Chem110MT1SolutionsFall2010

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Unformatted text preview: ©Prep101 Chem110 MT1 Instructor Chapter 8 Questions: Q 1.1 Consider the following approximate visible light spectrum: 7.0 x 10-5 Infrared 6.0 x 10-5 Red Orange 5.0 x 10-5 Yellow Green 4.0 x 10-5 Blue Violet cm Ultraviolet Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of 3.90 x 10-19 J, what colour of visible light is emitted? Solution: E= λ= hc λ → (6.626 x 10 λ= hc E )( Js 2.998 x 10 8 ms −1 3.90 x 10 -19 J -34 λ = 5.09 x 10 -5 cm → ) → λ = 5.09 x 10 -7 m λ = 509 nm Therefore, GREEN light will be emitted Solutions will be posted at www.prep101.com/solutions 15 ©Prep101 Q 1.2 Chem110 MT1 Instructor Ozone formation in the stratosphere is initiated by the following reaction: O2 + hν → 2O (1) Which is followed by the reaction: O + O2 → O3 (2) The energy needed to dissociate an oxygen molecule into oxygen atoms is 499 kJ/mol. What minimum wavelength (in nm) of light will initiate the reaction (1) ? Solution: We know the energy (499 kJ/mol). By using E = hν, we can solve for the frequency (ν). The frequency is related to wavelength (λ) by the equation ν = λ/c ( ) E = 499 kJ mol = ( ) = 8.286 × 10 499 × 10 3 J mol 6.022 × 10 23 ( mol −1 ) −19 (J ) () ( ) ( ) m c 2.998 × 10 ( s) c m = 2.397 × 10 ( )= 239.7 (nm) ν= ⇒λ= = ν 1.251 × 10 s λ 8.203 × 10 J E = 1.251 × 10 15 s−1 or Hz = h 6.626 × 10−34 J ⋅ s −19 E = hν ⇒ ν = 8 15 ( ) −7 −1 Solutions will be posted at www.prep101.com/solutions 16 ©Prep101 Chem110 MT1 Instructor Q 1.3 Calculate the energy in J of a photon of red light that has a frequency of 4.35 x 1014 s-1 Solution: E = hν Q 1.4 How many photons of frequency 1.50 x 1014 s-1 are needed to give 30.1 J of energy? Solution: First calculate the energy of a photon: E = hν → → E = (6.626 x 10-34 Js)(4.35 x 1014 s-1) = 2.88 x 10-19 J E = 6.626 x 10-34 Js (1.50 x 1014 s-1) = 9.94 x 10-20 J Now, let “a” be the number of photons (factor) to contribute 30.1 J since we know that 1 photon emits 9.94 x 1020 J of energy E(a) = 30.1 J → 9.94 x 10-20 J (a) = 30.1 J a = 3.03 x 1020 photons Solutions will be posted at www.prep101.com/solutions 17 ©Prep101 Chem110 MT1 Instructor Q 1.5 Calculate the wavelength in meters of microwave radiation that corresponds to an energy of 2.58 J·mol-1 photons. Solution: Divide by Avagadro’s number to give the energy per photon 2.58 J ⋅ mol-1 = 4.28 4 x10 −24 J 6.022 x 10 23 mol−1 Wavelength can now be determined by dividing the product of the speed of light and Plank’s constant by the energy per photon. E= hc λ → λ = 0.0464 m λ= hc E → λ= (6.626x10 -34 Js)(2.998 x10 8 ms -1 ) 4.28x10 -24 J or alternatively 4.64 x 10-2 m Q 1.6 The energy from the right type of radiation can be used to make the bonds of a molecule vibrate faster. Light of wavelength 3.34 μm can excite bonds in hydrogen chloride, HCl. What is the energy of a photon of this wavelength? Solution: First convert the wavelength from μm → m 3.34 μm x (1 m/1E6 μm) = 3.34 x 10-6 m Now energy can be calculated by the standard equation: E = hc λ E = (6.626x10-34 Js)(2.998x108 ms-1)/(3.34x10-6 m) = 5.95 x 10-20 J Solutions will be posted at www.prep101.com/solutions 18 ©Prep101 Chem110 MT1 Instructor Q 1.7 Define and show an example to compare wavelength and amplitude Solution: Amplitude is the maximum height of the wave above the centre line or the maximum depth below. Wavelength, λ – is the distance of one cycle of the wave. It is the distance between the tops of two successive crests (or the bottoms of two troughs). http://www.users.mis.net/~pthrush/lighting/wave.jpg Solutions will be posted at www.prep101.com/solutions 19 ©Prep101 Q 1.8 Chem110 MT1 Instructor Which of the waves below has the higher frequency and which has the higher energy? A) B) A, B C) B, A D) Solution: A, A B, B Correct answer: D B has a higher frequency because the distance between the peaks of