Chapter 6 Electronic Structure of Atoms

Chapter 6 Electronic Structure of Atoms 1. Electromagnetic radiation travels through vacuum at a speed of m/s. (a). 6.626 x 26 (b). 4186 (c). 3.00 x 8 (d). It depends on wavelength Explanation: The speed of light (electromagnetic radiation) through vacuum has a constant value of 3.00 x 8 m/s and is independent of the wavelength in vacuum. 2. Ham radio operators often broadcast on the 6-meter band. The frequency of this electromagnetic radiation is MHz. (a). 500 (b). 200 (c). 50 (d). 20 Explanation: The frequency and wavelength of electromagnetic radiation are related by the equation c = λν, where c is the speed of light (=3.00 x 8 m/s ), λ is the wavelength in m and ν is the frequency is s -1 or Hz. The frequency can be calculated by rearranging the above formula to get 8 " 1 c 3.00! m s í = = = 50 MHz ë 6 m 3. What is the frequency (s -1 ) of electromagnetic radiation that has a wavelength of 0.53 m? (a). 5.7 x 8 (b). 1.8 x -9 (c). 1.6 x 8 (d). 1.3 x -33 Explanation: The frequency and wavelength of electromagnetic radiation are related by the equation c = λν, where c is the speed of light (=3.00 x 8 m/s ), λ is the wavelength in m and ν is the frequency is s -1 or Hz. The frequency can be calculated by rearranging the above formula to get 8! 1 c 3.00 " m s 8! 1 í = = = 5.7 " s ë 0.53 m Copyright 2006 Dr. Harshavardhan D. Bapat 1

4. The energy of a photon of light is proportional to its frequency and proportional to its wavelength. (a). directly, directly (b). inversely, inversely (c). inversely, directly (d). directly, inversely Explanation: The energy is always directly proportional to the frequency while the frequency and wavelength of electromagnetic radiation are always inversely related to each other. 5. Of the following, radiation has the shortest wavelength. (a). X-ray (b). radio (c). microwave (d). ultraviolet Explanation: The wavelength and frequency are inversely related to each other, thus the radiation with the highest frequency (refer to figure 6.4 on page 219 from our text) will have the shortest wavelength. 6. What is the frequency of light (s -1 ) that has a wavelength of 1.23 x -6 cm? (a). 3.00 x 8 (b). 2.44 x 16 (c). 4. x -17 (d). 9.62 x 12 Explanation: The frequency and wavelength of electromagnetic radiation are related by the equation c = λν, where c is the speed of light (=3.00 x 8 m/s ), λ is the wavelength in m and ν is the frequency is s -1 or Hz. Since the wavelength is in cm it must be converted to meters before rearranging the above formula as follows: 1.23" -6 1m cm " = 1.23" 0 cm -8 m, í 8 c 3.00" m s = = -8 ë 1.23" m! 1 = 2.44" 16 s! 1 Copyright 2006 Dr. Harshavardhan D. Bapat 2

7. What is the frequency of light (s -1 ) that has a wavelength of 3.12 x -3 cm? (a). 3.69 (b). 2.44 x 16 (c). 9.62 x 12 (d). 4. x -17 Explanation: The frequency and wavelength of electromagnetic radiation are related by the equation c = λν, where c is the speed of light (=3.00 x 8 m/s ), λ is the wavelength in m and ν is the frequency is s -1 or Hz. Since the wavelength is in cm it must be converted to meters before rearranging the above formula as follows: 3.12 " -3 1m cm " = 3.12 " 0 cm -5 m, í 8 c 3.00" m s = = -5 ë 3.12 " m 8. The wavelength of a photon of energy 5.25 x -19 J is m. (a). 2.64 x 6 (b). 3.79 x -7 (c). 2.38 x 23 (d). 4.21 x -24! 1 = 9.62" Explanation: The wavelength and energy are related by the formula E = hc/λ, where h is Planck s constant (= 6.626 x -34 Js), c is the speed of light (3.00 x 8 m/s) and λ is the wavelength in meters. The wavelength can then be calculated by rearranging the above formula as follows:! 34 8 hc hc (6.626" Js) " (3.00" m/s)! 7 E = or ë = = = 3.79" m! 19 ë E 5.25" J 9. The energy of a photon that has a wavelength of 9.0 m is J. (a). 2.2 x -26 (b). 4.5 x 25 (c). 6.0 x -23 (d). 4.5 x -25 Explanation: The energy in Joules of any electromagnetic radiation whose wavelength is known can be calculated by the formula 12 s! 1 E = hc ë (6.626" =! 34 Js) " (3.00" 9.0 m 8 m/s) = 2.2"! 26 J When solving any of these problems pay attention to the units and ensure that they cancel. Some of these will be compound units (like m/s) that may appear problematic. Copyright 2006 Dr. Harshavardhan D. Bapat 3

