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Prep101 - Chemistry 201 Term Test 1 Booklet Solutions
Wave/Particle Duality and 𝐄 =
𝐡𝐜
𝛌
Practice Problems
1. For the hydrogen atom, the ionization energy is:
(a)
(b)
(c)
(d)
(e)
the energy it takes to promote an electron to n = infinity
the energy required to separate a proton and an electron
2.181018 J per atom
all of the above
none of the above
Answer: D
Solution: Statements (a), (b), and (c) are all true.
2.
(a)
(b)
(c)
(d)
(e)
What is the energy of one mole of photons with a wavelength of 285 nm?
6.981019 J
3.501018 J
3.15105 J
4.20105 J
2.11106 J
Answer: A
Solution:
hc (6.626  1034 Js)(3.0  108 m / s)
E

 6.98  1019 J
9

(285  10 m)
This is the energy per photon; we want the molar energy
Multiply by NAV to get 4.20105J
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3. The longest wavelength of radiation that will cause the emission of electrons from the surface of gold
is 257 nm. What is the energy per photon of this radiation?
(a)
(b)
(c)
(d)
(e)
2.5810-27 J
5.1110-32 J
1.2910-18 J
7.7310-19 J
8.5710-16 J
Answer: D
Solution: Energy per photon = h = hc/
(6.626  10-34 J s-1 )(2.998  108 m s-1 )

 7.73  10-19 J
257  10-9 m
4. What is the wavelength of light emitted when an electron in a hydrogen atom drops from the 4d
orbital to a 2s orbital?
(a)
(b)
(c)
(d)
(e)
4.09 x 10-19nm
486nm
6.17 x 1014nm
545nm
4.86 x 10-7nm
Answer: B
1
b  1
1 

 2
2

λ
hc  nio nhi 
1
2.18x 10-18
1 1


-34
8  2
λ
(6.626 x 10 )(3.0 x10 )  2 42 
1
= 2.06 x 106 m-1
λ
 = 4.86 x 10-7m = 486nm
Solution:
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5. A system can be described as having only 4 energy levels, schematically shown below. When an
emission spectrum is taken, how many lines will be observed, assuming that all possible transitions
occurred?
(i) 2
n=4
(ii) 4
n=3
(iii) 6
(iv) 8
n=2
(v) 10
n=1
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6. Referring to the previous question above: Which transition will yield the most energetic photons?
(i)
21
(ii)
31
(iii)
41
(iv)
32
(v)
43
a) Answer: (iii)
Solution: There are six possible transitionsn=4
n=3
n=2
n=1
b) Answer: (iii)
Solution: Going from n=4 to n=1 will have the largest E.
7. In 5.0 s, a 72-Watt (1 Watt = 1 Joule/second) light source emits 9.911020 photons of monochromatic
(single wavelength) radiation. The wavelength of the emitted light is closest to:
(a) 225 nm
(b) 550 nm
(c) 755 nm
(d) 1250 nm
(e) 2750 nm
Answer: B
Solution: 1Watt = 1J/sec,
therefore Energy for all photons = 72 J / sec 5sec  360J
360J
 3.63  1019 J
Energy for one photon =
20
9.91  10 photons
E
hc

, therefore  
hc (6.626  1034 )(3 108 )

 547nm
E
3.63  1019
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8. The first ionization energy of helium is 2370 kJ·mol1. What is the minimum frequency of
electromagnetic radiation that is required to ionize one helium atom?
(a)
(b)
(c)
(d)
(e)
4
1.98107 s1
5.941015 s1
3.581039 s1
2.801040 s1
4.641064 s1
Answer: B
Solution: E = h
For one atom of helium:
2370kJ
1mol

1st Ionization energy =
= 3.94 10-21kJ
mol
6.022  1023 atoms
E
3.94  1018 J
 
 5.95  1015
34
h 6.626  10 Js
9. A hydrogen atom in an excited state makes a transition to the state with n = 2. The photon emitted
has a frequency of v = 6.911014 sec1.
a) Calculate the wavelength  of this radiation in nanometres (nm).
b) Calculate the energy E of these photons in kJ mol1.
c) Calculate n for the excited state.
a) Answer:
C =   3.00108 m·s1 = 6.911014 s1   = 4.342107 m = 434.2 nm.
b) Answer:
E = hv = 6.2621034 J·s6.911014 s1 = 4.5791019 J/photon
= 4.57910196.0210231013 = 275.6 kJ·mol1.
c) Answer:
 1
1 
E   B  2  2  . For H, B = 2.18x10-18.x 6.022 x 1023 x 1/1000 kj/j
 nupper nlower 


