************************************************* Chemistry for Engineers Homework1 Atomic Structure Answer Key *************************************************

************************************************************************************************** Chemistry for Engineers Homework 1: Atomic Structure ************************************************************************************************** Hand in next Friday 13:00 Q1: Three lasers each emit monochromatic radiation of different wavelengths. Laser A’s radiation is found to have a frequency of 4.8 x Hz. Laser B emits photons of energy kJ/mol. Laser C had a wavelength specification of 2.8 microns. Show calculation(s) to support your answer to the following: a) Which laser emits radiation of frequency 107 million million Hertz? b) Which laser has a photon energy of 3.18 x J? c) Which laser emits a beam of visible light? d) What is the energy of each photon emitted from laser C? e) Which of the three lasers emits radiation of the shortest wavelength? f) Which of the three lasers emits infrared photons? Q2: Consider the following nuclides: mercury-196, thallium-205, osmium-189, iridium-193, lead-204 and gold-197 a) Which isotope has the lowest mass? b) Which isotope has the most neutrons per nucleus? c) Which isotope has the fewest protons per nucleus? d) Which nuclide has the highest proportion of protons per nucleus (as a percentage)? (as always, show calculations!) Q5: Using the Rydberg equation:  E/h = c/  = = R[(1/n L 2 ) – (1/n U 2 )] for the atomic spectrum of hydrogen a) Calculate the photon energy and wavelength corresponding to i) the three lowest energy lines of the Balmer series and ii) the convergence limit of the Balmer series. b) By calculating the wavelength of the highest energy line of the Paschen series, determine if the two series of lines overlap Useful constants: R = 3.29 x Hz h = 6.63 x Js c = 3 x 10 8 m/s N A = x ). (9 marks) Q3: In each case, show your reasoning. a) Identify the doubly-charged cation ion containing 18 electrons and a single nucleus that includes 24 neutrons b) Identify the singly-charged anion ion containing 36 electrons and a single nucleus that includes 46 neutrons? c) Identify the metal with the electronic configuration [Ar]4s 2 3d 8 d) Written in the same style, what is the electron configuration of arsenic? e) Written in the same style, what is the electron configuration of Mo 2+ ? f) What doubly charged anion is isoelectronic with the caesium cation? Q4: a) How many orbitals would you expect to find in the fourth electron shell (n = 4)? b) What orbital is defined by the quantum numbers n = 5 and l = 1? c) Which electrons experience a greater effective nuclear charge (Z eff ), the valence electrons of Mg 2+ or those of Al 3+ ? Explain.

Q1: Three lasers each emit monochromatic radiation of different wavelengths. Laser A’s radiation is found to have a frequency of 4.8 x Hz. Laser B emits photons of energy kJ/mol. Laser C had a wavelength specification of 2.8 microns. a) Which laser emits radiation of frequency 107 million million Hertz? b) Which laser has a photon energy of 3.18 x J? c) Which laser emits a beam of visible light? E / h = c /  = Laser C Laser A  = c/ E = h  = c /  = 3 x 10 8 / 2.8 x = 1.07 x s -1 = 6.63 x x 4.8 x = 3.18 x J = 3 x 10 8 / 4.8 x = 6.25 x m (625 nm) (visible range is nm)

d) What is the energy of each photon emitted from laser C (2 s.f.)? e) Which of the three lasers emits radiation of the shortest wavelength? f) Which of the three lasers emits infrared photons? E = hc / = 6.63 x x 3 x 10 8 / 2.8 x = 7.1 x J Laser BE mol = x 10 3 J/mol E = E mol /N A = x J/photon = hc / E = 6.63 x x 3 x 10 8 / x = 3.1 x m (or 310 nm) (compare Laser A = 625 nm and Laser C = 2800 nm) Laser C

Q2: Consider the following nuclides: mercury-196, thallium-205, osmium-189, iridium-193, lead-204 and gold-197 a) Which isotope has the lowest mass? b) Which isotope has the most neutrons per nucleus? c) Which isotope has the fewest protons per nucleus? d) Which nuclide has the highest proportion of protons per nucleus (as a percentage)? 189 Os (189 Da) ( marks) 205 Tl (124 n) 189 Os (76 p) 196 Hg (40.8%)

Q3: a) Identify the doubly-charged cation ion containing 18 electrons and a single nucleus that includes 24 neutrons b) Identify the singly charged anion ion containing 36 electrons and a single nucleus that includes 46 neutrons? c) Identify the metal with the electronic configuration [Ar]4s 2 3d 8 d) Written in the same style, what is the electron configuration of arsenic? e) Written in the same style, what is the electron configuration of Mo 2+ ? f) What doubly charged anion is isoelectronic with the caesium cation? [Kr]4d 4 ( marks) Te 2- (both are [Xe]) [Ar]4s 2 3d 10 4p 3 (the s electrons are lost first, don’t forget!) nickel (Ni) 44 Ca Br 

Q4: a) How many orbitals would you expect to find in the fourth electron shell (n = 4)? b) What orbital is defined by the quantum numbers n = 5 and l = 1? c) Which electrons experience a greater effective nuclear charge (Z eff ), the valence electrons of Mg 2+ or those of Al 3+ ? Explain. 5p orbital 16 ( = 4 2 ; one 4s, three 4p, five 4d, seven 4f) Al 3+ because the shielding is the same but the nuclear charge is greater than that of Mg 2+ (13 protons instead of 12)

Q5 : a) Calculate the photon energy and wavelength corresponding to i) the three lowest energy lines of the Balmer series and ii) the convergence limit of the Balmer series. Lowest energy line: n L = 2, n U = 3 so [(1/n L 2 ) – (1/n U 2 )] =  E = 3.03 x J  = nm second lowest energy line: n L = 2, n U = 4 so [(1/n L 2 ) – (1/n U 2 )] =  E = 4.09 x J  = nm third lowest energy line: n L = 2, n U = 5 so [(1/n L 2 ) – (1/n U 2 )] =  E = x J  = nm ii) the convergence limit: n L = 2, n U =  so [(1/n L 2 ) – (1/n U 2 )] = 0.25  E = x J  = nm

Q5 : b) By calculating the wavelength of the highest energy line of the Paschen series, determine if the two series of lines overlap Highest energy Paschen line corresponds to electron falling from infinity down to n = 3 state (i.e. convergence limit of Paschen series) n L = 3, n U =  so [(1/n L 2 ) – (1/n U 2 )] =  E = x J  = nm  no overlap between Balmer ( 821 nm) series