Introduction to inorganic chemistry ppt

a level inorganic chemistry notes pdf and basic organic chemistry ppt
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General Chemistry - Inorganic Chemistry Outline 1. Introduction 2. Substances and Separation 3. Atoms and Molecules 4. Atomic Structure 5. Hydrogen 6. Noble Gases 7. Oxygen 8. Water and Hydrogen Peroxide 9. Ionic Bond and Salts 10. Covalent Bond 11. Metallic Bond 12. The Chemical Equilibrium 13. Acids and Bases 14. Redox Reactions Slide 1 General Chemistry Prof. Dr. T. Jüstel References Basic • E. Riedel, Allgemeine und anorganische Chemie deGruyter, 7. Auflage 1999 • C.E. Mortimer, U. Müller, Chemie Thieme, 8. Auflage 2003 • P.W. Atkins, J.A. Beran, Chemie – einfach alles Wiley-VCH, 2. Auflage 1998 • M. Binnewies, M. Jäckel, H. Willner, G. Rayner-Canham, Allgemeine und Anorganische Chemie, Spektrum, 1. Auflage 2004 Advanced • E. Riedel, Anorganische Chemie deGruyter, 6. Auflage 2004 • A.F. Hollemann, N. Wiberg, Lehrbuch der anorganischen Chemie deGruyter, 102. Auflage 2007 • J.E. Huheey, E.A. Keiter, R. Keiter, Anorganische Chemie deGruyter, 3. Auflage 2003 Slide 2 General Chemistry Prof. Dr. T. Jüstel 1. Introduction What Is Chemistry? “Chemistry is the science of substances, their structure, their properties, and their conversion, which results in the formation of other substances“ (Linus Carl Pauling 1956, Nobel prizes: Chemistry 1954, Peace 1962) Example SiO + 2 Mg 2 MgO + Si (Powder) Chemical process 2 (e.g. reduction) Si (Single crystal, wafer) Physical process (e.g. recrystallization) Native quartz crystals 12‘‘ Wafer for semiconductor production Slide 3 General Chemistry Prof. Dr. T. Jüstel 1. Introduction Chemistry Has Been Exponentially Growing for Many Years Number of original publications traceable in Chemical Abstracts (CA) up to 2003 More than 30 mill. 26 mill. abstracts compounds and bio- sequences in CA database Development of the number are registered of publications per year Bio sequences 41% 1830 400 Polymers 3% 1930 55000 Alloys 2% Organics 47% 1985 460000 Inorganic 2% 1995 700000 Coordination 5% 2001 755000 compounds Slide 4 General Chemistry Prof. Dr. T. Jüstel 2. Substances and Separation Substances Are Compounds, which Chemical and Physical Properties Are Independent of Size and Shape Example: Stainless steel → driller, knife, frames, scissors, screw-drivers, etc. Substance Heterogeneous substance Homogeneous substance (non-uniform on a microscopic level) (uniform on a microscopic level) Solid-Solid Solids mixtures (granite) Alloy (brass) Solid-Liquid Suspension (milk of lime) Solution (NaCl-solution) Solid-Gaseous Aerosol (smoke) (Nanoscopic) Aerosol Liquid-Liquid Emulsion (milk) Liquid solution (ethanol in water) Liquid-Gaseous Aerosol (fog, foam) Gas solution (oxygen in water) Pure substances Slide 5 General Chemistry Prof. Dr. T. Jüstel 2. Substances and Separation Physical Methods for Separation of Heterogeneous Substances 1. Differences in density “lighter“ liquid Solid-Solid Re-slurry (washing of gold) more “heavy“ liquid Solid-Liquid sedimentation (1 G) 4 centrifugation (up to 10 G) filter Liquid-Liquid separation (separating funnel) crucible glass frit 2. Differences in particle size Solid-Solid sieve plug Solid-Liquid filtration (filter crucible) Solid-Gaseous filtration (air filter) suction port suction flask filtrate Slide 6 General Chemistry Prof. Dr. T. Jüstel 2. Substances and Separation Separation of Homogeneous Systems 1. Physical methods Vaporising and condensation: seawater→ rainwater Cooling: salt solutions→ salt crystals Condensation and vaporising: air→ N , O , noble gases 2 2 Adsorption and desorption Gas chromatography dissolution of vaporisable substances Liquid chromatography dissolution of solid substances Paper chromatography dissolution of solid substances (ß-carotine) 235/238 235 238 Centrifugation (gases) UF → UF + UF 6 6 6 2. Chemical methods 2+ 2+ 2- 2+ Precipitation Mg , Hg (aq) + S→ HgS↓ + Mg (aq) Gas purification drying of Ar or N via: 2 P O + 6 H O → 4 H PO 4 10 2 3 4 Slide 7 General Chemistry Prof. Dr. T. Jüstel 2. Substances and Separation Classification of Substances Heterogeneous substances system consists of different phases Homogeneous substances system consists of only one phase 1. Solutions phases consists of different types of molecules 2. Pure substances phase consists of a single type of molecules a. Compounds mol. structure based on different types of atoms b. Elements mol. structure based on a single type of atoms All substances can be cleaved into the corresponding elements through dissociation processes at sufficiently high temperatures: 400 °C 2 HgO(s) 2 Hg(s) + O (g) 2 6000 °C MgO(s) Mg(g) + O(g) (no formation of O , because oxygen exists 2 almost exclusively dissociated at 6000 °C) Slide 8 General Chemistry Prof. Dr. T. Jüstel 3. Atoms and Molecules Content 3.1 Law of mass preservation 3.2 Law of constant proportions 3.3 Law of multiple proportions 3.4 Law of equivalent proportions 3.5 Dalton‘s atom hypothesis 3.6 Volume ratio during chemical reactions 3.7 Relative mass of atoms 3.8 Molar masses 3.9 Absolute atomic masses Slide 9 General Chemistry Prof. Dr. T. Jüstel 3.1 Law of Mass Preservation During All Chemical Reactions the Total Mass of the Reactants Remains Constant (Antoine Lavoisier 1774) Experimental confirmation by thorough determination of the mass of the educts and products (Hans Landolt 1908) -5 ⇒ change of mass 10 % But: chemical reactions are subject to energy conversion ∆E 2 Energy/mass equivalent: E = mc (Albert Einstein 1915) -9 Highly exothermic reaction: ∆E = 500 kJ ⇒ change of mass 10 % ⇒ Change of mass during chemical reactions is beyond weighing accuracy Slide 10 General Chemistry Prof. Dr. T. Jüstel 3.2 Law of Constant Proportions The Mass Ratio of Two Chemical Elements Reacting to One Compound Is Constant (Joseph Louis Proust 1799) Hoffman‘s decomposition apparatus: Volume ratio H/O = 2:1 (H O) 2 Mass ratio Water electrolysis H/O = 1:7.936 Further examples for constant mass ratios Fe/S = 1:0.57 (FeS) Mg/O = 1:0.666 (MgO) Na/Cl = 1:1.542 (NaCl) H/N = 1:4.632 (NH ) 3 Slide 11 General Chemistry Prof. Dr. T. Jüstel 3.3 Law of Multiple Proportions The Mass Ratios of Two Chemical Elements Reacting to One Compound Are Related by One Simple Number (John Dalton 1803) Examples for multiple mass ratios Nitrogen oxides “known today“ • N/O = 1:0.571 = 1:1x0.571 N O Dinitrogen monoxides 2 • N/O = 1:1.142 = 1:2x0.571 NO Nitrogen monoxide • N/O = 1:1.713 = 1:3x0.571 N O Dinitrogen trioxide 2 3 • N/O = 1:2.284 = 1:4x0.571 NO Nitrogen dioxide 2 • N/O = 1:2.855 = 1:5x0.571 N O Dinitrogen pentoxide 2 5 Carbon oxides • C/O = 1:1.333 = 1:1x1.333 CO Carbon monoxide • C/O = 1:2.666 = 1:2x1.333 CO Carbon dioxide 2 Slide 12 General Chemistry Prof. Dr. T. Jüstel 3.4 Law of Equivalent Proportions Elements Are United to Chemical Compounds With Respect to Certain Masses or Integer Multiples Thereof (J.B. Richter 1791) By comparison of the mass ratios of nitrogen and oxygen in known nitrogen oxides with the corresponding ratios for the reaction of oxygen or nitrogen with hydrogen, it became clear that they only can be synthesized in certain integer ratios Nitrogen oxides, again NH : H O 3 2 1. N/O = 1:0.571 = (3x4.632):(1x7.936) 1.0 N:0.5 O 2. N/O = 1:1.142 = (3x4.632):(2x7.936) 1.0 N:1.0 O 3. N/O = 1:1.713 = (3x4.632):(3x7.936) 1.0 N:1.5 O 4. N/O = 1:2.284 = (3x4.632):(4x7.936) 1.0 N:2.0 O 5. N/O = 1:2.855 = (3x4.632):(5x7.936) 1.0 N:2.5 O ⇒ Concept of equivalent masses Slide 13 General Chemistry Prof. Dr. T. Jüstel 3.5 Dalton‘s Atom Hypothesis Atoms as Smallest Parts of Matter (John Dalton 1808) 1. Elements cannot be split indefinitely, since they consist of tiny non-cleavable particles, the so-called atoms 2. All atoms of an element are of one sort (mass and shape) 3. Atoms of different elements possess different properties 2 A + B → A B 2 A + B → AB 2 A + 3 B → A B 2 3 A + 2 B → AB 2 2 A + 5 B → A B 2 5 etc. Relative atom masses cannot be measured directly as long as the exact ratio of the atoms in the newly formed compound is not know Slide 14 General Chemistry Prof. Dr. T. Jüstel 3.6 Volume Ratios During Chemical Reactions Observations Regarding Gases Every quantity of a substance equals a certain gas volume at a certain pressure and temperature, if that quantity is gaseous or can be vaporized Stoichiometric law of mass ⇒ law of volume ⇒ The volume ratio of two gaseous elements reacting to one chemical compound is constant and can be expressed by simple integer numbers Examples 2 volumes of hydrogen + 1 volume of oxygen → 2 volumes of water vapour 1 volume of hydrogen + 1 volume of chlorine → 2 volumes of hydrogen chloride Slide 15 General Chemistry Prof. Dr. T. Jüstel 3.7 Relative Atom Masses Relative Atom Masses Can Be Derived by Experimentally Determined Mass Ratios During Chemical Reactions (See Chapter 3.2) Mass ratios in water: H/O = 1:7.936 Ratio of atomic numbers in water: H O ⇒ 1 O = 15.872 H 2 Definition of a point of reference needed: 12 The carbon isotope C was chosen by IUPAC in 1961 to be the reference point and exhibits a relative atom mass A = 12.000 r Element Rel. atom mass A r Definition of the atomic mass unit: Hydrogen 1.008 u 12 1 u = 1/12 m ( C-atom) Chlorine 35.453 u Oxygen 15.999 u Elements consist of a number of isotopes Nitrogen 14.007 u 13 14 Carbon for example also contains C and C Carbon 12.011 u A (C) 12 r Slide 16 General Chemistry Prof. Dr. T. Jüstel 3.8 Molar Masses The Amount of an Element in Gram, Which Equals the Numerical Value of the Relative Atom, Always Contains the Same Number of Atoms, i.e. N Atoms A The mass of one mole of a substance is called the molar mass, M. The amount of that substance is thus given by: M = Molar mass g/mol n = m/M m = Mass g The corresponding particle count is: n = Amount of substance mol . N = Avogadro-constant particles/mol A N = n N A N = Particle count Calculation of molar masses: M(H O) = 2 M(H) + M(O) = 21.008 g/mol + 15.999 g/mol = 18.015 g/mol 2 M(CO ) = M(C) + 2 M(O) = 12.011 g/mol + 215.999 g/mol = 44.009 g/mol 2 Slide 17 General Chemistry Prof. Dr. T. Jüstel 3.9 Absolute Atom Masses The Absolute Atom Masses Are Given by the Division of the Molar Mass by the Avogadro-Constant, N A Unit cell of copper Determination of the Avogadro-constant necessary (cubic-face-centered) m 4M(Cu) -3 Density ρ = 8.93 gcm == 3 V N a A 4M(Cu) . 23 -1 ⇒ N = = 6.02214 10 mol A 3 ρa . -8 a a = lattice constant of Cu = 3.62 10 cm = 3.62 Å Example 12 12 m( C) = M( C)/N A -1 = 12.0 gmol /N A -23 = 1.9926910 g Slide 18 General Chemistry Prof. Dr. T. Jüstel 3.9 Absolute Atom Masses The Absolute Atom Masses Can Be Calculated by Means of the Atomic Mass Unit, u . 12 . -24 Atomic mass unit 1 u = 1/12 m( C) = 1.66054 10 g -24 Element Rel. atom mass A Molar mass g/mol Abs. atom mass 10 g r Hydrogen 1.008 u 1.008 1.678 Chlorine 35.453 u 35.453 58.871 Oxygen 15.999 u 15.999 26.567 Nitrogen 14.007 u 14.007 23.259 Carbon 12.011 u 12.011 19.945 In day-to-day life only relative atom and molecule masses or atom and molecule weights are used. Strictly speaking, the term weight is inadmissible, because weight is dependent on the gravitational field, in contrary to mass. Slide 19 General Chemistry Prof. Dr. T. Jüstel 4. The Atomic Structure Content 4.1 Fundamental Particles 4.2 Atomic Nuclei and Chemical Elements 4.3 Isotopes 4.4 Mass Defect – Stability of Matter 4.5 Radioactive Decay 4.6 Nuclear Reactions 4.7 Origin and Abundance of the Elements 4.8 Quantum Theory According to Planck 4.9 Atomic Spectra 4.10 Bohr‘s Atomic Model 4.11 The Wave Character of Electrons 4.12 Eigen Functions of the Schrödinger-Equation 4.13 Quantum Numbers 4.14 Energies of the Orbitals 4.15 Structure of the Periodic Table Slide 20 General Chemistry Prof. Dr. T. Jüstel

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