Group 15 (the nitrogen family) contains nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi). The general valence-shell configuration is ns2np3 — a half-filled p sub-shell that gives extra stability and explains the resistance to losing or gaining electrons.
General trends
Down the group the atomic and ionic radii increase while ionisation enthalpy decreases; the +3 ion becomes more stable than +5 because of the inert-pair effect (the ns2 electrons become reluctant to bond). Common oxidation states are −3, +3 and +5. Nitrogen, being small and highly electronegative, shows the widest range (−3 to +5) and uniquely forms strong pπ–pπ multiple bonds, so it exists as the diatomic molecule N≡N. The heavier members cannot form pπ–pπ bonds effectively and so exist as larger single-bonded molecules (P4, As4).
Metallic character increases down the group: N and P are non-metals, As and Sb are metalloids, and Bi is a metal. The stability of the +5 state decreases down the group while the +3 state becomes more stable, so Bi(V) is a strong oxidiser.
Dinitrogen (N2)
Prepared in the lab by gently heating ammonium nitrite: NH4NO2 → N2 + 2H2O. The N≡N triple bond has a very high bond enthalpy (~941 kJ mol-1), making N2 chemically inert at room temperature.
Ammonia and the Haber process
Ammonia is manufactured by the Haber process: N2 + 3H2 ⇌ 2NH3 (ΔH = −92 kJ). Optimum conditions are ~200 atm, ~700 K and a finely divided iron catalyst with Mo as promoter. NH3 is a pyramidal Lewis base (lone pair on N) that turns moist red litmus blue and forms complexes such as [Cu(NH3)4]2+.
Oxides of nitrogen and nitric acid
Nitrogen forms many oxides: N2O, NO (neutral), N2O3, NO2, N2O5 — oxidation states +1 to +5. Nitric acid is made by the Ostwald process: catalytic oxidation of NH3 over Pt/Rh gauze gives NO, which is oxidised to NO2 and dissolved in water (4NO2 + O2 + 2H2O → 4HNO3). Concentrated HNO3 is a powerful oxidiser; with most metals it does not liberate H2 but gives NO2 (concentrated) or NO (dilute).
Phosphorus and its compounds
Phosphorus has allotropes: reactive white (P4 tetrahedra, stored under water), polymeric red, and layered black. Phosphine (PH3) is a weak base prepared by heating white P with NaOH. Halides PCl3 (pyramidal, sp3) and PCl5 (trigonal bipyramidal, sp3d) are made from P and Cl2; PCl5 is ionic in the solid as [PCl4]+[PCl6]-. The oxoacids include H3PO2 (1 OH, monobasic), H3PO3 (2 OH, dibasic) and H3PO4 (3 OH, tribasic) — basicity equals the number of P–OH groups, since P–H bonds are not ionisable.
Why does nitrogen exist as a diatomic gas (N2) whereas phosphorus exists as a tetra-atomic solid (P4)?
Solution- Nitrogen is a small, second-period element and can form strong pπ–pπ multiple bonds because its 2p orbitals overlap effectively side-on.
- Two N atoms therefore form a stable N≡N triple bond, giving the diatomic molecule N2.
- Phosphorus is larger; its 3p orbitals are too diffuse for effective pπ–pπ overlap, so multiple bonding is weak.
- P instead forms three single bonds per atom, giving a tetrahedral P4 molecule which catenates into a solid.
Answer: Effective pπ–pπ bonding in small N gives N≡N (gas); poor pπ overlap in larger P favours single-bonded P4 (solid).
State the conditions used in the Haber process and explain, using Le Chatelier's principle, why a moderately high temperature (~700 K) is used rather than a very low one.
Solution- Reaction: N2 + 3H2 ⇌ 2NH3, ΔH = −92 kJ (exothermic, Δn < 0).
- Conditions: pressure ~200 atm, temperature ~700 K, finely divided iron catalyst with molybdenum promoter.
- Being exothermic, a low temperature favours a higher equilibrium yield of NH3.
- But at low temperature the rate is very slow, so the reaction takes too long to be economical.
- A compromise (~700 K) is chosen to obtain an acceptable yield at an acceptable rate.
Answer: ~200 atm, ~700 K, Fe/Mo catalyst; the moderate temperature balances yield (favoured by low T) against reaction rate (favoured by high T).
Arrange NH3, PH3, AsH3, SbH3 and BiH3 in order of (a) increasing thermal stability and (b) decreasing basic strength.
Solution- Thermal stability of EH3 depends on the E–H bond strength, which falls as atomic size increases down the group.
- Therefore stability increases: BiH3 < SbH3 < AsH3 < PH3 < NH3.
- Basic strength depends on the availability/density of the lone pair, which is greatest on small, electronegative N.
- Hence basicity decreases: NH3 > PH3 > AsH3 > SbH3 > BiH3.
Answer: (a) BiH3 < SbH3 < AsH3 < PH3 < NH3; (b) NH3 > PH3 > AsH3 > SbH3 > BiH3.
Determine the basicity of H3PO2, H3PO3 and H3PO4 and explain the difference.
Solution- Basicity = number of ionisable P–OH groups; P–H bonds do not ionise.
- H3PO2 (hypophosphorous): one P–OH and two P–H → monobasic.
- H3PO3 (phosphorous): two P–OH and one P–H → dibasic.
- H3PO4 (phosphoric): three P–OH and no P–H → tribasic.
Answer: H3PO2 is monobasic, H3PO3 dibasic and H3PO4 tribasic, set by the number of P–OH groups.
Why does PCl5 behave as an ionic solid, and what is its structure in the vapour phase?
Solution- In the solid state PCl5 exists as ions [PCl4]+ (tetrahedral) and [PCl6]- (octahedral).
- This ionic lattice is more stable than the molecular form in the condensed phase, so solid PCl5 conducts when fused.
- In the vapour (and liquid) phase PCl5 is a covalent molecule with sp3d hybridisation.
- Its shape is trigonal bipyramidal, with two longer axial P–Cl bonds and three shorter equatorial bonds.
Answer: Solid PCl5 is ionic, [PCl4]+[PCl6]-; in the vapour it is a covalent, sp3d, trigonal-bipyramidal molecule.
Concentrated nitric acid renders iron passive. Explain this observation and write the balanced reaction of dilute HNO3 with copper.
Solution- Concentrated HNO3 is a strong oxidiser and quickly oxidises the iron surface to a thin, adherent, unreactive oxide film.
- This passivating oxide layer stops further attack, so iron becomes passive in concentrated HNO3.
- Dilute HNO3 oxidises copper but liberates NO (not H2).
- Balanced: 3Cu + 8HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O.
Answer: A protective oxide film passivates iron; with dilute acid, 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O.