Inorganic Chemistry II Notes PDF

INORGANIC CHEMISTRY II ATOMIC STRUCTURE INORGANIC CHEMISTRY II RUTHERFORD’S GOLD FOIL EXPERIMENT Most of the α particles passed through the gold leaf with little or no deflection A few were deflected at wide...

INORGANIC CHEMISTRY II ATOMIC STRUCTURE INORGANIC CHEMISTRY II RUTHERFORD’S GOLD FOIL EXPERIMENT Most of the α particles passed through the gold leaf with little or no deflection A few were deflected at wide angles occasionally a particle rebounded from the foil, c, and was detected by a screen or counter placed on the same side of the foil as the source. SOME LOW-MASS PARTICLES NUCLEAR STRUCTURE Bohr's picture of the hydrogen atom. A single electron of mass me moves in a circular orbit with velocity v at a distance r from a nucleus of mass mn. BOHR THEORY OF THE HYDROGEN ATOM For a stable orbit to exist the outward force exerted by the moving electron trying to escape its circular orbit must be opposed exactly by the forces of attraction between the electron and the nucleus. BOHR THEORY OF THE HYDROGEN ATOM The outward force, F0, is expressed as: BOHR THEORY OF THE HYDROGEN ATOM This outward force, F0, is opposed by: 1. Electrostatic force 2. Gravitational force BOHR THEORY OF THE HYDROGEN ATOM Assume that Gravitational force is negligible Electrostatic force >>> Gravitational force BOHR THEORY OF THE HYDROGEN ATOM The electrostatic attractive force, Fe , between an electron of charge -e and a proton of charge +e, is BOHR THEORY OF THE HYDROGEN ATOM The condition for a stable orbit is that F0 + Fe equals zero: BOHR THEORY OF THE HYDROGEN ATOM Now we are able to calculate the energy of an electron moving in one of the Bohr orbits. The total energy, E, is the sum of the kinetic energy, KE, and the potential energy, PE: BOHR THEORY OF THE HYDROGEN ATOM in which KE is the energy due to motion, and PE is the energy due to electrostatic attraction, BOHR THEORY OF THE HYDROGEN ATOM Now, the total energy is BOHR THEORY OF THE HYDROGEN ATOM To determine the total energy, we only need to determine the value of orbit radius r To keep the electron from falling into the nucleus Bohr proposed a model in which the angular momentum of the orbiting electron could have only certain values. BOHR THEORY OF THE HYDROGEN ATOM The result of this restriction is that only certain electron orbits are possible. According to Bohr's postulate the quantum unit of angular momentum is h/2π, in which h is the constant in Planck's famous equation E = hv. (E is energy in ergs, h = 6.626196×10-27 erg sec, and ν is frequency in sec-1.) 6.626196×10 -27 erg sec = -34 6.626196×10 J sec BOHR THEORY OF THE HYDROGEN ATOM In mathematical terms Bohr's assumption was that in which n = 1,2,3, · · · (all integers to ∞). Solving for v in previous equationwe can write BOHR THEORY OF THE HYDROGEN ATOM Substituting the value of ν from the previous equation into the condition for a stable orbit, we obtain: BOHR THEORY OF THE HYDROGEN ATOM gives the radius of the possible electron orbits for the hydrogen atom in terms of the quantum number n. The energy associated with each possible orbit now can be calculated by substituting the value of r from into the energy which gives - BOHR THEORY OF THE HYDROGEN ATOM Exercise. Calculate the radius of the first Bohr orbit. Substituting n = 1 and the values of the constants we obtain The Bohr radius for n = 1 is designated a0. BOHR THEORY OF THE HYDROGEN ATOM Exercise. Calculate the velocity of an electron in the first Bohr orbit of a hydrogen atom. Substituting n = 1 and r = a0 = 0.529177 × 10-8 cm we obtain AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. The electromagnetic absorption spectrum of hydrogen atoms. The lines in this spectrum represent ultraviolet radiation that is absorbed by hydrogen atoms as a mixture of all wavelengths is passed through a gas sample. ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN The pattern of absorption frequencies is called an absorption spectrum and is an identifying property of any particular atom or molecule. Because frequency is directly proportional to energy (E = hv), the absorption spectrum of an atom, such as hydrogen, shows that the electron can have only certain energy values, as Bohr proposed. ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN It is common practice to express positions of absorption in terms of the wave number, which is the reciprocal of the wavelength, λ: Since: λv = c Thus wave number and energy are directly proportional. