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Atomic radius as a function of charge and atomic number PowerPoint Presentation
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Atomic radius as a function of charge and atomic number

Atomic radius as a function of charge and atomic number

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Atomic radius as a function of charge and atomic number

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  1. Now that we have taken the trouble to learn about electronic configurations, let see if what we learned works as an effective model for understanding some of the chemistry of the periodic table.

  2. Atomic radius as a function of charge and atomic number

  3. A spherically symmetric charge distribution seems to be associated with chemical stability; filled or half filled p and d shells also have spherical symmetry 2s 3s 1s

  4. Atomic radius as a function of charge and atomic number

  5. Ionization Energy Ionization energy: the energy it takes to remove an electron from a neutral atom

  6. A blowup of the first few rows Be has a filled s level 1s2 2s2 N has half filled p levels 1s2 2s2 2p3 Mg has filled 3S2 1s2 2s2 2p6 3s2 P has half filled p levels 1s2 2s2 2p6 3s2 3p3

  7. How do you measure the energy it takes to remove electrons?

  8. How can you measure the ionization of an ion of Na from Na+3 to Na+4? • 9543 kJ mol-1 • 9543 kJ mol-1/6*1023 ions /mol = 1.6x10-20 kJ/ion = 1600 x10-20 J/ion • E = h* • E = 6.6*10-34 J s* • 1.600 x10-17 J/ion = 6.6*10-34 J s* • = 2.4*1016 s-1 c =  ;  = c/  • = 3 *108 m s-1/2.4*1016 s-1 • = 1.25x 10-8 m

  9. You probably haven’t noticed but we have been using positive numbers to reflect energy that we need to invest in order to achieve the desired results. If on the other hand energy is given off during the desired process, we will indicate this using a negative sign. X + e- X-

  10. Lets look at the energetics of forming NaCl from Na metal and chlorine gas * McFay Chapter 6

  11. How can we determine how much energy is release to measure the formation of NaCl from Na and Cl2 at T = 298.15 K? 1. We can measure it directly 2. Since energy cannot be created or destroyed, we can also measure it indirectly in steps using a cycle. What do we need to know? Na(s) Na (g) +107.3 kJ/mol Na(g) Na+ + e- +495.8 kJ/mol Cl2 (g) 2 Cl +244 kJ/mol Cl + e- Cl- -348.6 kJ/mol -787 kJ/mol Na+ (g) + Cl-(g) NaCl(s)

  12. 0 kcal mol-1

  13. Is sodium chloride ionic in the gas phase or is it covalent? Do we have Na+ and Cl - floating around?

  14. 376.5 kJ mol-1 0 kJ mol-1 107.3+122+495.8= 725.1 H = 725.1-348.6 = 376.5 kJ mol-1 If the covalent NaCl bond is > -376.5 kJ mol-1, then we can have gaseous NaCl formed as a molecule.

  15. How about the reaction of Al with Br2? Al + 3/2Br2 = AlBr3 Al is a metal at T = 298.15 K Br2 is a liquid at T = 298.15 K How much heat if any is given off when Al reacts with Br2? McFay Chapter6

  16. How do you measure the energy it takes to remove electrons?

  17. Sublimation enthalpy of Al: Al(s)  Al(g) H 324 kJ mol-1 Ionization enthalpy: Al(g)  Al+1 + e- H 578 kJ mol-1 Al+1 Al+2 + e- H 1817 kJ mol-1 Al+2 Al+3 + 3 e- H 2745 kJ mol-1 Ionization enthalpy: Al(g)  Al+3 + 3 e- H 5140 kJ mol-1 Vaporization Enthalpy: Br2 (liq)  Br2 (gas) H 31.2 kJ mol-1 Bond strength: Br2 2 Br forming Br atoms H 192 kJ mol-1 Electron affinity: Br + e-  Br- H -320 kJ mol-1 Lattice energy of AlBr3

  18. MaXb aM+z(g) + bX-y(g) Note that the numbers are reported as the amount of energy necessary to break apart the ionic lattice.

  19. Sublimation enthalpy of Al: Al(s)  Al(g) H 324 kJ mol-1 Ionization enthalpy: Al(g)  Al+3 + 3 e- H 5140 kJ mol-1 Vaporization Enthalpy: Br2 (liq) 1 Br2 (gas) H 31.2 kJ mol-1 Bond strength: Br2 2 Br H 192 kJ mol-1 Electron affinity: Br + e-  Br- H -320 kJ mol-1 Lattice energy of AlBr3 H -5361kJ mol-1 Al+ 1.5 Br2 AlBr3 324+5140+(31.2+192)*1.5 = 5799 energy invested -320*3 - 5361 = -6321 energy returned -522 kJ mol-1 return on investment

