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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 105 Infrared 6.0 x 105 Red Orange 5.0 x 105 Yellow Green 4.0 x 105 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 1019 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 s1 Solution: E = hν Q 1.4 How many photons of frequency 1.50 x 1014 s1 are needed to give 30.1 J of
energy? Solution: First calculate the energy of a photon:
E = hν → → E = (6.626 x 1034 Js)(4.35 x 1014 s1) = 2.88 x 1019 J E = 6.626 x 1034 Js (1.50 x 1014 s1) = 9.94 x 1020 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 1020 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·mol1 photons. Solution: Divide by Avagadro’s number to give the energy per photon
2.58 J ⋅ mol1
= 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 102 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 106 m Now energy can be calculated by the standard equation: E = hc
λ E = (6.626x1034 Js)(2.998x108 ms1)/(3.34x106 m) = 5.95 x 1020 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 1034Js)/[( 1.5 x 106 m/sec)(9.11 x 1031kg)
= 4.8 x 1010m = 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 E18 J 1
1 = 4.58 × 1019 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 E7 m
=
E
4.58 × 10 −19 J 4.34 E7 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 E9 m
Ephoton = hc/λ = (6.626 E34 Js)(2.998 E8 m)/(325 E9) = 6.11 E19 J/photon
6.11 E19 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 NH bonds. What is the
wavelength of the radiation needed to break one NH 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 E19 J/bond
λ = hc/E → (6.626 E34 Js)(2.998 E8 m/s) / (6.63 E19 J/bond) → 3.00 E7 m 3.0 E7 m (109 nm / 1 m) = 300.0 nm
The visible region is considered in between 400800 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 E19 J
We can now calculate the frequency by the equation, E = hν
ν = E/h = 4.086 E19 J / (6.626 E34 Js) = 6.166 E14 s1
Now calculate the wavelength from the frequency, λ = c/ν
λ = 2.998 E8 ms1 / (6.166 E14 s1) = 4.86 E7 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 E20 J We can now calculate the frequency by the equation, E = hν
ν = 4.576 E20 J / (6.626 E34 Js) = 6.906 E14 s1
Now calculate the wavelength from the frequency, λ = c/ν
λ = 2.998 E8 ms1 / (6.906 E14 s1) = 4.34 E7 m
4.23 E7 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
Xrays ni = 7, nf = 4 (Brackett series) 1
1 ∆E = 2.179 × 10 −18 J 2 − 2 7
4 → ∆E = 9.176 E20 J We can now calculate the frequency by the equation, E = hν
ν = 9.176 E20 J / (6.626 E34 Js) = 1.3848 E14 s1
Now calculate the wavelength from the frequency, λ = c/ν
λ = 2.998 E8 ms1 / (1.3848 E14 s1) = 2.16 E6 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 E6 m
We can now calculate the frequency by the from the wavelength, λ = c/ν
ν = (2.998 E8 m·s1) / (1.094 E6 m) = 2.74 E14 s1
Energy can be calculated from the frequency, ∆E = hν,
∆E = 6.626E34 Js (2.74 E14 s1) = 1.8158 E19 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 s1, ν = (3.0 E8 m·s1)/(5.1 E7 m) = 5.88 E14 s1
KEe= h(ννo) → KEe = 6.626 E34 Js ((5.88 E14 s1) – (5.57 E14 s1)) KEe = 6.626 E34 Js (3.12 E13 s1)
KEe = ½ mu2 → u= → 2 KEe
me KEe = 2.07 E20 J
→ u= 2 (2.07 E − 20 J )
(9.109 E − 31 kg ) u = 2.1 E5 m·s1 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·s1
B) 2.187 E22 kJ, 6.36 E9 m·s1
C) 2.187 E19 J, 3.3 E14 s1
D) 2.187 E19 J, 4.76 E11 m·s1
E) 2.187 E22 kJ, 6.9 E5 m·s1 Solution: Correct Answer  E
Calculate the KEe from the excess light (emitted – threshold)
KEe= h(ννo) → KEe = 6.626 E34 Js ((1.0 E15 s1) – (6.7 E14 s1)) KEe = 2.187 E19 J (or alternatively 2.187 E22 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·s1 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 1019 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 s1, 5.81 E5 m·s1
C) 5.18 E14 s1, 5.81 E5 m·s1
D) 5.18 E14 s1, 3.70 E25 m·s1
E) 7.5 E14 s1, 3.70 E25 m·s1 Solution: Correct answer C
KEe= h(ννo) where KEe = 1.54 E19 J
and
ν = c/λ = 3.0 E8 m·s1 / (4 E7 m) = 7.5 E14 s1 ( ) → → 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 s1 2 (1.54 E − 19 J )
(9.109 E − 31 kg ) u = 5.81 E5 m·s1 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 E19 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 E19 J → E = 3.09 eV KEe = h(EΦ) 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 E20 J
B) 1322 pm, 8.716 E20 J
C) 1.322 E7 m, 4.36 E20 J
D) 53 Å, 2.179 E18 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 E18 J / (52) = 8.716 E20 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 E11 m) / (Z) where Z = 2 protons for He+ rn = (2)2(5.29 E11 m) / (2) → Use de Broglie equation λ = h/mu rn = 1.058 E10 m = λ u = (6.626 E34 J·s) / ( 9.109 E31 kg)(1.058 E10 m) = 6.88 E6 m/s Q 1.26 In 1927, two American physicists (Clinton Davisson and Lester Germer)
conducted the famous DavissonGermer 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 hydrogenlike 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 834 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 hydrogenlike 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 AD 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 dorbital. The first five are
placed singly in each of the 5 degenerate dorbitals (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 S2 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) S2
The species with the least favorable electron affinity is S2 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) S2 < 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 S2 versus Ar because S2 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.
 Spring '10
 Frenser

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