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. The lowest-energy absorption corresponds to the line at 82,259 cm-1 that the absorption lines are crowded closer together as the limit of 109,678 cm-1 is approached. Above this limit absorption is continuous. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. The emission spectrum from heated hydrogen atoms. The emission lines occur in series named for their discoverers: Lyman, Balmer, Paschen. The Brackett and Pfund series are farther to the right in the infrared region. The lines become more closely spaced to the left in each series until they finally merge at the series limit. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. If atoms and molecules are heated to high temperatures, light of certain frequencies is emitted. For example, hydrogen atoms emit red light when heated. An atom that possesses excess energy (an "excited" atom) emits light in a pattern known as its emission spectrum. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. Although the emission spectrum of hydrogen appears to be complicated, Johannes Rydberg formulated a fairly simple mathematical expression that gives all the line positions. This expression, called the Rydberg equation, is: In the Rydberg equation n and m are integers, with m greater than n; Rh is called the Rydberg constant and is known accurately from experiment to be 109,677.581 cm-1. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. Exercise. Calculate for the lines in the Lyman series (n=1) and m = 2, 3, and 4. AN EXPERIMENTAL ARRANGEMENT FOR STUDYING THE EMISSION SPECTRA OF HYDROGEN ATOMS. a) The Lyman series of lines arises from transitions from the n = 2, 3, 4, · · · levels into the n = 1 orbit. b) The Balmer series of lines arises from transitions from the n = 3, 4, 5, · · · levels into the n = 2 orbit. c) The Paschen series of lines arises from transitions from the n = 4, 5, 6, · · · levels into the n = 3 orbit. d) The Brackett series of lines arises from transitions from the n = 5, 6, · · · levels into the n = 4 orbit. e) The Pfund series of lines arises from transitions from the n = 6, · · · levels into the n = 5 orbit. LBPBP Land Bank of the Philippines Bababa Po ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN For the excited hydrogen atom with an n = 6 electron, there are five possible modes of decay, each with a finite probability of occurrence. Thus five emission lines, corresponding to five emitted photon energies (hv), result. ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN The transition energy (ΔEH) of any electron jump in the hydrogen atom is the energy difference between an initial state, I, and a final state, II. That is, ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN - ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN is equivalent to the Rydberg equation, with nI = n, nII = m, and Rh = (2π2mee4)/ch3. If we use the value of 9.109558 × 10-28 g for the rest mass of the electron, the Bohr-theory value of the Rydberg constant is: IONIZATION ENERGY OF ATOMIC HYDROGEN The ionization energy (IE) of an atom or molecule is the energy needed to remove an electron from the gaseous atom or molecule in its ground state, thereby forming a positive ion. To calculate IE for hydrogen we can start with this Equation: For the ground state nI = 1, and for the state in which the electron is removed completely from the atom nII = ∞. IONIZATION ENERGY OF ATOMIC HYDROGEN Thus Recall that Therefore ABSORPTION AND EMISSION SPECTRA OF ATOMIC HYDROGEN Then substituting 1/2α0 into the expression for IE we have Ionization energies usually are expressed in electron volts (eV). Since 1 erg = 6.24145 × 1011 eV we calculate The experimental value of the IE of the hydrogen atom is 13.598 eV. GENERAL BOHR THEORY FOR A ONE-ELECTRON ATOM The problem of one electron moving around any nucleus of charge +Z is very similar to the hydrogen-atom problem. Since the attractive force is -Ze2/r2 the condition for a stable orbit is: Proceeding from this condition in the same way as with the hydrogen atom we find: GENERAL BOHR THEORY FOR A ONE-ELECTRON ATOM and Thus for the general case of nuclear charge +Z, for a transition from nI to nII we have or simply GENERAL BOHR THEORY FOR A ONE-ELECTRON ATOM Exercise. Calculate the third ionization energy of a lithium atom. First ionization energy, IE1 Second ionization energy, IE2 Third ionization energy, IE3 of lithium is the energy required to remove the one remaining electron from Li2+. For lithium, Z = 3 and IE3 = (3)2(2.1799 × 10-11 erg) = 1.96191 × 10-10 erg = 122.45 eV. The experimental value also is 122.45 eV.