the waves are smaller than A. B also has a higher energy because Energy is inversely proportional to wavelength. Since A has a longer wavelength, it will be of lower energy. Q 1.9 What is the wavelength of electrons moving at 0.5% of the speed of light? A) B) C) D) Solution: 0.48 nm 0.96 nm 4.8 nm 96 nm Correct answer: A Speed of light = 3.0 x 108 m/sec. Speed of electrons = (3.0 x 108 m/sec)0.005 = 1.5 x 106 m/sec λ = h/mv = (6.626 x 10-34Js)/[( 1.5 x 106 m/sec)(9.11 x 10-31kg) = 4.8 x 10-10m = 0.48 nm Solutions will be posted at www.prep101.com/solutions 20 ©Prep101 Q 1.10 Chem110 MT1 Instructor What is the wavelength of light emitted when an electron in a hydrogen atom undergoes a transition from the n = 5 to n =2 level? Solution: ∆E = Einitial − Efinal 1 1 ∆E = R H − n2 n2 i f RH = 2.179×10−18 J. ** your prof indicates that hcRH = 2.17 E-18 J 1 1 = 4.58 × 10-19 J ∆E = (2.179×10−18 J) − (2) 2 (5) 2 E= λ= hc λ , therefore hc (6.626 × 10 −34 Js)(3.0 × 108 m / s ) = 4.34 E-7 m = E 4.58 × 10 −19 J 4.34 E-7 m (1 E9 nm / 1 m) = 434 nm Q 1.11 One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. What is the energy (in kJ) of photons of this wavelength, expressed on a molar basis? Solution: 325 nm = 325 E-9 m Ephoton = hc/λ = (6.626 E-34 Js)(2.998 E8 m)/(325 E-9) = 6.11 E-19 J/photon 6.11 E-19 J/photon (6.022 E23 photon/mol) = 3.68 E5 J/mol = 368 kJ/mol Solutions will be posted at www.prep101.com/solutions 21 ©Prep101 Chem110 MT1 Instructor Q 1.12 400 kJ of energy is required to break a mole of N-H bonds. What is the wavelength of the radiation needed to break one N-H bond? Is this radiation in the visible region of the electromagnetic spectrum? Solution: E = 400 E3 J/mol (1 mol / 6.022 E23 bonds) = 6.64 E-19 J/bond λ = hc/E → (6.626 E-34 Js)(2.998 E8 m/s) / (6.63 E-19 J/bond) → 3.00 E-7 m 3.0 E-7 m (109 nm / 1 m) = 300.0 nm The visible region is considered in between 400-800 nm. Therefore, 300 nm is within the ultraviolet (UV) region. Q 1.13 Determine the wavelength of the line in Balmer series of hydrogen corresponding to the transition from n = 4 to n = 2. Solution: First calculate the energy ∆E from the transition from nf = 5 and ni = 2. 1 1 ∆E = 2.179 ×10 −18 J 2 − 2 → n i nf 1 1 ∆E = 2.179 ×10 −18 J 2 − 2 4 2 ∆E = -4.086 E-19 J We can now calculate the frequency by the equation, E = hν ν = E/h = 4.086 E-19 J / (6.626 E-34 Js) = 6.166 E14 s-1 Now calculate the wavelength from the frequency, λ = c/ν λ = 2.998 E8 ms-1 / (6.166 E14 s-1) = 4.86 E-7 m Solutions will be posted at www.prep101.com/solutions 22 ©Prep101 Q 1.14 Chem110 MT1 Instructor Calculate the wavelength in nanometers when an electron in a hydrogen atom returns from n = 5 to n = 2. Solution: 1 1 ∆E = 2.179 ×10 −18 J 2 − 2 5 2 → ∆E = -4.576 E-20 J We can now calculate the frequency by the equation, E = hν ν = 4.576 E-20 J / (6.626 E-34 Js) = 6.906 E14 s-1 Now calculate the wavelength from the frequency, λ = c/ν λ = 2.998 E8 ms-1 / (6.906 E14 s-1) = 4.34 E-7 m 4.23 E-7 m → 434 nm Solutions will be posted at www.prep101.com/solutions 23 ©Prep101 Q 1.15 Chem110 MT1 Instructor The Brackett series describe emission lines of the hydrogen atom where nf = 4. Calculate the energy, frequency and wavelength of the associated radiation when the excited state is n = 7 and the Brackett series of emission lines are observed. Describe what type of radiation is emitted according to the table below: Frequency 106 1010 1014 1016 1018 Solution: Radiation Radio Microwaves Infrared Ultraviolet X-rays ni = 7, nf = 4 (Brackett series) 1 1 ∆E = 2.179 × 10 −18 J 2 − 2 7 4 → ∆E = -9.176 E-20 J We can now calculate the frequency by the equation, E = hν ν = 9.176 E-20 J / (6.626 E-34 Js) = 1.3848 E14 s-1 Now calculate the wavelength from the frequency, λ = c/ν λ = 2.998 E8 ms-1 / (1.3848 E14 s-1) = 2.16 E-6 m Radiation is emitted in the infrared region (because the frequency is 1014) Solutions will be posted at