. The energy of a photon that has a wavelength of 13.2 nm is J. (a). 9.55 x -25 (b). 1.51 x -17 (c). 1.99 x -25 (d). 4.42 x -23 Explanation: The wavelength is given in nm here which needs to be converted to meters and then the energy can be calculated as follows:! 34 8 1m! 8 (6.626" Js) " (3.00" m/s)! 13.2nm" = 1.32" m, E = = 1.51" 9! 8 nm 1.32" m 11. What is the frequency (s -1 ) of a photon of energy 4.38 x -18 J? (a). 438 (b). 1.45 x -16 (c). 6.61 x 15 (d). 2.30 x 7 Explanation: The frequency ν of this photon can be calculated by rearranging the equation E = h ν where E is the energy, h = Planck s constant and ν = frequency in s -1.! 18 E 4.38" J 15! 1 E = hí or í = = = 6.61" s! 34 h 6.626" Js 12. What is the wavelength (angstroms) of a photon of energy of 4.38 x -18 J? (a). 438 (b). 4.54x -8 (c). 454 (d). 1.45 x 8 Explanation: The wavelength calculated here will be in meters which can be easily converted to angstroms as follows: " 34 8 hc hc (6.626! Js)! (3.00! m/s) E = or ë = = = 4.538! " 18 ë E 4.38! J " 8 1angstrom 4.538! m! = 453.8 = 454 angstrom " 8 m " 8 m 17 J Copyright 2006 Dr. Harshavardhan D. Bapat 4

13. A mole of red photons of wavelength 755 nm has kj of energy. (a). 2.63 x -19 (b). 6.022 x 23 (c). 6.05 x -3 (d). 158.6 Explanation: The energy of red photons of a wavelength of 755 nm can be easily calculated using the formula E = hc/λ. This will be the energy of only ONE photon and will have to be multiplied by Avogadro s number to convert it to the energy of one mole of red photons. The wavelength needs to be converted to meters before performing this calculation: " 34 (6.626! Js)! (3.00! E = 1m 755.0 nm! 9 nm 158550.08 J! 1kJ 00 J = 158.6 kj 8 m/s) = 2.632! " 19 J photon! 6.022! 23 photons mole = 14. Of the following, radiation has the longest wavelength and radiation has the greatest energy. Gamma, ultraviolet or visible (a). ultraviolet, gamma (b). visible, ultraviolet (c). gamma, gamma (d). visible, gamma Explanation: From figure 6.4 in our text you can see that the visible radiation will have the longest wavelength while the gamma will have the highest frequency. Since energy is directly proportional to frequency, gamma will be the most energetic. 15. What color of visible light has the longest wavelength? (a). blue (b). violet (c). red (d). yellow 16. What color of visible light has the highest energy? (a). violet (b). yellow (c). red (d). green Copyright 2006 Dr. Harshavardhan D. Bapat 5

Explanation: From figure 6.4 in our text you can see that violet colored light would have the highest frequency in the visible region. Since energy is directly proportional to the frequency, violet light will have the highest energy. 17. An electron is a Bohr hydrogen atom has energy of -1.362 x -19 J. The value of n for this electron is. (a). 1 (b). 2 (c). 3 (d). 4 Explanation: The energy of an electron in a particular energy state in the hydrogen atom can be calculated by using the formula E = (-2.18 x -18 J)/n 2, where n is the principal quantum number for the energy state. The value of n can be found by rearranging the above formula as follows: " " 2.18! " 1.362! 18 n = " 19 J = 4 J 18. A spectrum containing only specific wavelengths is called a spectrum. (a). continuous (b). line (c). visible (d). Bohr 19. The n = 2 to n = 6 transition in the Bohr hydrogen atom corresponds to the of a photon with a wavelength of nm. (a). emission, 411 (b). absorption, 411 (c). absorption, 657 (d). emission, 389 Explanation: There are 2 parts to this question. Since the electron is moving from a smaller value of n (n i ) to a larger value of n (n f ), it must be absorbing energy. The wavelength responsible for this transition can be calculated by using the formula: 1 ' 1 1 $ 7 ( 1' 1 1 $ 7 = R H 1.097 m 1.097 m ë % ( 2 2 n n " =! % ( 2 2 " =! & i f # & 2 6 # 1 7 ( 1 ( 1 1 = 1.097! m ( 0.222) = 2435340m and ë = ë 2435340 m = 411nm ( 1 ( 1 ' 1 % ( & 4 1 36 $ " # = 4.11! ( 7 m Copyright 2006 Dr. Harshavardhan D. Bapat 6

