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Practice Exercise 1 The sulfate ion, SO,2 , can be drawn in many ways. If you minimize formal charge on the sulfur, how many S=0

Practice Exercise 1 The sulfate ion, SO,2 , can be drawn in many ways. If you minimize formal charge on the sulfur, how many S=0

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Practice excercise 1

**Section 8.6 Resonance Structures** **Page 319** The electronegativity of nitrogen is greater than that of carbon or sulfur. Therefore, we expect any negative formal charge to reside on the nitrogen atom. For these reasons, the middle Lewis structure is the dominant one for NCS⁻. **Practice Exercise 1** The sulfate ion, SO₄²⁻, can be drawn in many ways. If you minimize the formal charge on sulfur, how many S=O double bonds should you draw in the Lewis structure? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 **Practice Exercise 2** The cyanate ion, NCO⁻, has three possible Lewis structures. (a) Draw these three structures and assign formal charges in each. (b) Which Lewis structure is dominant? --- **A Closer Look: Oxidation Numbers, Formal Charges, and Actual Partial Charges** In Chapter 4, we introduced the rules for assigning oxidation numbers to atoms. Electronegativity is the basis for these numbers. An atom’s oxidation number is the charge the atom would have if its bonds were completely ionic. That is, in determining the oxidation number, all shared electrons are counted with the more electronegative atom. For instance, a more electronegative Cl atom might be expected to have roughly 3.0 / (3.0 + 2.1) = 0.59 of the electrical charge in the bonding pair, whereas the H atom would have 2.1 / (3.0 + 2.1) = 0.41 of the charge. Because the bond consists of two electrons, the Cl atom's share is 0.59 × 2e = 1.18e, or 0.18e more than the neutral Cl atom.
Transcribed Image Text:**Section 8.6 Resonance Structures** **Page 319** The electronegativity of nitrogen is greater than that of carbon or sulfur. Therefore, we expect any negative formal charge to reside on the nitrogen atom. For these reasons, the middle Lewis structure is the dominant one for NCS⁻. **Practice Exercise 1** The sulfate ion, SO₄²⁻, can be drawn in many ways. If you minimize the formal charge on sulfur, how many S=O double bonds should you draw in the Lewis structure? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 **Practice Exercise 2** The cyanate ion, NCO⁻, has three possible Lewis structures. (a) Draw these three structures and assign formal charges in each. (b) Which Lewis structure is dominant? --- **A Closer Look: Oxidation Numbers, Formal Charges, and Actual Partial Charges** In Chapter 4, we introduced the rules for assigning oxidation numbers to atoms. Electronegativity is the basis for these numbers. An atom’s oxidation number is the charge the atom would have if its bonds were completely ionic. That is, in determining the oxidation number, all shared electrons are counted with the more electronegative atom. For instance, a more electronegative Cl atom might be expected to have roughly 3.0 / (3.0 + 2.1) = 0.59 of the electrical charge in the bonding pair, whereas the H atom would have 2.1 / (3.0 + 2.1) = 0.41 of the charge. Because the bond consists of two electrons, the Cl atom's share is 0.59 × 2e = 1.18e, or 0.18e more than the neutral Cl atom.
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