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Concept 06 / 09The p-Block Elements

Group 15: Nitrogen, Ammonia, Nitric Acid and Oxides of Nitrogen

General trends of group 15 elements, anomalous nitrogen, trends in hydrides, preparation and properties of dinitrogen, ammonia by the Haber process, nitric acid by the Ostwald process and its oxidising reactions with metals, the brown ring test, and the structures and properties of oxides of nitrogen.

Figure 1
Group 15 trends. Nitrogen, phosphorus, arsenic, antimony, bismuth; configuration ns2np3ns^{2}np^{3} (half-filled, extra stable, giving high ionisation enthalpies). Oxidation states range from −3-3 to +5+5; the stability of +5+5 decreases and of +3+3 increases down the group (inert pair effect), so Bi5+\mathrm{Bi^{5+}} compounds are strong oxidants (NaBiO3\mathrm{NaBiO_{3}}) and Bi is mainly +3+3. Metallic character increases down the group. Hydrides EH3\mathrm{EH_{3}}: basic strength NH3>PH3>AsH3>SbH3>BiH3\mathrm{NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3}}; thermal stability decreases and reducing character increases in the same direction; bond angles fall from 107.8∘107.8^{\circ} (NH3\mathrm{NH_{3}}) to about 90∘90^{\circ}. Boiling points: PH3<AsH3<NH3<SbH3<BiH3\mathrm{PH_{3} < AsH_{3} < NH_{3} < SbH_{3} < BiH_{3}} (NH3\mathrm{NH_{3}} is raised by hydrogen bonding). Anomalous nitrogen. Small size, high electronegativity and no d orbitals. It forms pπp\pi-pπp\pi multiple bonds (N≡N\mathrm{N \equiv N}, bond enthalpy 941.4 kJ mol−1941.4\,\mathrm{kJ\,mol^{-1}}), which makes N2\mathrm{N_{2}} very inert; its maximum covalency is 44; the N-N single bond is weaker than P-P (lone pair repulsion). Phosphorus and the rest form single-bonded tetrahedral E4\mathrm{E_{4}} units and can expand their octets (PCl5\mathrm{PCl_{5}}). Dinitrogen. Laboratory: NH4Cl(aq)+NaNO2(aq)→N2+2H2O+NaCl\mathrm{NH_{4}Cl(aq) + NaNO_{2}(aq) \rightarrow N_{2} + 2H_{2}O + NaCl}; also (NH4)2Cr2O7→N2+4H2O+Cr2O3\mathrm{(NH_{4})_{2}Cr_{2}O_{7} \rightarrow N_{2} + 4H_{2}O + Cr_{2}O_{3}} (on heating). Very pure N2\mathrm{N_{2}} from Ba(N3)2→Ba+3N2\mathrm{Ba(N_{3})_{2} \rightarrow Ba + 3N_{2}}. Industrially by fractional distillation of liquid air. Inert at room temperature; combines with Li, Mg (3Mg+N2→Mg3N2\mathrm{3Mg + N_{2} \rightarrow Mg_{3}N_{2}}), with O2\mathrm{O_{2}} in lightning (NO), and with H2\mathrm{H_{2}} in the Haber process. Ammonia. Laboratory: 2NH4Cl+Ca(OH)2→2NH3+2H2O+CaCl2\mathrm{2NH_{4}Cl + Ca(OH)_{2} \rightarrow 2NH_{3} + 2H_{2}O + CaCl_{2}}. Haber process: N2+3H2⇌2NH3\mathrm{N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}}, ΔfH∘=−46.1 kJ mol−1\Delta_{f}H^{\circ} = -46.1\,\mathrm{kJ\,mol^{-1}}; about 200×105 Pa200 \times 10^{5}\,\mathrm{Pa}, 700 K700\,\mathrm{K}, iron oxide catalyst with K2O\mathrm{K_{2}O} and Al2O3\mathrm{Al_{2}O_{3}}. Trigonal pyramidal (107.8∘107.8^{\circ}), hydrogen-bonded, very soluble, weakly basic. As a Lewis base it forms complexes: Cu2++4NH3→[Cu(NH3)4]2+\mathrm{Cu^{2+} + 4NH_{3} \rightarrow [Cu(NH_{3})_{4}]^{2+}} (deep blue); AgCl+2NH3→[Ag(NH3)2]Cl\mathrm{AgCl + 2NH_{3} \rightarrow [Ag(NH_{3})_{2}]Cl} (dissolves). Nitric acid. Ostwald process (see figure): catalytic oxidation of ammonia to NO, oxidation to NO2\mathrm{NO_{2}}, absorption in water. A strong acid and powerful oxidising agent; the products