20. The n =5 to n = 3 transition in the Bohr hydrogen atom corresponds to the of a photon with a wavelength of nm. (a). absorption, 657 (b). absorption, 1280 (c). emission, 657 (d). emission, 1282 Explanation: There are 2 parts to this question. Since the electron is moving from a larger value of n (n i ) to a smaller value of n (n f ), it must be emitting energy. The wavelength responsible for this emission can be calculated by using the formula: 1 ë ( 1 & 2 ' ni % # = 1.097 " $ 1 1 7! 1( 1 1 % 7! = R m 1.097 m 1 H! 2 &! = " 2 2 # ë nf ' 5 3 $ (! 0.0711) =! 780088.88 m,ignore the negative sign here and then ë = 1282 nm 21. The debroglie wavelength of a particle is given by. (a). h + mv (b). hmv (c). h/mv (d). mv/c 22. What is the de Broglie wavelength (m) of a 2.0 kg object moving at a speed of 50 m/s? (a). 6.6 x -36 (b). 1.5 x 35 (c). 5.3 x -33 (d). 2.6 x -35 Explanation: The de Broglie wavelength of an object can be calculated by remembering that 1 J = 1 kg m 2 s -2, which changes h = 6.626 x -34 kg m 2 s -1. Now using the formula that relates the mass and velocity of an object to its wavelength, the wavelength can be calculated as follows:! 34 2! 1 h 6.626" kg # m s! 36 ë = = = 6.626 " m mv 2.0 kg " 50 m/s 23. At what speed (m/s) must a 3.0 mg object be moving in order to have a de Broglie wavelength of 5.4 x -29 m? (a). 1.6 x -28 (b). 3.9 x -4 (c). 2.0 x 12 (d). 4.1 Copyright 2006 Dr. Harshavardhan D. Bapat 7

Explanation: The speed of this object can be calculated by remembering that 1 J = 1 kg m 2 s -2, which changes h = 6.626 x -34 kg m 2 s -1. Another change that must be made in this question is converting the mass from mg to kg. 1kg " 6 3.0 mg! = 3! kg 6 1! mg " 34 2 6.626! kg m s v = " 6 " (3! kg)! (5.4! " 1 29 = 4.1 m/s m/s) 24. The de Broglie wavelength of an electron is 8.7 x -11 m. The mass of an electron is 9.1 x -31 kg. The velocity of this electron is m/s. (a). 8.4 x 3 (b). 1.2 x -7 (c). 6.9 x -5 (d). 8.4 x 6 Explanation: The speed of this electron can be calculated by remembering that 1 J = 1 kg m 2 s -2, which changes h = 6.626 x -34 kg m 2 s -1. " 34 2 " 1 6.626! kgm s 6 v = = 8.4! m/s " 31 " 11 (9.1! kg)! (8.7! m) 25. The quantum number defines the shape of an orbital. (a). spin (b). magnetic (c). principal (d). azimuthal 26. There are orbitals in the third shell (a). 25 (b). 4 (c). 9 (d). 16 Explanation: The number of orbitals in a shell is easily calculated by the formula # of orbitals = n 2 where n = principal quantum number, which is 3 in this case. 27. The azimuthal quantum number is 3 in orbitals. (a). s (b). p (c). d (d). f Copyright 2006 Dr. Harshavardhan D. Bapat 8

