CHEM 231 Inorganic Chemistry Lecture 01 PDF

CHEM 231 Inorganic Chemistry I, 3.0 Credits Dr. Mohamed Qenawy Ass. Prof. @ Chemistry Department, UAEU  Inorganic Chemistry I (CHEM231) This presentation contains pictures copyrighted by Prentice Ha...

CHEM 231 Inorganic Chemistry I, 3.0 Credits Dr. Mohamed Qenawy Ass. Prof. @ Chemistry Department, UAEU  Inorganic Chemistry I (CHEM231) This presentation contains pictures copyrighted by Prentice Hall from the textbook ‘Inorganic Chemistry’ by Housecroft & Sharpe.’ 1.1 Introduction 1.2 Fundamental particles of an atom 1.3 Atomic number, mass number and isotopes 1.4 Successes in early quantum theory  The Modern View of Atomic Structure The atom is mostly an empty space: -Nucleus - Protons and Neutrons. -Electron Cloud -Where you might find an electron. Table 1.1 Properties of the proton, electron and neutron.  Atomic Number, Mass Number and Isotopes ❑Atomic number (Z) = number of protons in the nucleus. The atomic number defines the element. ❑Mass number (A) = total number of nucleons in the nucleus (i.e., protons and neutrons). ❑By convention, for element E, we write (No. of N = A-Z) (No. of P + N) Mass number A (No. of P = No. of E) Atomic number Z E Elemental symbol ❑Isotopes vary in number of neutrons and hence mass number. ❑In a neutral atom, the number of electrons = number of protons  Atomic Mass Units (amu)/Relative Atomic Mass (Ar)-The Atomic Mass Scale ❑One 12C atom weighs 1.9926x10-23g ❑We define: mass of one 12C atom = exactly 12 amu. ❑(1amu = 1/12 of the mass of one 12C atom) Hence: ❑1 amu = 1.66054 x 10-24 g (1/12 of 1.9926x10-23g) ❑1 g = 1/1.66054x10-24 amu = 6.02214 x 1023 amu ❑The atomic weights (average atomic masses) of the elements are listed on the periodic table in units of amu ❑Values in amu are the same in units of g/mole.  Isotopes ❑Isotopes are atoms of the same element with different numbers of neutrons and consequently different masse (mixtures of isotopes). ❑If an element has only one nuclide (e.g., F-19, P-31), then the element is monotopic. ❑Isotopes can be separated and quantified by mass spectrometry.  Mass Spectrometry ❑The relative signal intensities of the peaks in the mass spectrum of Cl indicate the fractional abundances of the isotopes Cl-35 and Cl-37. NOTE THERE IS NO PEAK AT 35.45 AMU!  Relative Atomic Mass Calculate the value of Ar for naturally occurring chlorine if the distribution of isotopes is 75.77% 1735Cl and 24.23% 1737Cl. Accurate masses for 35Cl and 37Cl are 34.97 and 36.97. The relative atomic mass of chlorine is the weighted mean of the mass numbers of the two isotopes: Relative atomic mass, Ar = (75.77/100 x 34.97) + (24.23/100 x 36.97) Ar = 35.45  Allotropes ❑Allotropes of an element are different structural modification of that element. Figure 16.6 Representations of the structures of some allotropes of sulfur: (a) S6 , (b) S7, (c) S8 and (d) catena-S (the chain continues at each end).  