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22 Chemistry of the Nonmetals Solutions to Exercises 22.31 Analyze/Plan. Consider the periodic properties of Xe and Ar. Solve. Xenon is larger, and can more readily accommodate an expanded octet. More important is the lower ionization energy of xenon; because the valence electrons are a greater average distance from the nucleus, they are more readily promoted to a state in which the Xe atom can form bonds with fluorine. 22.32 Your friend cannot be correct. In general, noble gas elements have very stable electron configurations with complete S and p subshellls. They have very large positive ionization energies; they do not lose electrons easily. They have positive electron affinities; they do not attract electrons to themselves. They do not easily gain, lose or share electrons, so they do not readily form the chemical bonds required to create compounds. To date, the only known compounds of noble gases involve Xe and Kr bound to other nonmetals. Specifically, there are no known compounds of Ne. 22.33 Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve. (a) Ca(OBr)₂, Br, +1 (b) Br, +5 (c) Xe, +6 (d) +7 (e) I, +3 (f) I, +5; F, -1 22.34 (a) +5 (b) HI, I, -1 (c) ICl₃; I, +3; -1 (d) NaOCl, Cl, +1 (e) +7 (f) Xe, +4; F, -1 22.35 Analyze/Plan. Review the nomenclature rules and ion names in Section 2.8, as well as the rules for assignming oxidation numbers in Section 4.4. Solve. (a) iron(III) chlorate, Cl, +5 (b) chlorous acid, Cl, +3 (c) xenon hexafluoride, F, -1 (d) bromine pentafluoride; Br, +5; F, -1 (e) xenon oxide tetrafluoride, F, -1 (f) iodic acid, I, +5 22.36 (a) potassium chlorate, Cl, +5 (b) calcium iodate, I, +5 (c) aluminum chloride, Cl, -1 (d) bromic acid, Br, +5 (e) paraperiodic acid, I, +7 (f) xenon tetrafluoride, F, -1 22.37 Analyze/Plan. Consider intermolecular forces and periodic properties, including oxidizing power, of the listed substances. Solve. (a) Van der Waals intermolecular attractive forces increase with increasing numbers of electrons in the atoms. (b) reacts with water: + 2HF(aq) + 1/2 O₂(g). That is, fluorine is too strong an oxidizing agent to exist in water. (c) HF has extensive hydrogen bonding. (d) Oxidizing power is related to electronegativity. Electronegativity decreases in the order given. 22.38 (a) The more electronegative the central atom, the greater the extent to which it withdraws charge from oxygen, in turn making the bond more polar, and enhancing ionization of 675