depend on concentration and on the metal: 3Cu+8HNO3(dil)→3Cu(NO3)2+2NO+4H2O\mathrm{3Cu + 8HNO_{3}(dil) \rightarrow 3Cu(NO_{3})_{2} + 2NO + 4H_{2}O} Cu+4HNO3(conc)→Cu(NO3)2+2NO2+2H2O\mathrm{Cu + 4HNO_{3}(conc) \rightarrow Cu(NO_{3})_{2} + 2NO_{2} + 2H_{2}O} 4Zn+10HNO3(dil)→4Zn(NO3)2+5H2O+N2O\mathrm{4Zn + 10HNO_{3}(dil) \rightarrow 4Zn(NO_{3})_{2} + 5H_{2}O + N_{2}O} Zn+4HNO3(conc)→Zn(NO3)2+2H2O+2NO2\mathrm{Zn + 4HNO_{3}(conc) \rightarrow Zn(NO_{3})_{2} + 2H_{2}O + 2NO_{2}} Non-metals are oxidised to their highest oxoacids (I2→HIO3\mathrm{I_{2} \rightarrow HIO_{3}}, C→CO2\mathrm{C \rightarrow CO_{2}}, S8→H2SO4\mathrm{S_{8} \rightarrow H_{2}SO_{4}}, P4→H3PO4\mathrm{P_{4} \rightarrow H_{3}PO_{4}}). Al, Cr and Fe are rendered passive by concentrated HNO3\mathrm{HNO_{3}} (oxide film). Brown ring test for nitrate: add FeSO4\mathrm{FeSO_{4}} solution, then concentrated H2SO4\mathrm{H_{2}SO_{4}} down the side: NO3−+3Fe2++4H+→NO+3Fe3++2H2O\mathrm{NO_{3}^{-} + 3Fe^{2+} + 4H^{+} \rightarrow NO + 3Fe^{3+} + 2H_{2}O}, then [Fe(H2O)6]2++NO→[Fe(H2O)5(NO)]2++H2O\mathrm{[Fe(H_{2}O)_{6}]^{2+} + NO \rightarrow [Fe(H_{2}O)_{5}(NO)]^{2+} + H_{2}O} (brown ring at the interface). Oxides of nitrogen. N2O\mathrm{N_{2}O} (+1+1, colourless, neutral, laughing gas); NO (+2+2, colourless, neutral, paramagnetic); N2O3\mathrm{N_{2}O_{3}} (+3+3, blue solid, acidic); NO2\mathrm{NO_{2}} (+4+4, brown, acidic, paramagnetic, dimerises to colourless N2O4\mathrm{N_{2}O_{4}}); N2O5\mathrm{N_{2}O_{5}} (+5+5, colourless solid, acidic, anhydride of HNO3\mathrm{HNO_{3}}). Key Takeaways: - Basicity and stability of hydrides decrease from NH3\mathrm{NH_{3}} to BiH3\mathrm{BiH_{3}}; reducing power increases. - N≡N\mathrm{N \equiv N} (941 kJ mol−1941\,\mathrm{kJ\,mol^{-1}}) makes N2\mathrm{N_{2}} inert. - Ostwald: NH3→NO→NO2→HNO3\mathrm{NH_{3} \rightarrow NO \rightarrow NO_{2} \rightarrow HNO_{3}} over Pt/Rh. - Dilute HNO3\mathrm{HNO_{3}} with Cu gives NO; concentrated gives NO2\mathrm{NO_{2}}. - NO and NO2\mathrm{NO_{2}} are odd-electron, paramagnetic molecules. - Pitfall: NH3\mathrm{NH_{3}} is not the highest-boiling group 15 hydride; SbH3\mathrm{SbH_{3}} and BiH3\mathrm{BiH_{3}} boil higher. Brain-Teaser 1: How many moles of NO are produced when 1.5 mol1.5\,\mathrm{mol} of copper reacts with excess dilute HNO3\mathrm{HNO_{3}}? Solution: 3 Cu3\,\mathrm{Cu} give 2 NO2\,\mathrm{NO}, so 1.5 mol1.5\,\mathrm{mol} Cu gives 1.0 mol1.0\,\mathrm{mol} NO. (Electron check: 1.5×2=3 mol1.5 \times 2 = 3\,\mathrm{mol} electrons; each N goes +5→+2+5 \to +2, taking 33 electrons, so 1 mol1\,\mathrm{mol} NO.) Brain-Teaser 2: Why does NO2\mathrm{NO_{2}} dimerise but NO does not readily do so? Solution: NO2\mathrm{NO_{2}} has its unpaired electron largely on nitrogen, and pairing it forms a stable N-N bond in N2O4\mathrm{N_{2}O_{4}} (colourless, diamagnetic). In NO, the unpaired electron is in a delocalised π∗\pi^{*} orbital over both atoms, so dimerisation gains little; NO dimerises only in the liquid and solid states. Brain-Teaser 3: What is the oxidation state of iron and the charge on NO in the brown ring complex [Fe(H2O)5(NO)]2+\mathrm{[Fe(H_{2}O)_{5}(NO)]^{2+}}? Solution: Conventionally the ligand is taken as NO+\mathrm{NO^{+}} (nitrosonium), after an electron transfer from Fe2+\mathrm{Fe^{2+}}, so iron is +1+1: x+0+1=+2x + 0 + 1 = +2 gives x=+1x = +1.