www.prep101.com/solutions 24 ©Prep101 Q 1.16 Chem110 MT1 Instructor In the Paschen series of the hydrogen spectrum, there is a line with a wavelength of 1094.1 nm from the emission of an electron. What are the transitions involved in this series? Lyman series Balmer series Paschen series Brackett series Pfund series A) B) C) D) E) Answer n=1 n=2 n=3 n=4 n=5 ni = 3, nf = 2 ni = 2, nf = 3 ni = 6, nf = 3 ni = 4, nf = 3 ni = 4, nf = 2 Correct Answer: C Paschen refers to nf = 3. We are also told the wavelength is from the emission of an electron so ni > 3 (for this question). Eliminate answers A, B, E 1094.1 nm 1.094 E-6 m We can now calculate the frequency by the from the wavelength, λ = c/ν ν = (2.998 E8 m·s-1) / (1.094 E-6 m) = 2.74 E14 s-1 Energy can be calculated from the frequency, ∆E = hν, ∆E = 6.626E-34 Js (2.74 E14 s-1) = -1.8158 E-19 J ** Energy is negative because energy was emitted 1 1 − 1.82 ×10 −19 J = 2.179 ×10 −18 J 2 − 2 x 3 → − 0.0833 + 1 1 = 9 x2 → → 0.02777 = 1 1 − 0.0833 = 2 − 2 x 3 1 → x2 x2 = 36 x = ni = 6 Solutions will be posted at www.prep101.com/solutions 25 ©Prep101 Q 1.17 Chem110 MT1 Instructor Which of the following statements about red light and blue light is false? COLOUR Blue Green Orange Red WAVELENGTH 450–475 nm 495–570 nm 590–620 nm 620–750 nm A) Red light and blue light have the same speed in a vacuum B) Red light is refracted less by a glass prism than blue light C) Red light has a longer wavelength than blue light D) Blue light has a smaller frequency than red light E) Both red and blue light come from emissions in the visible region of the electromagnetic spectrum. Solution: D Red has the longest wavelength. Wavelength is inversely proportional to frequency. Therefore, red has a small frequency. Blue has short wavelength, so frequency is larger than red. Q 1.18 Which of the following statements is true? A) The distance from orbit n = 1 to n = 2 is the same distance from orbit n = 2 to n = 3. B) Absorption occurs when an electron goes from a lower energy state to a higher energy state. C) Emission occurs when ∆E is positive. D) An absorption spectrum is a black background with discrete coloured lines relating to wavelengths. E) Hydrogen, helium and sodium all have identical emission spectra. Solution: B A is false (see Bohr equation for orbit radius). C is false – emission occurs when ∆E is negative. Solutions will be posted at www.prep101.com/solutions 26 ©Prep101 Q 1.19 Chem110 MT1 Instructor Potassium metal must absorb radiation with a minimum frequency of 5.57 x 1014 Hz before it can emit electrons from its surface via the photoelectric effect. If K(s) is irradiated with light of wavelength 510 nm, what is the velocity of an emitted electron? Solution: Calculate the KEe from the excess light: (emitted – threshold) νo = 5.57 E14 s-1, ν = (3.0 E8 m·s-1)/(5.1 E-7 m) = 5.88 E14 s-1 KEe= h(ν-νo) → KEe = 6.626 E-34 Js ((5.88 E14 s-1) – (5.57 E14 s-1)) KEe = 6.626 E-34 Js (3.12 E13 s-1) KEe = ½ mu2 → u= → 2 KEe me KEe = 2.07 E-20 J → u= 2 (2.07 E − 20 J ) (9.109 E − 31 kg ) u = 2.1 E5 m·s-1 Solutions will be posted at www.prep101.com/solutions 27 ©Prep101 Q 1.20 Chem110 MT1 Instructor (A) What the KE of one electron ejected when light of ν = 1.0 E15 Hz strikes a metal with a threshold frequency of νo = 6.7 E14 Hz? (B) What would be the velocity of this ejected electron? A) 3.3 E14 J, 6.9 E5 m·s-1 B) -2.187 E-22 kJ, 6.36 E-9 m·s-1 C) -2.187 E-19 J, 3.3 E14 s-1 D) 2.187 E-19 J, 4.76 E11 m·s-1 E) 2.187 E-22 kJ, 6.9 E5 m·s-1 Solution: Correct Answer - E Calculate the KEe from the excess light (emitted – threshold) KEe= h(ν-νo) → KEe = 6.626 E-34 Js ((1.0 E15 s-1) – (6.7 E14 s-1)) KEe = 2.187 E-19 J (or alternatively 2.187 E-22 kJ) KEe = ½ mu2 → u= 2 KEe me → u= 2 (2.2 E − 19 J ) (9.109 E − 31 kg ) u = 6.9 E5 m·s-1 Solutions will be posted at www.prep101.com/solutions 28 ©Prep101 Q 1.21 Chem110 MT1 Instructor Light with a wavelength