chapter5 - density The bonding in SnCl 2 will be...

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Chapter 5 Homework 5.1 a) The distortion of an anion from a spherical shape. b) The holes between anions in the crystal packing in which cations can fit. c) The diagram used to show that many elements and compounds possess combinations of the three boding categories: metallic, covalent, and ionic. 5.3 Hard and brittle crystals, high melting points; electrically conducting in liquid phase and in aqueous solutions. 5.5 a) K + , because the radius will be determined by the inner orbitals (2s and 2p) while the radius of K is determined by the 3s orbital. b) Ca 2+ , because the ions are isoelectronic but calcium has one more proton, and hence a higher Z eff and a smaller ionic radius. c) Rb + , because again the ions are isoelectronic, with rubidium having two more protons than bromide, and hence a higher Z eff and a smaller ionic radius. 5.11 Tin(II) chloride has a higher melting point because tin(II) has a fairly low charge
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Unformatted text preview: density. The bonding in SnCl 2 will be predominantly ionic. The high charge density of the tin(IV) ion would result in covalent bonding for SnCl 4 and therefore a low melting point. 5.15 Magnesium chloride, because the dipositive smaller magnesium ion has a significantly higher charge density than the monopositive larger sodium ion. The magnesium ion would therefore be likely to have a “shell” of strongly attracted water molecules to diffuse the charge. 5.18 a) Ions are assumed to be charged, incompressible, nonpolarizable spheres. b) Ions try to surround themselves with as many ions of the opposite charge as possible and as closely as possible. c) the cation-to-anion ratio must reflect the chemical composition of the compound. 5.25 a) Metallic and a lesser contribution of ionic. b) Covalent and a lesser contribution of ionic....
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