SUGGESTED ANSWER FOR SK015 TUTORIAL 2.0 2023/2024 2.1: Bohr’s Atomic Model 1. a) Based on Bohr’s atomic model of the hydrogen atom, i) explain the existence of energy levels in an atom. Bohr’s postulate imply that the electron moves in circular permitted orbit that has a specific amount of energy (quantized energy). The nearest orbit to the nucleus is the ground state which has the smallest value of energy and as the orbit getting farther from the nucleus, the energy of the orbit getting higherl. This explain the existence of energy levels in an atom. ii) state the difference between ground state and excited state. Ground state: : The lowest energy state of an atom, which corresponds to the most stable energy state and the nearest to the nucleus of the atom. Excited state: : Level that is higher in energy than the ground state. The higher the excited state, the farther the electron from the nucleus. b) Distinguish a continuous spectrum and a line spectrum Line spectrum Continuous spectrum A spectrum consists of discontinuous and discrete lines with specific wavelength. A spectrum consists of radiation distributed over all wavelengths without any blank space/area. Form definite lines and each line si separated by a blank area. No definite lines and no blank areas formed. Example : emission spectrum of the atoms of an element. Example : rainbow, light from glowing objects 2. a) Describe the formation of line spectrum of hydrogen atom. When energy is supplied to hydrogen atom, the electron absorbed the energy and is excited from a lower to a higher-energy level . This electron is unstable, thus it will fall to lower energy level. When the electron fall to a lower energy level, photon energy is emitted in the form of light with specific wavelength and frequency, thus detected as a line spectrum in the line spectrum.. b) Compare and contrast the Lyman and the Balmer series. Lyman series Balmer series Formed when electrons transit from higher energy levels to n=1 and emit photons with specific wavelengths and frequencies that fall in the ultraviolet region. Formed when electrons transit from higher energy level to n=2 and emit photons with specific wavelengths and frequencies that fall in the visible region.
3. The emission spectrum of hydrogen atom in the visible region is shown below: λ (nm) line spectrum W X Y Z a) State the transition of electrons that produce line W and Y respectively. Line W ,from n = 3 to n = 2 ; Line Y, from n=5 to n = 2 b) Which of the above lines has the lowest frequency? Line W c) Calculate the wavelength that corresponds to line X. 1 = ( 1 1 2 − 1 2 2 ) , 1 < 2 = 1.097 × 107−1 ( 1 2 2 − 1 5 2 ) λ = 434 nm 4. One of the lines in the Balmer has a frequency of 4.57 x 1014 Hz a) Calculate its wavelength. ѵ = 4.57 x 1014 s -1 = 3.00 x 108 ms-1 λ = 6.56 x 10-7 m c) Calculate the energy emitted. ∆E = hυ = (6.63 x 10-34 Js-1 ) x 4.57 x 1014 s -1 = 3.03 x 10-19 J c) Determine the transition of electron involved. ∆ = ( 1 2 − 1 2) 3.03 x 10-19 J = 2.18 × 10−18 ( 1 2 − 1 2 2 ) ni = 3
5. Calculate the wavelength (in nm) and frequency of the: a) third line in the Paschen series. Transition from ni=6 to nf=3 Energy of photon emitted : ∆ = ( 1 2 − 1 2) = 2.18 × 10−18 ( 1 6 2 − 1 3 2 ) = -1.82 x 10-19 J Wavelength ∆E = hc λ 1.82 x 10-19 J = (6.63 x 10-34 Js-1 )(3.00 x 108 ms-1 ) λ λ = 1092.9 nm Frequency ∆E = hυ 1.82 x 10-19 J = (6.63 x 10-34 Js-1 ) υ υ = 2.75 x 1014 s -1 b) second line in the Brackett series. Transition from ni=6 to nf=4 Energy of photon emitted : ∆ = ( 1 2 − 1 2) = 2.18 × 10−18 ( 1 6 2 − 1 4 2 ) = -7.57 x 10-20 J Wavelength ∆E = hc λ 7.57 x 10-20 J = (6.63 x 10-34 Js-1 )(3.00 x 108 ms-1 ) λ λ = 2627.5 nm Frequency ∆E = hυ 7.57 x 10-20 J = (6.63 x 10-34 Js-1 ) υ υ = 1.14 x 1014 s -1 6. Calculate the ionization energy of hydrogen atom from Lyman series in kJ mol−1 .