of 400 nm strikes a surface of cesium in a photocell and the maximum kinetic energy of the electrons ejected is 1.54 x 10-19 J. Determine (A) the threshhold frequency of the electron in cesium and, (B) the velocity of this ejected electron from that metal. A) 400 nm, cannot be determined B) 7.5 E14 s-1, 5.81 E5 m·s-1 C) 5.18 E14 s-1, 5.81 E5 m·s-1 D) 5.18 E14 s-1, 3.70 E-25 m·s-1 E) 7.5 E14 s-1, 3.70 E25 m·s-1 Solution: Correct answer C KEe= h(ν-νo) where KEe = 1.54 E-19 J and ν = c/λ = 3.0 E8 m·s-1 / (4 E-7 m) = 7.5 E14 s-1 ( ) → → u= 1.54 ×10 −19 J = 6.626 ×10 −34 Js 7.5 ×1014 s −ν o KEe = ½ mu2 → u= 2 KEe me νo = 5.18 E14 s-1 2 (1.54 E − 19 J ) (9.109 E − 31 kg ) u = 5.81 E5 m·s-1 Solutions will be posted at www.prep101.com/solutions 29 ©Prep101 Q* 1.22 Chem110 MT1 Instructor The workfunction is 2.70 eV for Na, 3.66 eV for Mg, and 2.14 eV for Cs. Assume we shine light with a wavelength of 400 nm on the surface of the each of these metals. Which of the following statements is/are TRUE about the observed results? (Note 1 eV = 1.609 E-19 J) A) Electrons are not ejected from Mg, but would be if the (incident) wavelength of light were increased. B) Electrons are ejected from both Cs and Na, and they would have a greater velocity from Cs. C) No electrons are ejected from Na, Mg, or Cs but would be if the light frequency were increased. D) Electrons are ejected from Mg and they have a velocity of 4.4 E5 m/s E) Two of the four statements are TRUE. Solution: B Convert wavelength to energy (J → eV) E = hc/λ → E = 4.97 E-19 J → E = 3.09 eV KEe = h(E-Φ) We know that electrons are emitted if the energy exceeds the threshold frequency (or work function, 3.09 eV). So in this case, electrons are emitted for Na and Cs and not Mg. (Eliminate C, D). If the incident wavelength was increased, the energy would be smaller and electrons would still not be ejected. (Eliminate A) KEe = ½ mu2 KEe is proportional to velocity. KE for Cs > Ke for Na, so velocity of Cs particle is greater. Solutions will be posted at www.prep101.com/solutions 30 ©Prep101 Q 1.23 Chem110 MT1 Instructor Calculate the diameter and energy of the 5th Bohr orbit for a hydrogen atom. A) 2645 pm, -8.716 E-20 J B) 1322 pm, -8.716 E-20 J C) 1.322 E-7 m, 4.36 E-20 J D) 53 Å, 2.179 E-18 J E) 264.5 m, energy cannot be determined Solution: A is correct rn = n2ao where ao = 52.9 pm r5 = (5)2 (52.9 pm) = 1322.5 pm This is the radius! The diameter is 2r. Therefore the diameter is 2645 pm En = -2.179 E-18 J / (52) = -8.716 E-20 J Q 1.24 For the Bohr hydrogen atom, determine whether there is an orbit having a radius of 4.00 Å? Explain. Solution: rn = (n)2(0.529 Å) → 4.00/0.529 = n2 → n = 2.75 Since n ≠ integer; then therefore such an orbit does not exist. Solutions will be posted at www.prep101.com/solutions 31 ©Prep101 Q* 1.25 If the wavelength of an electron is equal to the radius of the n = 2 orbital in the He+ atom, what must the velocity of the electron be? A) B) C) D) E) Solution: Chem110 MT1 Instructor 6.88 E6 m/s 1.38 E7 m/s 3.44 E6 m/s 5.47 E5 m/s The velocity cannot have an exact value A rn (n)2 (5.29 E-11 m) / (Z) where Z = 2 protons for He+ rn = (2)2(5.29 E-11 m) / (2) → Use de Broglie equation λ = h/mu rn = 1.058 E-10 m = λ u = (6.626 E-34 J·s) / ( 9.109 E-31 kg)(1.058 E-10 m) = 6.88 E6 m/s Q 1.26 In 1927, two American physicists (Clinton Davisson and Lester Germer) conducted the famous Davisson-Germer experiment that was awarded the Nobel Prize for Physics in 1937. Which of the following is TRUE concerning their experiment? A) They observed diffraction patterns photons confirming that light had wave properties. B) They observed diffraction patterns of electrons confirming that matter had wave properties. C) Their experiment contradicted de Broglie’s equation D) They observed standing wave patterns from electrons confirming the existence of orbitals. E) They proved emission spectra were characteristic