Problems for Chapter 1 & 2 ANSWERS 1. Draw an appropriate Lewis electron dot structure for the compound HONNH. H N N O H

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1 Problems for hapter 1 & 2 ASWERS 1. Draw an appropriate Lewis electron dot structure for the compound. Valence electrons 2x = 2 2x = 10 = 6 TTAL = Show the formal charge, if one exists, on each atom in the Lewis electron dot structure at below: F = 6-!/2(4) - 4 = F = 4-!/2(8) - 0 = F = 6-!/2(4) - 4 = ) F = 6-!/2(2) - 6 = Rewrite the structural formula, below, using a Kekule structure ( 3 ) 2 ( 3 )=() Show the three-dimensional structure for the compound, AlBr 3. If the compound has an overall dipole moment, show the direction of that dipole with an arrow, +. Br Br Al Br Trigonal Planar Shape (flat, like a 3 bladed propellar) These two dipoles, have vectors which resolve into a vector equal in magnitude to, but opposite direction to the one pointing up... ET DIPLE 5. Show the three-dimensional structure for the compound, 3. If the compound has an overall dipole moment, show the direction of that dipole with an arrow, +. Pyramidal shape, based on sp 3 nitrogen ET DIPLE

2 6. Draw a Lewis dot structure for the following compounds, show formal charge if any a. 3 = b Draw the Kekule structures for the above condensed structures. 8. Draw the 3-D shape of each molecule, in question (6) above, about the 3 Trigonal Planar Tetrahedral 9. ame the 3-D shape of each molecule in question (6), about the: Trigonal Planar Tetrahedral 10. ame the orbital hybridization for the compounds in question (6) above, about the: sp 2 sp What will be the approximate bond angles in the molecules in question (6) above, about the: arbon can exist in a trivalent state (three bonds) but neutral in charge. Predict (show) the Lewis structure for such a carbon with all bonds to hydrogen.

3 13. What orbital hybridization is likely for the carbon in the above species? Remember that hybridized electrons are always paired (bonded or nonbonded). Draw the carbon showing all orbitals. arbon is sp 2 hybridized, with one p orbital left for the unpaired electron to reside in 14.Draw the Kekulé structure for the skeletal drawing shown below Draw the above compound in a condensed structure. 3 2 [( 3 ) 2 ]( 3 ) Show, with an arrow, the polar bonds and the direction of polarity in the following molecules. l What will be the numerical value of the K a whose pk a is 4.2? (You may use two different powers of ten as your answer.) pka = - Log Ka, pka = 4.2 Ka = = If a compound is % carbon and % hydrogen, what is its empirical formula? If there is % carbon and % hydrogen, there must be % oxygen in sample. In 100 g of sample there is: g, g and g. In 100 g of sample there is: g/ = 5.16 moles of arbon, g/1.008 = moles of ydrogen g/ = 1.72 moles of xygen Molar ratio : 5.16, 10.33, 1.72, divide all by smallest ratio i.e. 1.72, formula becomes 3 6

4 19. Identify the acids and bases in the following equation Li acid base conjugate base conjugate acid 20. The ion is the conjugate base of 2. If water has a pka of 15.7, which of the following acids would be completely neutralized by a? ircle them. (Bold is correct answer) Acid pka A 14 B K eq A- + - A pka =14 pka = 15.7 K eq = K eq = Pka(conj acid) pka (acid) B- + pka = pka =33 K eq - B pka = 15.7 K eq pka = 15.7 K eq = K eq = Draw the following molecules in their three dimensional form and show, on the structure, the following: (1) The polarity of each polar bond ( >) and (2) The overall polarity direction of the molecule (+ >). If there is no molecular dipole moment, circle the structure 3 2 BF 3 et Dipole F F B F Dipole moments ancell 22. ompounds A and B, at right, have the same polar bond. owever, A has a dipole moment of 1.31 D while that of B is 2.11 D. Explain why B has almost double the dipole moment of A. Illustrate your explination. (int: Look at formal charges in structure B) et Dipole Extremely large dipole oriented toward negative charge

5 23. In each of the following chemical reactions, circle the acid and place an X over the conjugate base a a X All 3 + l All X X All All 3 Lewis Acid 24. ombustion analysis of an organic compound shows it to have 64.3% carbon and 7.2% hydrogen. What is the emperical formula for the compound? If there is 64.3% carbon and 7.2% hydrogen, there must be 28.5% oxygen in sample. In 100 g of sample there is: 64.3 g, 7.2 g and 28.5 g. In 100 g of sample there is: 64.3g/ = 5.35 moles of arbon, 7.2g/1.008 = 7.14 moles of ydrogen 28.5g/ = 1.78 moles of xygen Molar ratio : 5.35, 7.14, 1.78, divide all by smallest ratio i.e. 1.78, formula becomes Identify the Lewis acids and bases in the reactions of question (23). Lewis acids are electron acceptors Lewis bbases are electron donors a a + 4 Lewis Acid Lewis Base Lewis Acid Lewis Base + All 3 + l All 4 + Lewis Acid Lewis Base All All 3 Lewis Acid Lewis Base

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