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Chapter 8

Chapter 8. Sections 4 & 5 Ions. Atomic Size. Size goes UP on going down a group.

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Chapter 8

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  1. Chapter 8 Sections 4 & 5 Ions

  2. Atomic Size • Size goes UP on going down a group. • Because electrons are added further from the nucleus, there is less attraction. This is due to additional energy levels and the shielding effect. Each additional energy level “shields” the electrons from being pulled in toward the nucleus. • Size goes DOWN on going across a period.

  3. Atomic Size Size decreases across a period owing to increase in the positive charge from the protons. Each added electron feels a greater and greater + charge because the protons are pulling in the same direction, where the electrons are scattered. Large Small

  4. Which is Bigger? • Na or K ? • Na or Mg ? • Al or I ?

  5. Which is Bigger? • Na or K ? K • Na or Mg ? Na • Al or I ? I

  6. Ion Sizes Does the size go up or down when losing an electron to form a cation?

  7. + + Li , 78 pm 2e and 3 p Ion Sizes Forming a cation. • CATIONS are SMALLER than the atoms from which they come. • The electron/proton attraction has gone UP and so size DECREASES. Li,152 pm 3e and 3p

  8. Ion Sizes Does the size go up or down when gaining an electron to form an anion?

  9. - - F, 71 pm F , 133 pm 9e and 9p 10 e and 9 p Ion Sizes Forming an anion. • ANIONS are LARGER than the atoms from which they come. • The electron/proton attraction has gone DOWN and so size INCREASES. • Trends in ion sizes are the same as atom sizes.

  10. Trends in Ion Sizes Figure 8.13

  11. Which is Bigger? • Cl or Cl- ? • K+ or K ? • Ca or Ca+2 ? • I- or Br- ?

  12. Which is Bigger? • Cl or Cl- ? Cl- • K+ or K ? K • Ca or Ca+2 ? Ca • I- or Br- ? I-

  13. Electron Configurations All you have to do is add or subtract electrons from the atom’s configuration! Mg Mg2+ O O2-

  14. Trends in Ionization Energy • IE increases across a period because the positive charge increases. • Metals lose electrons more easily than nonmetals. • Nonmetals lose electrons with difficulty (they like to GAIN electrons).

  15. Trends in Ionization Energy • IE decreases down a group • Because size increases (Shielding Effect) • Reducing ability generally increases down the periodic table.

  16. Which has a higher 1st ionization energy? • Mg or Ca ? • Al or S ? • Cs or Ba ?

  17. Which has a higher 1st ionization energy? • Mg or Ca ? Mg • Al or S ? S • Cs or Ba ? Ba

  18. Electronegativity,   is a measure of the ability of an atom in a molecule to attract electrons to itself. Concept proposed by Linus Pauling 1901-1994

  19. Periodic Trends: Electronegativity • In a group: Atoms with fewer energy levels can attract electrons better (less shielding). So, electronegativity decreases DOWN a group of elements. • In a period: More protons, while the energy levels are the same, means atoms can better attract electrons. So, electronegativity increases RIGHT in a period of elements.

  20. Electronegativity

  21. Energy Effects • The change in energy when separated gaseous ions are packed together to form an ionic solid. • M+(g) + X-(g)  MX(s) • Lattice energy is negative (exothermic) from the point of view of the system.

  22. Lattice Energy • To determine which compound will have the highest lattice energy, take into consideration the following: • The size of the ions in the compound • The smaller the size, the greater the lattice energy • The charge of the ions in the compound • The greater the charge, the greater the lattice energy

  23. Calculating ∆Hf • We can take advantage of the fact the energy is a state function and break the reaction into steps, the sum of which is the overall reaction. • Let’s do #41 Na(s) + ½ Cl2 (g)  NaCl(s) Given the following: Lattice energy -786 kJ/mol Ionization energy for Na 495 kJ/mol Electron affinity for Cl -349 kJ/mol Bond energy of Cl2 239 kJ/mol Enthalpy sublimation for Na 109 kJ/mol

  24. Process Step 1: Sublimation of Na Na(s)  Na(g) 109 kJ/mol Step 2: Ionization of Na Na (g)  Na+ (g) + e- 495 kJ/mol Step 3: Dissociation of Cl2 ½ Cl2 (g)  Cl(g) 119.5 kJ/mol Step 4: Formation of Cl- Cl (g) + e-  Cl-(g) -349 kJ/mol Step 5: Formation of NaCl Na+(g) + Cl-(g)  NaCl(s) -786 kJ/mol Na(s) + ½ Cl2 (g)  NaCl(s) -411.5 kJ/mol

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