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8 Chemical Bonding Solutions to Exercises (c) In order to reduce the formal charge of X to zero, X must have more bonding electrons. This is accomplished by changing the appropriate number of lone pairs on O to multiple bonds between X and O. 3- 2- :O: :O: Lewis Symbols (section 8.1) 8.9 (a) Valence electrons are those that take part in chemical bonding, those in the outermost electron shell of the atom. This usually means the electrons beyond the core noble-gas configuration of the atom, although it is sometimes only the outer shell electrons. (b) N A nitrogen atom has 5 valence electrons. Valence electrons (c) The atom (Si) has 4 valence electrons. [Ne] valence electrons 8.10 (a) Atoms will gain, lose, or share electrons to achieve the nearest noble-gas electron configuration. Except for H and He, this corresponds to eight electrons in the valence shell, thus the term octet rule. (b) S: A sulfur atom has six valence electrons, so it must gain two electrons to achieve an octet. (c) = [He]2s²2p³ The atom (N) has five valence electrons and must gain three electrons to achieve an octet. 8.11 Si: The 3s and 3p electrons are valence electrons; the 1s, 2s and 2p electrons are nonvalence or core electrons. Valence electrons are involved in chemical bonding, while nonvalence or core electrons are not. 8.12 (a) Ti: Ti has four (4) valence electrons. These valence electrons are available for chemical bonding, while core electons do not participate in chemical bonding. (b) Hf: (c) If Hf and Ti both behave as if they have four (4) valence electrons, the 6s and 5d orbitals in Hf behave as valence orbitals and the 4f behaves as a core orbital. This is reasonable because 4f is complete and 4f electrons are, on average, closer to the nucleus than 5d or 6s electrons. 8.13 (a) (b) : Br: (c) :Ar: (d) Sr 8.14 (a) (b) As (c) :Sn (d) 197