for each element. Solution: B Solutions will be posted at www.prep101.com/solutions 32 ©Prep101 Chem110 MT1 Instructor Q 1.27 The Bohr model applies to which of the following species? A) He+ B) C5+ C) Be3+ D) Li2+ E) All of the above species are applicable. Solution: E The Bohr model applies to all hydrogen-like atoms (1 electron systems). Q 1.28 Which of the following represents the orbit with the largest radius? A) n = 4 orbit for He+ B) n = 1 orbit for H C) n = 2 orbit for Be3+ D) n = 3 orbit for Li2+ E) n = 3 orbit for H Solution: Answer E Recall that rn = n2(ao)/Z where Z = # protons So as ↑Z, then radius decreases; eliminate D Examine E → For n = 3, Z = 1, n2/Z = 9/1 = 9(ao) Examine A → For n = 4, Z = 2, n2/Z = 16/2 = 8(ao) E>A Solutions will be posted at www.prep101.com/solutions 33 ©Prep101 Q* 1.29 Chem110 MT1 Instructor Which of the following statements is/are true? A) The wavefunction, Ψ, is a solution to the Schrodinger equation B) The square of the wavefunction, Ψ2, is the total probability of finding the electron in a spherical shell at distance (r) from the nucleus. C) Orbitals with the same ‘n’ value cannot overlap in space D) For the same orbital type (same ‘ℓ’ quantum number), orbitals with the higher ‘n’ value will have a more negative energy E) Two of the four statements are true Solution: A Orbitals of same ‘n’ value can overlap in space (See Fig 8-34 in text). The Ψ2 is the finding the electron in space. Q* 1.30 Which statement(s) is/are FALSE in comparing the electronic transition from ni = 5 to nf = 2 in hydrogen with that of the electronic transition from ni = 2 to nf = 5 in Be3+? A) The photon involved in the H transition lies in the visible spectrum while that involved in the Be3+ transition lies in the UV region of the spectrum. B) A photon of the same frequency is involved in both transition, but is emitted in the hydrogen electron transition and absorbed in the Be3+ electronic transition. C) The wavelength of the photon emitted by hydrogen is 16 times the wavelength of the photon absorbed by Be3+. D) The transition in hydrogen is a Balmer emission series line E) Two of the four statements are FALSE. Solution: B The frequency is different for a hydrogen atom than a hydrogen-like atom, by the equation ∆E = Z2hcRH((1/n12)-(1/nf2)) and Z = # protons. The photon involved in the H transition lies in the visible spectrum (recall the emission lines). For Be3+ (Z = 4), so the photon emitted by hydrogen is 16 times the wavelength absorbed by Be3+ because Z2 = 16 and energy is inversely proportional to wavelength. When nf = 2, this is a Balmer emission series. Q 1.31 Define the Pauli Exclusion Principle Solution: Each electron in an atom has a unique set of quantum numbers. When one electron in an atom is described by a particular set of quantum numbers, no other electron in the atom is described by the same set. • No two electrons can have the same set of four quantum numbers Solutions will be posted at www.prep101.com/solutions 34 ©Prep101 Q 1.32 Give all possible sets of four quantum numbers for an electron in n = 2, ℓ =1 Solution: n 2 2 2 2 2 2 Q 1.33 Chem110 MT1 Instructor ℓ 1 1 1 1 1 1 mℓ -1 -1 0 0 1 1 ms +½ -½ +½ -½ +½ -½ Complete the chart • Only put a designation if the set of quantum numbers is allowed Set of quantum #’s (n, l, ml, ms) 1, 0, 0, ½ 3, 1, 2, ½ 3, 2, 1, 0 4, 4, 3, ½ 5, 0, 0, ½ Allowed? (yes or no) Designation Set of quantum #’s (n, l, ml, ms) 1, 0, 0, ½ -3, 1, 2, 0 3, 2, 1, ½ 4, 4, -3, ½ 5, 0, 0, ½ Allowed? (yes or no) Y N Y N Y Designation Solution: 1s 3d 5s Q 1.34 An electron can have ms = ± 1/2. What is the difference? Solution: An electron can either be spin up or spin down, designated by either ms = + ½ or – ½ respectively Solutions will be posted at www.prep101.com/solutions 35 ©Prep101 Chem110 MT1 Instructor Q 1.35 A 3s orbital can hold a maximum of 2 electrons. Write out both possible sets of four quantum