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Concept 09 / 09The p-Block Elements

Group 18: Noble Gases and Compounds of Xenon

Electronic configuration, ionisation enthalpy and physical properties of noble gases, Bartlett's discovery of noble gas compounds, preparation, structures (VSEPR) and hydrolysis of xenon fluorides and oxides, and uses of helium, neon, argon and xenon.

Figure 1
Group 18: helium, neon, argon, krypton, xenon, radon. Configuration ns2np6ns^{2}np^{6} (He: 1s21s^{2}), a closed shell. Properties. Very high ionisation enthalpies (decreasing down the group), large positive electron gain enthalpies, monatomic gases held only by weak dispersion forces. Boiling points are very low and increase down the group with polarisability (He 4.2 K4.2\,\mathrm{K}, the lowest of any substance). Solubility in water also increases down the group. Helium diffuses through most materials and cannot be solidified at ordinary pressure. Discovery of compounds. In 1962 Neil Bartlett prepared O2+[PtF6]−\mathrm{O_{2}^{+}[PtF_{6}]^{-}} by oxidising O2\mathrm{O_{2}} with PtF6\mathrm{PtF_{6}}. Noticing that the first ionisation enthalpy of O2\mathrm{O_{2}} (1175 kJ mol−11175\,\mathrm{kJ\,mol^{-1}}) is almost the same as that of xenon (11701170), he made the first noble gas compound, a red solid formulated as Xe+[PtF6]−\mathrm{Xe^{+}[PtF_{6}]^{-}}. Xenon, with the lowest ionisation enthalpy among the stable noble gases, forms the most compounds (with F and O); krypton forms KrF2\mathrm{KrF_{2}}; He, Ne and Ar form no stable compounds under ordinary conditions. Xenon fluorides. Made by direct reaction, the product depending on the Xe : F2\mathrm{F_{2}} ratio: Xe+F2→XeF2\mathrm{Xe + F_{2} \rightarrow XeF_{2}} (xenon in excess, 673 K673\,\mathrm{K}, 1 bar1\,\mathrm{bar}); Xe+2F2→XeF4\mathrm{Xe + 2F_{2} \rightarrow XeF_{4}} (ratio 1:51 : 5, 873 K873\,\mathrm{K}, 7 bar7\,\mathrm{bar}); Xe+3F2→XeF6\mathrm{Xe + 3F_{2} \rightarrow XeF_{6}} (ratio 1:201 : 20, 573 K573\,\mathrm{K}, 6060 to 70 bar70\,\mathrm{bar}); also XeF4+O2F2→XeF6+O2\mathrm{XeF_{4} + O_{2}F_{2} \rightarrow XeF_{6} + O_{2}}. They are colourless crystalline solids and powerful fluorinating agents. Structures (VSEPR, see figure): XeF2\mathrm{XeF_{2}} linear (sp3dsp^{3}d, three equatorial lone pairs); XeF4\mathrm{XeF_{4}} square planar (sp3d2sp^{3}d^{2}, two trans lone pairs); XeF6\mathrm{XeF_{6}} distorted octahedral (sp3d3sp^{3}d^{3}, one lone pair). Fluoride donor and acceptor behaviour: XeF2+PF5→[XeF]+[PF6]−\mathrm{XeF_{2} + PF_{5} \rightarrow [XeF]^{+}[PF_{6}]^{-}}; XeF6+MF→M+[XeF7]−\mathrm{XeF_{6} + MF \rightarrow M^{+}[XeF_{7}]^{-}} (M = Na, K, Rb, Cs). Hydrolysis and oxides. 2XeF2+2H2O→2Xe+4HF+O2\mathrm{2XeF_{2} + 2H_{2}O \rightarrow 2Xe + 4HF + O_{2}}. 6XeF4+12H2O→4Xe+2XeO3+24HF+3O2\mathrm{6XeF_{4} + 12H_{2}O \rightarrow 4Xe + 2XeO_{3} + 24HF + 3O_{2}} (disproportionation). XeF6+3H2O→XeO3+6HF\mathrm{XeF_{6} + 3H_{2}O \rightarrow XeO_{3} + 6HF} (complete). Partial hydrolysis of XeF6\mathrm{XeF_{6}}: XeF6+H2O→XeOF4+2HF\mathrm{XeF_{6} + H_{2}O \rightarrow XeOF_{4} + 2HF}; XeF6+2H2O→XeO2F2+4HF\mathrm{XeF_{6} + 2H_{2}O \rightarrow XeO_{2}F_{2} + 4HF}. XeO3\mathrm{XeO_{3}} is a colourless explosive solid with a pyramidal shape (one lone pair); XeOF4\mathrm{XeOF_{4}} is square pyramidal; XeO2F2\mathrm{XeO_{2}F_{2}} has a see-saw shape. Uses. Helium: filling balloons and airships (light and non-inflammable), mixed with oxygen for deep-sea diving (much less soluble in blood than nitrogen, preventing the bends), cryogenics and superconducting magnets (MRI). Neon: discharge signs (orange-red glow), beacon lights. Argon: inert atmosphere for welding and metallurgy, filling light bulbs. Xenon and krypton: high-intensity lamps and photographic flash tubes. Radon: radiotherapy (radioactive). Key Takeaways: - Closed shells: high IE, positive electron gain enthalpy, monatomic, weak dispersion forces. - Bartlett: O2\mathrm{O_{2}} and Xe have almost equal IE, leading to Xe+[PtF6]−\mathrm{Xe^{+}[PtF_{6}]^{-}}. - XeF2\mathrm{XeF_{2}} linear, XeF4\mathrm{XeF_{4}} square planar, XeF6\mathrm{XeF_{6}} distorted octahedral, XeO3\mathrm{XeO_{3}} pyramidal, XeOF4\mathrm{XeOF_{4}} square pyramidal. - Complete hydrolysis of XeF6\mathrm{XeF_{6}} gives XeO3\mathrm{XeO_{3}} and HF. - Pitfall: XeF4\mathrm{XeF_{4}} hydrolysis is a disproportionation that also releases Xe and O2\mathrm{O_{2}}. Brain-Teaser 1: Find the number of lone pairs on Xe and the shape of XeOF4\mathrm{XeOF_{4}}. Solution: Xe has 88 valence electrons; it forms one Xe=O (using 22) and four Xe-F (using 44), leaving 22 electrons, one lone pair. Steric number 66 (55 bonded atoms plus 11 lone pair): square pyramidal, with O at the apex. Brain-Teaser 2: How many moles of HF form when 1 mol1\,\mathrm{mol} of XeF6\mathrm{XeF_{6}} is completely hydrolysed, and what is the oxidation state of Xe in the other product? Solution: XeF6+3H2O→XeO3+6HF\mathrm{XeF_{6} + 3H_{2}O \rightarrow XeO_{3} + 6HF}: 6 mol6\,\mathrm{mol} HF. Xenon stays at +6+6 in XeO3\mathrm{XeO_{3}} (this hydrolysis is not a redox reaction). Brain-Teaser 3: Why does helium not form compounds while xenon does? Solution: Helium's 1s1s electrons are extremely tightly held (IE 2372 kJ mol−12372\,\mathrm{kJ\,mol^{-1}}) and it has no low-lying empty orbitals to use in bonding. Xenon's valence electrons are far from the nucleus and well shielded (IE 1170 kJ mol−11170\,\mathrm{kJ\,mol^{-1}}), so the very electronegative F and O can oxidise it and form bonds, with xenon expanding its valence shell.