Ionization energy ni = 1, nf = ∞ ∆ = ( 1 2 − 1 2) = 2.18 × 10−18 ( 1 1 2 − 1 ∞2 ) = . × − 1 electron ≡ 2.18 × 10−18 1 mol electron ≡ (6.02 x 1023) (2.18 × 10−18) = 1312 kJ 7. If an emission spectrum of hydrogen atom emit photons with the average kinetic energy of 10370 J mol−1 isin the infrared region, calculate the frequency of IR wave formed. 6.023 x 1023 electron - 10370 J 1 electron - 1 electron x 10370 J 6.023 x 1023 electron = 2.276 x 10-24 J ∆E = hυ 2.276 x 10-24 J = (6.63 x 10-34 Js-1 ) υ υ = 3.43 x 109 s -1 8. Two important concepts that related to the behavior of electrons in an atom are the Heisenberg uncertainty principle and the wave-particle duality of matter. (a) State the Heisenberg uncertainty principle as it related to the determining the position and momentum of an object. For any moving mass, Heisenberg uncertainty principle states that it is impossible to know at the same time, both momentum p (defined as mass times velocity) and position of a particle with certain. Since electron is a moving mass, so it is impossible to know both momentum and position of an electron with certain as Bohr’s claims in his postulate. (b) What characteristic of the Bohr theory of the atom is considered unsatisfactory as a result of the Heisenberg uncertainty principle? Bohr’s postulate states that electrons are circling in fixed orbits of a nucleus of an atom. This shows that Bohr was certain about the position and momentum of the electrons in their orbits around the nucleus of an atom. On the other hand, Heisenberg uncertainty principle is uncertain of the position and momentum of a moving object, i.e the moving electron.
(c) Explain why the uncertainty principle or the wave nature of particles is not significant when describing the behavior of macroscopic objects, but it is very significant when describing the behavior of electrons. For macroscopic objects, the uncertainty principle is not significant since it is possible and easy to determine the position and momentum of the objects. However, due to electrons as the microscopic objects and the light mass of electrons, make the uncertainty principle become significant. ATAU 8. (a) Heisenberg Uncertainty Principle: • It is impossible to determine (or measure) both the position and the momentum if any particle (or object or body) simultaneously. • The more exactly the position of a particle is known, the less exactly the momentum or velocity of the particle can be known. • ∆x ∆p ≥ h 4π where h = Plank’s constant, x = uncertainty in position, p = uncertainty in momentum. (b) Bohr postulated that the electron in an H atom travels about the nucleus in a circular orbit and has a fixed angular momentum. With a fixed radius of orbit and a fixed momentum (or energy), ∆x ∆p < h 4π , hence the Heisenberg principle is violated. (c) The wavelength of a particle is given by the De Broglie relation λ = h mv . For masses of macroscopic objects, h m is so small for any v that is too small to be detectable. For an electron, m is so small that h mv yields a detectable .