  20. Production of the Alkali Metals 1 Na; Cl2 : produced by passing an electric current through molten NaCl; Lithium is produced in the same way. Molten NaCl + - Na+ Cl- At the anode Cl- is attracted and oxidized to Cl atoms which combine to form Cl2 At the cathode, Na+ is is attracted and reduced to Na metal which floats on top of the molten NaCl

  21. anions are attracted to the anode (+) cations are attracted to the cathode (-)

  22. Li metal is produced similarly Why use molten NaCl; molten LiCl? Why not use aqueous solutions of NaCl and LiCl?

  23. Production of the Alkali Metals In aqueous media, Na+ and Cl- carry the current and depending on the concentration, either Cl- is oxidized to Cl2 or water is oxidized to oxygen, but at the cathode water is reduced to H2 aqueous NaCl anode + - cathode Na+ Cl- At the anode Cl- is oxidized to Cl2; 2Cl- = Cl2 + 2 e- and/or 2 H2O = O2 + 4 H+ + 4 e- At the cathode: 2 e- + 2 H2O = H2 + 2 OH-

  24. Production of Potassium Na(l) + KCl (l) K + NaCl 850 °C Na (boiling temperature): 883 °C K (boiling temperature): 759 °C Le Chatelier’s principle: If a system at equilibrium is disturbed, it will react in a manner to restore equilibrium Rb (bp 688 °C) and Cs (bp 671 °C) prepared similarly using metallic Ca (bp 1484 °C)

  25. Aluminum, the most abundant element in the earths crust gets its name from alum: KAl(SO4)2.12H2O. The mineral it comes from is called Bauxite, from a town in France called Le Baux.

  26. Corundum Al2O3; Aluminum Oxide This is not a mineral you often hear about, but it is the second hardest natural mineral known to man. When trace amounts of titanium and iron get into the Al2O3 crystal lattice during its formation a beautiful blue sapphire is formed The red color is caused mainly by the presence of the element chromium. Its name comes from ruber, Latin for red. Sapphire Ruby

  27. Yellow Sapphire Electrolysis of Al2O3 in Na3AlF6 produces Al metal at about 1000 ° C. The aluminum has a greater density than the molten reactants and sinks to the bottom. The production of Al metal is the single largest consumer of electricity in the US.

  28. Relative reactivity of the halogens cca3 halogens Br- + Cl2 = ? 1 2Br- + Cl2 = Br2 + 2 Cl- I- + Cl2 = ? 2 2I- + Cl2 = I2 + 2 Cl- I- + Br2 = ? 4 I- + Br2 = I2 + 2Br-

  29. Production of the halogens Cl2 andF2:: electrolysis 2 HF + in the presence of KF = F2 + H2 Br - + Cl2 = Br2 + Cl- I - + Cl2 = I2 + Cl-

  30. Uses of the halogens F used to make –(CF2)n- which is called Teflon; used in some refrigerants called Freons Fluorotrichloromethane FCCl3 Dichlorodifluoromethane F2CCl2 Trichlorotrifluoroethane, Cl2FCCF2Cl Bromochlorodifluoromethane, BrClCF2 Dibromotetrafluoroethane, BrCF2CCF2Br Chlorodifluoromethane ClF2CH Cl2 used in water purification as a disinfectant Br2 major use in in making AgBr used in photography; this may change

  31. Iodine: an essential for normal thyroid function (usually injested as NaI or as organic iodide; used as a contrasting agent for medical purposes; Radioactive iodine used to treat hyperactive thyroid problems. Reactivity of the halogens F2 > Cl2> Br2> I2 cca3 halogen

  32. Noble gases He, Ne Ar undergo no chemistry Xe reacts with F2 to form various adducts XeF2, XeF4, XeF6 Ar, Kr Xe differ from Ne and He in they have unfilled d levels reasonbly close in energy Uses: He used for deep sea diving; in Magnetic resonance imaging; boiling point is 4 K; as a carrier gas in scientific instruments; Ar: used in chemistry to provide an inert atmosphere; in light bulbs

  33. A 1.0 g sample of an alkaline earth metal M reacts completely with 0.8092 g of chlorine gas to yield and ionic salt with the formula MCl2. In the process 9.46 kJ of heat is released. What is the molecular mass and identity of the metal M? M + Cl2 = MCl2 1 mol of M reacts with 1 mol Cl2 0.8092g/71g/mol = 0.0114 mol 1 g M/gAt Wt M = 0.0114 mol M = 87.7 g/mol A check of the periodic table reveals that M is? Sr If 9.46 kJ/mol of heat was released, how much heat would be released if 1 mol of SrCl2 was formed? -9.46 kJ/g *87.7 g/mol = -829.6 kJ/mol