IE EA IE EA502 EN text 2006502 40 F 2006 Cl 1604 Br

Ie ea ie ea502 en text 2006502 40 f 2006 cl 1604 br

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15. (IE EA) (IE EA)/502 EN (text) 2006/502 = 4.0 F 2006 kJ/mol 4.0 Cl 1604 kJ/mol 3.2 Br 1463 kJ/mol 2.9 I 1302 kJ/mol 2.6 The values calculated from IE and EA show the same trend as (and agree fairly closely) with the values given in the text. 16. a. There are two attractions of the form 4.0 3.0 2.8 2.5 J The J 17. Using the periodic table, we expect the general trend for electronegativity to be: 18. The most polar bond will have the greatest difference in electronegativity between the two atoms. From positions in the periodic table, we would predict: F Br
CHAPTER 13 BONDING: GENERAL CONCEPTS 529 19. The general trends in electronegativity used in Exercises 13.17 and 13.18 are only rules of thumb. In this exercise we use experimental values of electronegativities and can begin to see several exceptions. The order of EN using Figure 13.3 is: a. C (2.5) < N (3.0) < O (3.5) same as predicted b. Se (2.4) < S (2.5) < Cl (3.0) same c. Si (1.8) = Ge (1.8) = Sn (1.8) different d. Tl (1.8) = Ge (1.8) < S (2.5) different e. Rb (0.8) = K (0.8) < Na (0.9) different f. Ga (1.6) < B (2.0) < O (3.5) Most polar bonds using actual EN values: a. SiF and GeF (GeF predicted) b. PCl (same as predicted) c. SF (same as predicted) d. TiCl (same as predicted) e. SiH and SnH (SnH predicted) f. AlBr (TlBr predicted) 20. Use the electronegativity trend to predict the partial negative end and the partial positive end of the bond dipole (if there is one). To do this, you need to remember that H has electro-negativity between B and C and identical to P. Answers b, d, and e are incorrect. For d (Br same 21. See Exercise 20 for a discussion on bond dipoles. We will use arrows to indicate the bond dipoles. The arrow always points to the partial negative end of the bond dipole, which will always be to the more electronegative atom. The tail of the arrow indicates the partial positive

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