28. The n = 1 shell contains p orbitals. All the other shells contain p orbitals. (a). 3, 6 (b). 0, 3 (c). 6, 2 (d). 3, 3 Explanation: If n = 1, then the only possible value of l is 0 which means that n = 1 can contain only s orbitals. When n > 1, the value of l = 1 is possible making the existence of 3 p orbitals possible. 29. The principal quantum number of the first d subshell is. (a). 1 (b). 2 (c). 3 (d). 4 Explanation: The value of l for a d subshell is 2 which means that the first shell in which one would come across this value will be n = 3. 30. The total number of orbitals in a shell is given by. (a). L 2 (b). n 2 (c). 2n (d). 2n + l 31. Each p-subshell can accommodate a maximum of electrons. (a). 6 (b). 2 (c). (d). 3 Explanation: There are 3 different p orbitals: p x, p y and p z. Each of these can contain 2 electrons leading to the maximum number of electrons as 6. 32. How many quantum numbers are necessary to designate a particular electron in an atom? (a). 3 (b). 4 (c). 2 (d). 1 Copyright 2006 Dr. Harshavardhan D. Bapat 9

33. The 3p subshell in the ground state of atomic xenon contains electrons. (a). 2 (b). 6 (c). 36 (d). Explanation: Since Xe is a noble gas, its subshells will be completely filled regardless of their principal quantum number. Thus the 3p subshell will contain 6 electrons. 34. [Ar]4s 2 3d 4p 3 is the electron configuration of a(n) atom. (a). As (b). V (c). P (d). Sb Explanation: The easiest way to answer this question is to count the total number of electrons and find which element that number corresponds to. The total number of electrons is = 18 (for the Ar) + 2 + + 3 = 33 which corresponds to As. 35. There are unpaired electrons in a ground state phosphorus atom. (a). 0 (b). 1 (c). 2 (d). 3 Explanation: You will need the electronic configuration of the phosphorus atom to answer this question which is 1s 2 2s 2 2p 6 3s 2 3p 3. According to Hund s rule the 3 electrons in the 3p subshell will have the same spin and be unpaired. 36. In a ground-state manganese atom, the subshell is partially filled. (a). 3s (b). 4s (c). 4p (d). 3d Explanation: The electronic configuration of manganese will help you answer this question which is [Ar]4s 2 3d 5, this is an example of an anomalous electron configuration where the 3d subshell is half-filled. The 3d subshell can accommodate a total of electrons but here has only 5. Copyright 2006 Dr. Harshavardhan D. Bapat

37. The principal quantum number for the outermost electrons in a Br atom in the ground state is. (a). 2 (b). 3 (c). 4 (d). 5 Explanation: The electronic configuration of bromine is [Ar]3d 4s 2 4p 5 shows that the outermost electrons are in the s and p orbitals in the 4 th energy level making the principal quantum number = 4. 38. The azimuthal quantum number for the outermost electrons in a nitrogen atom in the ground state is. (a). 0 (b). 1 (c). 2 (d). 3 Explanation: The electronic configuration of nitrogen is [He]2s 2 2p 3, showing that the outermost electrons are in the 2p orbital. The azimuthal quantum number (l) for any p orbital is 1. 39. All of the have a valence shell electron configuration ns 1. (a). noble gases (b). halogens (c). chalcogens (d). alkali metals 40. The elements in the period of the periodic table have a core-electron configuration that is the same as the electron configuration of neon. (a). sixth (b). second (c). third (d). fifth Explanation: The core electrons are electrons that are not in the valence shells. The electron configuration of elements in the 3 rd period can be written with neon s configuration as their core electrons. Copyright 2006 Dr. Harshavardhan D. Bapat 11

41. Elements in group have a np 6 electron configuration in the outer shell. (a). 4A (b). 6A (c). 7A (d). 8A 42. Which group in the periodic table contains elements with the valence electron configuration of ns 2 np 1? (a). 1A (b). 2A (c). 3A (d). 4A Explanation: The electron configurations of elements in group 3A are written with the valence electron configuration of ns 2 np 1. 43. Which one of the following is correct? (a). ν + λ = c (b). ν/λ = c (c). λ = cν (d). νλ = c 44. In the Bohr model of the atom,. (a). electrons travel in circular paths called orbitals (b). electrons can have any energy (c). electron energies are quantized (d). electron paths are controlled by probability 45. According to the Heisenberg Uncertainty Principle, it is impossible to know precisely both the position and the of an electron. (a). mass (b). color (c). momentum (d). speed Copyright 2006 Dr. Harshavardhan D. Bapat 12