numbers associated with each electron. Solution: Electron #1 n = 3, ℓ = 0, mℓ = 0, ms = +½ Electron #1 n = 3, ℓ = 0, mℓ = 0, ms = - ½ Q 1.36 Write the order of relative energies for atomic orbitals from n = 1 up to n = 6 Solution: Q 1.37 Rank the energies of the four electrons by the following sets of quantum numbers. Identify the set(s) of quantum numbers that are not possible. (n, ℓ, mℓ, ms) Solution: (1, 0, 0, ½) (5, 1, 2, ½) (5, 1, 0, ½) (4, 4, 0, ½) (4, 3, 0, ½) Those not possible are: (5, 1, 2, ½) & (4, 4, 0, ½) The remaining three, of increasing order, are: (1, 0, 0, ½), (4, 3, 0, ½), (5, 1, 0, ½) Solutions will be posted at www.prep101.com/solutions 36 ©Prep101 Chem110 MT1 Instructor Q 1.38 Define degenerate and give an example in relation to atomic orbitals. Solution: Degenerate refers to energy equivalent atomic orbitals (equal in energy). The 2px, 2py, 2pz are the 3 atomic orbitals of the 2p level. These are all equivalent in energy and thus termed degenerate. Q 1.39 Write the electronic configuration for A) F C) Sm D) Ac E) Os F) Rh G) Cu Solution: B) K H) Lu I) Cr J) Mn K) Sn L) Lr A) [He]2s22p5 B) [Ar]4s1 C) [Xe]6s24f6 D) [Rn]7s26d1 E) [Xe]6s24f145d6 F) [Kr]5s14d8 Important exception G) [Ar]4s13d10 Important exception H) [Xe]6s24f145d1 I) [Ar]4s13d5 J) [Ar]4s23d5 K) [Kr]5s24d105p2 L) [Rn]7s25f146d1 Solutions will be posted at www.prep101.com/solutions 37 ©Prep101 Q 1.40 Chem110 MT1 Instructor Identify the neutral element corresponding to the electronic configuration A) [Kr]5s24d105p1 B) [Ne]3s23p6 C) [Kr]5s24d105p4 Solution: A) Indium B) Argon C) Tellurium (Te) Q 1.41 Determine the number of unpaired electrons in Si, F, Mo and Co Solution: Si = [Ne]3s23p2 2 unpaired electrons F = [He]2s22p5 1 unpaired electron Mo = [Kr]5s14d5 6 unpaired electrons (1s+5d) Co = [Ar]4s23d7 3 unpaired electrons Q 1.42 Which is not a valid set of quantum numbers for an electron? A) n = 6, ℓ = 2, mℓ = 0, ms = ½ B) n = 3, ℓ = 2, mℓ = -2, ms = ½ C) n = 7, ℓ = 0, mℓ = 0, ms = ½ D) n = 17, ℓ = 16, mℓ = 16, ms = -½ E) n = 4, ℓ = -2, mℓ = 2, ms = -½ Solution: E ℓ cannot be negative. Series A-D are all acceptable quantum numbers Solutions will be posted at www.prep101.com/solutions 38 ©Prep101 Q* 1.43 Chem110 MT1 Instructor Which of these species has the greatest number of unpaired electrons? A) Mn (Z = 35) B) P (Z = 15) C) Rb (Z = 37) D) Mo (Z = 42) E) Cu (Z = 29) Solution: D Mn → 4s2d5 (5 unpaired electrons) P → 3s23p3 (3 unpaired electrons) Rb → 5s1 (1 unpaired electron) Mo → 5s14d5 (6 unpaired electrons) Cu → 4s13d10 (1 unpaired electron) Q* 1.44 What would be a valid set of quantum numbers for the last electron filled in the electronic configurations for osmium (Os, Z = 76)? A) n = 6, ℓ = 0, mℓ = 0, ms = ½ B) n = 5, ℓ = 3, mℓ = -3, ms = ½ C) n = 6, ℓ = 1, mℓ = 0, ms = ½ D) n = 6, ℓ = 2, mℓ = 0, ms = ½ E) n = 5, ℓ = 2, mℓ = -2, ms = ½ Solution: E Look up Os (Z = 76) → d6 → ℓ must equal 2; eliminate A, B, C We have 6 electrons that have to be situated in the d-orbital. The first five are placed singly in each of the 5 degenerate d-orbitals (by Hund’s Rule). The sixth (and last electron) is of opposite spin and starts at the -ℓ orbital. Solutions will be posted at www.prep101.com/solutions 39 ©Prep101 Q 1.45 Chem110 MT1 Instructor [Xe]f145d106s1 is the ground state electronic configuration for which species? A) Hg B) Ag C) Au D) Pt E) Cd Solution: Count the number of electrons [Xe]f145d106s1 = 54 + 14 + 10 + 1 = 79 Au has Z = 79 (neutral state) Q 1.46 How many valence electrons are found in tellurium (Te, Z = 52)? A) 1 B) 2 C) 4 D) 6 E) 16 Solution: D Te → [Kr]5s24d105p4 → 2 + 10 + 4 = 16 Q 1.47 What would be a permitted value for the principle quantum number if we knew the magnetic quantum number (mℓ) was 7? A) 5 B) 6 C) 7 D) 15 E) None of the above Solution: D Principle quantum number = n If mℓ = 7, then ℓ ≥ 7; therefore minimum ‘n’ must be 8. Solutions will be