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Chapter08rw-final - 8 8.1(a(b(c 8.2 Basic Concepts of...

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190 8 Basic Concepts of Chemical Bonding Visualizing Concepts 8.1 Analyze/Plan . Count the number of electrons in the Lewis symbol. This corresponds to the ‘A’-group number of the family. Solve . (a) Group 14 or 4A (b) Group 2 or 2A (c) Group 15 or 5A (These are the appropriate groups in the s and p blocks, where Lewis symbols are most useful.) 8.2 Analyze . Given the size and charge of four different ions, determine their ionic bonding characteristics. Plan . The magnitude of lattice energy is directly proportional to the charges of the two ions and inversely proportional to their separation. E el = κ Q 1 Q 2 /d. Apply these concepts to A, B, X and Y. (a) AY and BX have a 1:1 ratio of cations and anions. In an ionic compound, the total positive and negative charges must be equal. In order to form a 1:1 compound, the magnitude of positive charge on the cation must equal the magnitude of negative charge on the anion. A 2+ combines with Y 2 and B + combines with X to form 1:1 compounds. (b) AY has the larger lattice energy. The A Y and B X separations are nearly equal. (A is smaller than B, but X is smaller than Y, so the differences in cation and anion radii approximately cancel.) In AY, Q 1 Q 2 = (2)(2) = 4, while in BX, Q 1 Q 2 = (1)(1) = 1. (c) BX has the smaller lattice energy. To recap the arguments in part (b), the d values in the two compounds are similar and BX has the smaller Q 1 Q 2 , so it has the smaller lattice energy. 8.3 Analyze/Plan . Count the valence electrons in the orbital diagram, take ion charge into account, and find the element with this orbital electron count on the periodic chart. Write the complete electron configuration for the ion. Solve . (a) This ion has seven 3d electrons. Transition metals, or d-block elements, have valence electrons in d-orbitals. Transition metal ions first lose electrons from the 4s orbital, then from 3d if required by the charge. This 2+ ion has lost two electrons from 4s, none from 3d. The transition metal with seven 3d-electrons is cobalt, Co. (b) The electron configuration of Co is [Ar]4s 2 3d 7 . (The configuration of Co 2+ is [Ar]3d 7 ).
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8 Chemical Bonding Solutions to Exercises 191 8.4 Analyze/Plan . This question is a ”reverse” Lewis structure. Count the valence electrons shown in the Lewis structure. For each atom, assume zero formal charge and determine the number of valence electrons an unbound atom has. Name the element. Solve . A: 1 shared e pair = 1 valence electron + 3 unshared pairs = 7 valence electrons, F E: 2 shared pairs = 2 valence electrons + 2 unshared pairs = 6 valence electrons, O D: 4 shared pairs = 4 valence electrons, C Q: 3 shared pairs = 3 valence electrons + 1 unshared pair = 5 valence electrons, N X: 1 shared pair = 1 valence electron, no unshared pairs, H Z: same as X, H Check . Count the valence electrons in the Lewis structure. Does the number correspond to the molecular formula CH 2 ONF? 12 e pair in the Lewis structure. CH 2 ONF = 4 + 2 + 6 + 5 + 7 = 24 e , 12 e pair. The molecular formula we derived matches the Lewis structure.
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