), ΔH° = −907 kJ mol⁻¹, over a platinum–rhodium gauze at 850–900 °C. Step 2 — Further oxidation: 2NO(g) + O₂(g) → 2NO₂(g). Step 3 — Absorption in water: 3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g); the NO is recycled. Nitric acid produced (50–70% by mass) is used in fertilisers (ammonium nitrate), explosives (TNT, nitroglycerin), and dyes.

Phosphorus exists in several allotropic forms with dramatically different properties. White phosphorus (P₄) consists of four phosphorus atoms at the corners of a tetrahedron, each forming three single P–P bonds. The P–P–P bond angle is only 60°, far from the ideal 109.5°, creating intense angular (ring) strain and making white phosphorus extremely reactive, spontaneously flammable in air (pyrophoric), and highly toxic. Red phosphorus is a polymeric solid in which P₄ units have opened and linked into chains; it is much less reactive, non-toxic, and is used in safety-match heads. Black phosphorus, the most stable allotrope, has a layered structure resembling graphite and can conduct electricity along its layers. The conversion of white to red phosphorus requires heat (≈ 250 °C) and is irreversible under normal conditions.

• \ce{N2} — Nitrogen (mol): Triple-bonded diatomic molecule; the bond enthalpy of N≡N is ~945 kJ mol⁻¹, making it very difficult to break.
• \ce{H2} — Hydrogen (mol): Diatomic reactant; supplied industrially from steam reforming of natural gas.
• \ce{NH3} — Ammonia (mol): Product; a trigonal pyramidal molecule with a lone pair on nitrogen; used as fertiliser and as feedstock for nitric acid.
• \Delta H^\circ — Standard enthalpy change (kJ mol⁻¹): = −92 kJ mol⁻¹; the negative sign indicates the forward reaction is exothermic, releasing heat to surroundings.

Problem. Explain why increasing pressure increases the yield of ammonia in the Haber process, and calculate the percentage change in moles of gas when one mole of N₂ reacts completely with 3 mol H₂ to form 2 mol NH₃.

Answer. Increasing pressure shifts equilibrium toward products because Δn(gas) = −2 (moles decrease by 50%), favouring the side with fewer gas molecules.

• Le Chatelier's Principle — The Haber process is the canonical example of applying Le Chatelier's principle — understanding how pressure, temperature, and concentration shifts affect equilibrium yield of ammonia.
• Group 16 (Oxygen Family) — Oxygen and sulfur are key reaction partners for Group 15 elements (e.g., formation of oxoacids HNO₃ and H₃PO₄), and Group 16 shows analogous trends in oxidation states and allotropy.
• VSEPR Theory and Molecular Geometry — NH₃ (trigonal pyramidal), NO₂ (bent), and PCl₅ (trigonal bipyramidal) are classic VSEPR examples whose shapes directly explain Group 15 reactivity and properties.
• Catalysis and Activation Energy — Both the Haber process (iron catalyst) and Ostwald process (Pt–Rh catalyst) illustrate heterogeneous catalysis; understanding how catalysts lower activation energy is essential to interpreting these industrial processes.
• Allotropy in Non-Metals — Phosphorus allotropes (white, red, black) provide a rich case study alongside carbon allotropes (diamond, graphite, fullerene) for understanding how bonding arrangements determine physical and chemical properties.