Self-Assessment: Structure of the Atom, Quiz Solutions

Self-Assessment: Structure of the Atom Weekly Homework Quiz - Solution Outlines Problem #1 140 142 (a) Cerium has many isotopes (8 to be exact), but only Ce and Ce are present in substantial amounts. Which isotope of cerium is the most abundant? from the periodic table, you see that the atomic mass of Ce is 140.115, which must be the weighted sum of the isotope masses 140 142 here we assume that it is necessary to consider only Ce and Ce 142 140 so, (x mass of Ce) + (1-x) mass of Ce = 140.115 for the purposes of this decision, we can approximate the atomic masses of the isotopes 140 142 as ~140 for Ce and ~142 for Ce and solve for x x = 0.94 which means that Ce is the most abundant isotope of cerium 140 Production of chromium in an electric arc furnace would involve the reaction of carbon with chromium sesquioxide according to the following reaction: Cr2O3 + a C = b CO + c Cr (i) Balance the equation, i.e., specify the values of a, b, and c. Insert the correct values below. Cr2O3 + 3C = 3 CO + 2Cr (ii) Calculate the minimum amount of chromium (in kg) produced if the reaction consumed 333 kg C and produced the stoichiometric amount of Cr. Assume 100% efficiency. 333 kg C = 333000/12.011 = 27725 moles C the stoichiometric amount of Cr is 2/3 amount of carbon on a molar basis. Therefore, amount of Cr = 27725 x 2/3 moles of Cr = (27725 x 2/3) x 51.996 = 961 kg Cr Problem #2 (a) Antimony has two isotopes, 121Sb and 123Sb. Which isotope has the higher natural abundance? from the periodic table, you see that the atomic mass of Sb is 121.757, which must be the weighted sum of the isotope masses so, (x mass of 121Sb) + (1-x) mass of 123Sb = 121.757 for the purposes of this decision, we can approximate the atomic masses of the isotopes as ~121 for 121Sb and ~123 for 123Sb and solve for x x = 0.62 which means that 121Sb is the more abundant isotope of antimony (b) Production of hafnium by the Kroll Process would involve the reaction of magnesium with hafnium tetrachloride according to the following reaction: HfCl4 + a Mg = b MgCl2 + c Hf (i) Balance the equation, i.e., specify the values of a, b, and c. Insert the correct values below. HfCl4 + 2 Mg = 2 MgCl2 + Hf (ii) Calculate the minimum amount of magnesium (in kg) needed to convert 111 kg HfCl4 into elemental hafnium. 111 kg HfCl4 = 111000/[178.49 + (4  35.45)] = 347 moles HfCl4 the stoichiometric amount of Mg is twice the amount of HfCl4 on a molar basis
amount of Mg = 347  2 moles of Mg = (347  2)  24.305 = 16.9 kg Mg Problem #3 (a) Show by means of a calculation that blue light of wavelength,  = 444 nm, is not capable of exciting electrons in Li2+(g) from the state n = 2 to n = 4. let’s equate the energy required to excite electrons in Li2+(g) from the state n = 2 to n = 4 with the minimum energy needed from an incident photon to cause the excitation hc   1 1 KZ 2  2
2  n f ni =   = hc  1 1 KZ 2  2
2  n f ni = 6.6  10
34  3.00  108  1 1 2.18  10
18  32  2
2 2 4 = 5.38  10-8 m = 53.8 nm < 444 nm  since E scales with 1/, blue light of wavelength  = 444 nm does not have enough energy per photon to cause the excitation (b) Is the value of the energy of transition from the state n = 2 to n = 4 in Li2+,
E24, greater than or less than the value of the energy of transition from the state n = 1 to n = 2 in Li2+,
E12? Explain with the use of an energy level diagram. There is no need to calculate the values of the two quantities. Problem #4 (a) In a gas discharge tube what is the minimum frequency () of a photon capable of ionizing ground-state electrons in Li2+? here is the central concept: the energy of the incident photon must be at least as great as the ionization energy (I.E.) Li2+ is a one-electron atom, so we can calculate the I.E. using the Bohr Model KZ 2 ) , where Z = 3 and n = 1 for the ground-state of Li2+ and K is n2 the ground-state energy of atomic hydrogen I .E. = E
E1 = 0
(
energy of incident photon is given by E = h
 = KZ 2 h (2.1810
18 J )(3)2 6.6  10
34 = 2.97 1016 Hz (b) Explain with reference to the relevant physical forces why the value of the 1st ionization energy of Li is less than the 3rd ionization energy of Li. the 1st ionization represents the removal of one of the 3 electrons from neutral Li the 3rd ionization represents the removal of the single electron from the Li2+ ion in the second case, the single electron alone feels the pull of the positive charge of the nucleus in the first case the same positive charge is felt by three electrons; hence, each electron feels a weaker pull than is the case with a lone electron under the influence of the same positive charge