posted at www.prep101.com/solutions 40 ©Prep101 Chem110 MT1 Instructor Chapter 9 Questions Q 2.1 Arrange the following in order of increasing radius: Sr2+, Br-, V5+, Ti4+ Solution: V5+ < Ti4+ < Sr2+ < BrThe electron configurations are... # of protons 38 Sr+2: [Kr] Br : [Kr] 35 V+5: [Ar] 23 Ti+4: [Ar] 22 Sr+2 and Br_ have the same number of electrons; however, Zeff is greater for Sr+2 due to a greater # of protons, resulting in a greater + charge that pulls the electrons closer to the nucleus. Thus, Sr+2 is smaller than Br-. V+5 and Ti+4 have the same number of electrons; however, Zeff is greater for V+5 due to a greater # of protons. Thus V+5 is smaller than Ti+4. Sr+2 & Br_ are larger than V+5 & Ti+4 because there are more electrons held in a larger subshell. Q 2.2 Arrange the following in order of decreasing electron affinity: Cl, Br, I. Solution: Cl > Br > I Electron affinity decreases from top to bottom. EA (Cl) = -349 kJ/mol EA (Br) = -324 kJ/mol EA (I) = -295 kJ/mol Chlorine is most likely to gain an electron. Solutions will be posted at www.prep101.com/solutions 46 ©Prep101 Chem110 MT1 Instructor Q 2.3 Which element, S or Cl, has the more negative electron affinity? Why? Solution: Electron affinity becomes more negative from left to right because Cl has higher Zeff than S. EA(Cl) = -349 kJ/mol, EA (S) = -200.4 kJ/mol Q 2.4 Which element, Li or K, has the larger atomic radius? Why? Solution: For K, valence electron is in 4s orbital while valence electron in Li is in 2s orbital. 4s orbital is larger than 2s. Q 2.5 Which atom, Kr or Xe, has the lower ionization energy? Why? Solution: Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron. 1 Ionization energy ∝ 2 n Solutions will be posted at www.prep101.com/solutions 47 ©Prep101 Chem110 MT1 Instructor Q 2.6 Arrange the following species in the orders requested: A) Ca2+, Cl−, S2−, K+ in order of increasing size (smallest first). B) Be, B, N, O in order of increasing first ionization energy (smallest first). Solutions: A) Ca2+ < K+ < Cl− < S2− B) B < Be < O < N Q 2.7 Circle which of the following has the highest ionization energy: Na Solution: Ca Mg Correct Answer: Mg Ionization energy increase as you move up a group and it also increases as you move right across a period. Q 2.8 Circle which of the following has the largest radius: N Solution: F P Mg Correct answer: Mg Radius increase as you move left along a period and as you move down a group. Solutions will be posted at www.prep101.com/solutions 48 ©Prep101 Q 2.9 Chem110 MT1 Instructor The species Ar, Ca+2, and S-2 have the same electron configuration. A) Give the electron configuration in the standard "spdf" notation. B) Which species has the lowest electron affinity? C) Which species has the smallest radius? D) Arrange the species in order of increasing ionization energy? Solutions: A) [Ar] or [Ne]3s23p6 B) S-2 The species with the least favorable electron affinity is S-2 because it has the smallest Zeff (a smaller positive charge to pull electrons towards the nucleus). C) Ca+2 Ca+2 has the greatest Zeff because it has the greatest number of protons pulling on the 18 electrons, resulting in the electrons being closer to the nucleus. D) S-2 < Ar < Ca+2 Ionization energy is the energy required to remove an electron. It is most difficult to remove an electron from Ca+2, as this will involve the removal of a core electron instead of a valence electron, and Ca+2 has the greatest Zeff. It is easier to remove an electron from S-2 versus Ar because S-2 has a smaller Zeff. Solutions will be posted at www.prep101.com/solutions 49 ©Prep101 Q 2.10 Chem110 MT1 Instructor Given the following data: K (g) : I1 = 419 kJ/mol Ca(g) : I1 = 590 kJ/mol I2=3050 kJ/mol I2=1140 kJ/mol A) Explain why the first ionization energy for potassium is smaller than the first ionization energy for calcium, but the second ionization energy for potassium is larger than the second