This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 1 Homework #6 Chapter 13 Bonding: General Concepts 13. Electronegativity increases from left to right across periods and from low to high along groups. a) C<N<O b) Se<S<Cl c) Sn<Ge< Si d) Tl<Ge<S e) Rb<K<Na f) Ga<B<O 14. The most polar bond will be the bond that will be between elements that have the greatest difference in electronegativity. a) Ge-F b) P-Cl c) S-F d) Ti-Cl e) Sn-H f) Tl-Br 16. The greater the difference in electronegativity the more polar the bond F-H>O-H>N-H>C-H>P-H 17. The more polar the bonds the more ionic character the bond has. Br-Br<N-O<C-F<Ca-O<K-F 19. When an element forms an anion (- charged ion) it gains an electron. Since there are more electrons, there is more electron-electron repulsion. In addition, the number of protons does not change, causing the protons to have a greater negative charge to contain, resulting in a larger atomic radius. When an element forms a cation (+ charged ion) it loses an electron. Since there are fewer electrons, there is less electron-electron repulsion. In addition, the number of protons does not change, causing the protons to have a smaller negative charge to contain, resulting in a smaller atomic radius. Isoelectronic: Ions containing the same number of electrons. In isoelectronic ions the ion with the largest atomic number (Z) will have the smallest radius because the ion with the largest Z will have the most protons. The ion with the smallest atomic number (Z) will have the largest radius. 20. Sc 3+ p=21 e- =18 Cl- p=17 e- =18 K + p=19 e- =18 Ca 2+ p=20 e- =18 S 2- p=16 e- =18 From smallest to largest Sc 3+ <Ca 2+ < K + < Cl- <S 2- Therefore in the order of the picture K + , Ca 2+ , Sc 3+ , S 2- , and Cl- 21. a) Cu> Cu + >Cu 2+ b) Pt 2+ >Pd 2+ >Ni 2+ c) O 2- >O- >O d)* La 3+ >Eu 3+ >Gd 3+ > Yb 3+ e) Te 2- >I- >Cs + >Ba 2+ >La 3+ * La 3+ is probably the smallest ion because it loses its entire 6s shell 2 22. a) Mg 2+ =1s 2 2s 2 2p 6 b) N 3- =1s 2 2s 2 2p 6 c) Be 2+ =1s 2 Sn 2+ =[Kr]5s 2 4d 10 O 2- =1s 2 2s 2 2p 6 Rb + =[Ar]4s 2 3d 10 4p 6 K + =1s 2 2s 2 2p 6 3s 2 3p 6 F- =1s 2 2s 2 2p 6 Ba 2+ =[Kr]5s 2 4d 10 5p 6 Al 3+ =1s 2 2s 2 2p 6 Te 2- =[Kr]5s 2 4d 10 5p 6 Se 2- =[Ar]4s 2 3d 10 4p 6 Tl + =[Xe]6s 2 4f 14 5d 10 I- =[Kr]5s 2 4d 10 5p 6 As 3+ =[Ar]4s 2 3d 10 23. a) Cs + =[Xe] b) Sr 2+ =[Kr] c) Ca 2+ =[Ar] d) Al 3+ =[Ne] S 2- =[Ar] F- =[Ne] N 3- =[Ne] Br- =[Kr] 24. a) Sc 3+ b) Te 2- c) Ti 4+ and Ce 4+ d) Ba 2+ 27. a) Al 2 S 3 aluminum sulfide b) K 3 N potassium nitride c) MgCl 2 magnesium chloride d) CsBr cesium bromide 37. H D bonds broken D bonds formed a) 2 H H Cl Cl H Cl H D D D 432 239 2 427 183 kJ kJ kJ kJ mol mol mol mol H b) 3 6 N N H H N H H D D D 941 3 432 6 391 109 kJ kJ kJ kJ mol mol mol mol H c) 2 2 2 C N H H C H C N N H H D D D D D...
View Full Document
This note was uploaded on 01/18/2012 for the course CHEM 1B taught by Professor Watts during the Winter '08 term at UCSB.
- Winter '08