Bohr’s theory of the atomic spectrum of hydrogen ❑Electrons determine the properties of atoms, ions, molecules. ❑Electron treated as a particle (old theories) and as a wave (recent models) ❑Bohr model – only directly applicable to 1-electron systems ❑Wave model – applicable to all systems but only 1885-1962 solvable exactly for 1-electron systems  Hydrogen Emission Spectrum ❑ Atoms emit and absorb characteristic wavelengths of light ❑ Wavelength of light is related (inversely) to energy (E=hυ=hc/λ) Spectral mission lines Hydrogen Emission Spectrum − 1  1 1  Wavenumber, cm -1 = = RH  2 − '2   n n  ' n>n Where RH is the Rydberg e 4 me constant RH = 8 0 h 3c 2 for hydrogen, 1.097 x 105 cm-1  Hydrogen Emission Spectrum by Lyman ❑ Transitions made from higher levels to the first orbit form the Lyman Series The formula for Lyman Series is: − 1 1 1  = = RH  2 − '2 ,  1 n  n = 2,3,4,..... and RH is Rydberg constant. These transitions emit UV photons  Hydrogen Emission Spectrum by Balmer ❑ Transitions made from higher levels to the second orbit form the Balmer Series The formula for Balmer Series is: − 1 1 1  = = RH  2 − '2 ,  1 n  n = 2,3,4,..... and RH is Rydberg constant. These transitions emit Visible photons  Hydrogen Emission Spectrum by Paschen ❑ Transitions made from higher levels to the third orbit form the Paschen Series The formula for Paschen Series is: − 1 1 1  = = RH  2 − '2 ,  1 n  n = 2,3,4,..... and RH is Rydberg constant. These transitions emit IR photons Bohr’s theory of the atomic spectrum of hydrogen ❑ Bohr Explained the line Emission spectra of hydrogen with a model of the atom where the energy of the electron is quantized. ❑ Bohr’s model explains much chemistry but does not correctly predict emission spectra for atoms other than hydrogen. ❑The atom can absorb light resulting in the formation of and excited state. When the electron is in the lowest available orbit, this is the ground state. Bohr Model: the radius of each allowed circular orbit r n  0h n2 2 e me Z 4 2 Z 2 rn = ; En = − = −13.6 2 eV me Ze 2 8 o h n 2 2 2 n ❑  0 = permittivity of free space = 8.854x10-12 J-1C2m-1 ❑ h = Planck constant = 6.626x10-34 Js ❑ n = principal quantum number (1, 2, 3..) ❑ me = mass of the electron = 9.109x10-31kg ❑ e = charge on the electron = 1.602x10-19C ❑ Z= nuclear charge (=1 for H) ❑ NOTE THAT THE ONLY VARIABLES ARE Z AND n THESE EQUATIONS ARE APPLICABLE TO 1-ELECTRON SYSTEMS ONLY  Bohr Model  0h n2 2 4 2 e me Z rn = En = − 8 o h n 2 2 2 me Ze 2 e 4 me En = − ❑r varies with n2 8 o h n 2 2 2 ❑r is inversely proportional to Z 4 e me RH = ❑E varies inversely with n2 8 0 h c 2 3 ❑E varies with Z2 1 En = − RH hc 2 For hydrogen Z = 1 n  Bohr Model Q. Compare the ground state radii of of H, He+ and Li2+ ? Ground state radius of H is:  0h n 2 2 rn =  0 h 212 −11 me Ze 2 r1 = = 5.29 x10 m = a0 me (1)e 2 The ground state radius of the H atom (52.9pm) is often used as a fundamental constant, a0 (Bohr Radius for hydrogen atom) + 1 1 −11 r1 ( He ) = r1 ( H )  = r1 ( H )  = 2.65 x10 m Z 2 1 1 r1 ( Li ) = r1 ( H )  = r1 ( H )  = 1.76 x10 −11 m 2+ Z 3  Bohr Model Q. Calculate the energy needed to promote an electron from n=1 to n=4 for the H atom. What wavelength of light would be absorbed for this transition? e 4 me (1) 2 e 4 me (1) 2 e 4 me  1 −18 E = E4 − E1 = − + =  1 −  = 2.04 x10 J 8 o h (4) 8 o h (1) 8 o h  16  2 2 2 2 2 2 2 2 E = 2.04 x10 −18 J  6.022 x10 23 mol −1 = 1.23x106 Jmol −1 = 1230kJmol −1 hc E = h =  hc 6.626 x10 −34 Js  3.00 x108 m / s = = −18 = 9. 74 x10 −8 m = 97.4nm E 2.04 x10 J This absorption/emission is part of the Lyman series – UV photons.