ionization energy for calcium. B) Write out the reactions associated to I2 for both species C) Rank the following from the smallest radius to largest radius: Se2-, Rb+, Br− Solutions: A) I1(K) < I1 (Ca) because Zeff increases from left to right across a period, so Ca has a higher Zeff and it is therefore harder to remove an electron. I2 (K) > I2 (Ca) because the ionization process for Ca leads to the formation of the stable noble gas configuration (Ca2+) while the ionization reaction for K requires the destruction of a stable noble gas configuration. The former will always be lower in energy. B) K+(g) → K2+ (g) + e− Ca+(g) → Ca2+(g) + e− C) The larger the negative charge, the greater the repulsion between electrons, the larger the radius. These are all atoms that have the configuration of Kr, so only charge influences radius. Therefore, Rb+ < Br− < Se2− The ionization energy decreases as you go down Group 1 because as n increases it is easier to remove the outer electron. Radius increases as you go down Group A because as n increases, the radius increases. Solutions will be posted at www.prep101.com/solutions 50 ©Prep101 Q 2.11 Chem110 MT1 Instructor Consider neutral atoms with the following electron configurations: A) 1s22s22p6 B) 1s22s22p63s1 C) 1s22s22p63s6 Which atom has the largest first ionization energy AND which one has the smallest second ionization energy? Explain. 1s22s22p6 = Ne 1s22s22p63s1 = Na 1s22s22p63s2 = Mg Solution: Ne has the largest first ionization energy as an electron is removed from the n = 2 shell, compared to n = 3 shell for Na and Mg. An electron from Na or Mg requires less energy to remove because the n =3 shell is further from the nucleus, and is better shielded by core electrons in the n = 2 shell. Mg has a smaller second ionization energy because the second electron to be removed is a valence electron (n = 3), whereas the second electron removed from Na is taken from a core electron (n = 2). Q 2.12 Arrange the following species in order of decreasing radius: Br-, Y3+, Rb+, Se2-, Kr? A) Se2-, Br-, Kr, Rb+, Y3+ B) Y3+, Rb+, Kr, Br-, Se2C) Kr, Se2-, Br-, Rb+, Y3+ D) Kr, Y3+, Rb+, Br-, Se2E) Rb+,Y3+, Kr, Se2-, Br- Solution: A All isoelectronic, so therefore greater number of electrons gives more pull towards nucleus (smaller radius). Anion < neutral < cation Solutions will be posted at www.prep101.com/solutions 51 ©Prep101 Q 2.13 Chem110 MT1 Instructor Arrange the following species in order of increasing radius: Se, Pb, Sn, Ge, As? Solution: Se < As < Ge < Sn < Pb Left to right: Top to bottom: Q 2.14 Ge > As > Se Pb > Sn > Ge Arrange the following species in order of increasing ionization energy: Al, In, Ga, Si, S? A) In, Ga, Si, S, Al B) Si, S, Al, Ga, In C) In, Al, Ga, Si, S D) In, Ga, Al, Si, S E) S, Si, Al, Ga, In Solution: Correction Answer: We know that IE is the ability (energy required) to remove an electron. Therefore, IE will increase from L → R (as the atom becomes more electronegative). It will be easier to remove an electron from a bigger atom than a smaller atom, so IE will decrease from top to bottom. L→R T→B Al < Si < S Al > Ga > In Therefore (smallest) In < Ga < Al < Si < S (largest) Solutions will be posted at www.prep101.com/solutions 52 ©Prep101 Q 2.15 Chem110 MT1 Instructor Indicate which species is paramagnetic A) FB) Ca2+, given the two electrons lost are 4s C) Fe2+, given the two electrons lost are 4s D) S2E) Ne Solution: Answer – C Paramagnetic – has unpaired electrons Ne is noble gas – Eliminate E F- → 1s22s22p6 Ca2+ and S2- → [Ne]3s23p6 Fe2+ → [Ar]3d6 (all paired) (all paired) (contains unpaired electrons) Solutions will be posted at www.prep101.com/solutions 53 ...
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This note was uploaded on 09/29/2010 for the course SCIENCE 120 taught by Professor Frenser during the Spring '10 term at McGill.

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