46. All of the orbitals in a given electron shell have the same value of the quantum number. (a). principal (b). azimuthal (c). magnetic (d). spin 47. Which one of the following is not a valid value for the magnetic quantum number of an electron in a 5d subshell? (a). 2 (b). 3 (c). 0 (d). 1 Explanation: For an electron in the 5d subshell the value of l = 2 and the magnetic quantum number m l can have values from l, 0, +l, meaning m l could not have a value = 3. 48. Which of the subshells below do not exist due to the constraints upon the azimuthal quantum number? (a). 2s (b). 2d (c). 2p (d). none of the above Explanation: The values of the azimuthal quantum number l are decided by the values of the principal quantum number n. The values of l will only be from 0 n 1. Thus for n = 2, only the values of 0 and 1 will be possible for l, which means that only the 2s and 2p orbitals will be possible. 49. An electron cannot have the quantum numbers n =, l =, m l =. (a). 2, 0, 0 (b). 2, 1, -1 (c). 3, 1, -1 (d). 1, 1, 1 Explanation: The values of 1, 1, 1 would be impossible since if n = 1, the only value of l would be = 0. Copyright 2006 Dr. Harshavardhan D. Bapat 13

50. An electron cannot have the quantum numbers n =, 1 =, m l =. (a). 6, 1, 0 (b). 3, 2, 3 (c). 3, 2, -2 (d). 1, 0, 0 Explanation: The values of 3, 2, 3 will be impossible since n and m l cannot have the same values. 51. Which quantum number determines the energy of an electron in a hydrogen atom? (a). n (b). n and l (c). m l (d). l 52. In a p x orbital, the subscript x denotes the of the electron. (a). energy (b). spin of the electrons (c). axis along which the orbital is aligned (d). size of the orbital Explanation: There are in all 3 degenrate p orbitals possible that are typically distinguished by the axis along which they are oriented. This is achieved by including the letter denoting the axis as a subscript. 53. Which electron configuration represents a violation of the Pauli exclusion principle? (a). 1s 2s 2p (b). 1s 2s 2p (c). 1s 2s 2p (d). 1s 2s 2p Explanation: According to the Pauli exclusion principle no two electrons in an atom cannot have the same 4 quantum numbers. The2 electrons in the 2s orbital have the same value for their m s which is not allowed. Copyright 2006 Dr. Harshavardhan D. Bapat 14

54. Which electron configuration represents a violation of the Pauli exclusion principle? (a). 1s 2s 2p (b). 1s 2s 2p (c). 1s 2s 2p (d). 1s 2s 2p Explanation: According to the Pauli exclusion principle no two electrons in an atom cannot have the same 4 quantum numbers. The2 electrons in the 2p orbital have the same value for their m s which is not allowed. 55. Which of the following is a valid set of four quantum numbers? (n, l, m l, m s ) (a). 2, 0, 0, + ½ (b). 2, 2, 1, - ½ (c). 1, 0, 1, + ½ (d). 2, 1, +2, + ½ Explanation: Here is why only option (a) is the correct answer: In option (b), n =l which is not allowed, in (c) m l 1 since l = 0 and in (d) m l > l which are all not allowed. 56. Which of the following is not a valid set of four quantum numbers? (n, l, m l, m s ) (a). 2, 0, 0, + ½ (b). 2, 1, 0, - ½ (c). 1, 1, 0, + ½ (d). 1, 0, 0, + ½ Explanation: Since n can never be equal to l, option c is the only set that is not valid. 57. Which one of the following configurations depicts an excited oxygen atom? (a). 1s 2 2s 2 2p 2 (b). 1s 2 2s 2 2p 2 3s 2 (c). 1s 2 2s 2 2p 1 (d). 1s 2 2s 2 2p 4 Copyright 2006 Dr. Harshavardhan D. Bapat 15

Explanation: The ground state electron configuration of the oxygen atom is shown in option (d), while option (b) shows that two electrons from the 2p orbital have been promoted to the 3s orbital. 58. Which one of the following configurations depicts an excited carbon atom? (a). 1s 2 2s 2 2p 1 3s 1 (b). 1s 2 2s 2 2p 3 (c). 1s 2 2s 2 2p 2 (d). 1s 2 2s 2 3s 1 Explanation: The ground state electron configuration is shown in (c), while option (a) shows that one of the electrons has been excited from the 2p to the 3s orbital. Copyright 2006 Dr. Harshavardhan D. Bapat 16