2.2: Quantum Mechanical Model & 2.3: Electronic Configuration 1. a) Define orbit and orbital. Orbit: The pathway where the electron moves around the nucleus. Orbital: An orbital is a three-dimensional region in space around the nucleus whereby the probability to find an electron is highest. b) What do these quantum numbers represent? i. n - principal quantum number that describe energy level and size of the orbitasl where n = 1, 2, 3..... ii. l - angular momentum quantum number describe the shape of the orbitasl where l = 0, 1, 2.....(n – 1) iii. m -- magnetic quantum number describe spatial orientation the shape of the orbitasl where m = ( - l..... 0......+l ) iv. s - electron spin quantum number describe the spin of electrons as clockwise or anti- clockwise where s= -½ or s= +½ c) Give one set of possible quantum numbers for an electron that occupy the 3p, 4d and 5s orbitals respectively. 3p : n=3, l=1, m=0, s=+1/2 or any appropriate set 4d : n=4, l=2, m=0, s=+1/2 or any appropriate set 5s : n=5, l=0, 1, m=0, s=+1/2 or any appropriate set 2. For the following quantum numbers of n and l below, n l name of subshell number of orbitals maximun no of electron I 3 1 3p 3 6 II 4 2 4d 5 10 III 5 0 5s 1 2 a) name the subshell and give the number of orbitals involved for each subshell. b) determine the maximum number of electrons that occupy each subshell.
3. Explain the following subshell as allowed or not allowed. a) 6s b) 4p c) 2d d) 3f a) 6s ( allowed – value of l= 0 which is smaller than n ) b) 4p ( allowed – value of l= 1 which is smaller than n) c) 2d (not allowed – value of l= 2 which is bigger than n) d) 3f (not allowed – value of l= 3 which is identical as n) 4. Draw the shape of the following orbitals: 1s, 2s, 3py, 3dxz, 3 2 2 x −y d and 3 2 z d 1s x y z 2s x y z 3py x y z 3dxz x y z 3dx 2 -y 2 x y z 3dz 2 x y z 5. a) State Aufbau principle. Electrons must occupy available orbitals of lower energy first before they filled orbitals of higher energy. b) Arrange the following orbitals in the order of increasing energy. 4dxy, 3dxy, 3dyz, 4pz, 3pz, 3py, 2py, 3s, 2s, 1s, 4s 1s < 2s < 2py < 3s < 3pz, 3py < 4s < 3dxy, 3dyz < 4pz < 4dxy 6. Atom X has 15 electrons. a) State Hund’s rule and Pauli Exclusion Principle
Hund’s rule: When electrons are filled into the orbital the orbital of equivalent energy (degenerate orbitals), each orbital is filled singly with electron of the same spin before it is paired. Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers. b) Draw the orbital diagram for X. 15 X : 1s2 2s2 2p6 3s2 3p3 1s 2s 2p 3s 3p c) Give a set of quantum numbers for the highest energy electrons. Highest energy electron are electrons that occupy the 3p orbitals n=3, l=1, m=-1, s=+1/2 or n=3, l=1, m=-1, s=-1/2 or n=3, l=1, m=0, s=+1/2 or n=3, l=1, m=0, s=-1/2 or n=3, l=1, m=+1, s=+1/2 or n=3, l=1, m=+1, s=-1/2 Any 3 sets. 7. Give the electronic configuration of the following species as orbital diagram and spdf notation. a) O2− O2− : 1s22s22p6 1s 2s 2p b) Al3+ Al3+: 1s22s22p6 1s 2s 2p c) Ni2+ Ni2+: 1s22s22p63s23p63d8 1s 2s 2p 3s 3p 3d 8. a) Write the electronic configuration of iron(II) and iron(III) ions. Fe2+: 1s2 2s2 2p6 3s2 3p6 3d6 Fe3+: 1s2 2s2 2p6 3s2 3p6 3d5
b) Which of these two species is more stable? Explain. Fe3+ is more stable than Fe2+ because of its half-filled 3d orbital is more stable than the partially-filled of 3d orbital in Fe2+ . c) Write the sets of quantum numbers for the electrons in the outermost orbital of iron(III) ions. n=3, l=2, m=+2, s=+1/2 n=3, l=2, m=-1 s=+1/2 n=3, l=2, m=+1, s=+1/2 n=3, l=2, m=-1, s=+1/2 n=3, l=2, m=0, s=+1/2