 1
1
1314
275.6  1314  1 2  2   275.6  328.5   2
 nupper 2 
nupper


1314
2
 nupper

= 24.84  n = 5
52.9
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Quantum Numbers and Electron Filling Practice Problems
Problem 1. The square of the wave function, 2 for an electron in an atom
(a)
(b)
(c)
(d)
(e)
describes the energy of the electron
specifies the momentum of the electron.
gives the probability of finding the electron in a region of space.
is proportional to the velocity of the electron.
is inversely proportional to the distance between the electron and the nucleus.
Answer: C
Solution:
This gives the probability of finding an electron in a region of space.
Problem 2. Which neutral species is isoelectronic with O2 ?
Answer: O2- has 10 electrons therefore Ne is isoelectronic with it.
The electron configuration is 1s22s22p6
Problem 3. The orbital diagram that best describes the nitrogen atom in its ground state electronic
configuraton is?
1s
2s
2p
(a)
()
()
()()()
(b)
()
()
()()()
(c)
()
()
()()()
(d)
()
()
()()()
(e)
()
()
()()( )
Answer: D
Solution: The electron configuration for N is 1s22s22p3. The 1s and 2s levels will have be full, each consisting
of 2 electrons with opposite spins (which rules out diagram A). The 2p level will have 3 electrons, one in each
orbital (which rules out diagram E) and C). The best diagram will show all three electrons in the 2p orbitals
with the same spin (i.e. either all facing up or down). Therefore, diagram D) is the best representation of an N
atom in its ground state.
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Problem 4. Which of the following configurations is not the ground-state configuration of the atom?
(a) He: 1s2
(b) Ne: 1s2 2s2 2p6
(c) Na: 1s2 2s2 2p6 3p1
(d) P: 1s2 2s2 2p6 3s2 3p3
(e) N: 1s2 2s2 2p3
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Answer: C
Solution: Sodium, Na, should be 1s2 2s2 2p6 3s1
Problem 5.
a) Write the electronic configuration for Ti3+
b) List the values of the four quantum numbers for a 3d electron. Where more than one value is allowed, list
all possible allowed values.
c) Sketch the radial probability function (distribution) for the 3s orbital.
Answer:
a) [Ar]3d1. The electronic configuration of Ti is [Ar]4s23d2. Electrons are removed from the s subshell first
when transition metals, such as Ti, changes to an ionic state. Thus, the electron configuration of Ti3+ is [Ar]3d1.
b) n = 3, l = 2, ml = -2,-1,0,1,2, ms = + ½,  ½
c)
Radial
probability
r
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Problem 6. Which of the following is a correct set of quantum numbers for a valence electron of the element
As?
n
l
ml
ms
(a)
3
1
-1
-1/2
(b)
3
3
+1
+1/2
(c)
4
2
-2
+1/2
(d)
4
1
0
+1/2
(e)
4
1
+2
-1/2
Answer: D
The electron configuration for As is 1s22s2p63s2p6d104s24p3.
The valence electrons for the element As all reside in the 4th level so n = 4. The valence electrons only exist in
s or p sublevels so l = 0 or 1. In the s sublevel there is only one orbital while in the p sublevel, there are 3
orbitals, therefore the value of ml = 0 for the s sublevel and ml = -1, 0, +1 for the p sublevel. ms = +1/2 or –1/2
since the spin of the electron in any orbital can only be up or down. The only answer that falls in range of all
the possible choices is D.
Problem 7. Which of the following sets of quantum numbers is not possible?
A.
B.
C.
D.
E.
N
1
4
3
2
2
l
0
0
3
1
0
ml
0
0
3
1
0
ms
+1/2
+1/2
1/2
1/2
+1/2
Answer: C
Solution: l = 0, 1, … (n-1), therefore if n = 3, then l cannot equal 3.
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Problem 8. Which of the following sets of quantum numbers describe an allowed state of an electron in an
atom?
n
l
ml
i.
3
2
-2
ii.
2
1
-2
iii.
4
2
3
iv.
1
1
0
v.
4
0
0
a)
b)
c)
d)
e)
1, 4 and 5
2, 3, and 4
2 and 4
3 and 4
1 and 5
Answer: E
n is the principal quantum number and can be any whole integer number. l is the secondary quantum number
= 0 to n-1. ml is the third quantum number = -l to +l. Set II) is invalid because ml cannot equal -2 if l is 1. Set III)
is invalid because ml cannot equal 3 if l is 2. Set IV) is invalid because l cannot equal 1 if n is 1.
Problem 9. One of the valence electrons in a ground state atom has the quantum numbers
n = 4, l = 1, ml = 0, ms = -1/2. The atom could be
(a) Na
(b) Cl
(c) Rb
(d) Br
(e) Sn
Answer: D
Solution: These quantum numbers correspond to 4p
The atom could be Br.

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Problem 10.
a) Give the electron configuration of each of the following atoms and ions using the standard “spdf” notation.
K
V3+
Mo
Ru
Y3+
b) What element does the following configuration correspond to?
[Kr]5s24d105p2
Answer:
a)
K
V3+
Mo
Ru
Y3+
[Ar]4s1
[Ar]3d2
[Kr]5s24d4
[Kr]5s24d6
[Kr]
b) Sn = Tin
Problem 11. Identify the specific element with the following electron configurations by symbol and name.
[Ne]3s2
[Ne]3s23p1
[Ar]4s13d5
[Kr]5s24d105p4
Answer:
[Ne]3s2
[Ne]3s23p1
[Ar]4s13d5
[Kr]5s24d105p4
Magnesium (Mg)
Aluminum (Al)
Chromium (Cr)
Tellurium (Te)
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Problem 12.
a) The number of possible ml values for a 5f electron are .
b) The number of unpaired electrons in the ground state of Cr are
.
c) What is the maximum number of electrons permitted in the 5th major energy level?
Answers
a) 7, ml = -l…0…+l where l= 3 as we are dealing with the f shell and ml=-3,-2,-1,0,1,2,3.
b) 6, [Ar] 4s 3d    
c) 50, l=0,1,2, to a maximum of n-1 i.e.4. Therefore 5s=2e- (l=0), 5p=6e- (l=1), 5d=10e- (l=2), 5f=14e(l=3), 5g= 18e- (l=4): total = 50.
Problem 13.
a) Give an anion and a cation that are isoelectronic with Cl-:
anion
cation
b) Give the electronic distribution for Fe2+:
c) Does the electronic distribution 1s22s22p63s23p64s13d10 represent a ground state, an excited state or an
impossible state?
Answer
a) Many solutions. Some examples are: anion = S2- and cation K+
K+ is [Ar] and S2- is [Ar]
b) Fe=26, for Fe: [Ar] 4s23d6, for Fe2+: [Ar] 3d6
c) It is the ground state for copper.
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Problem 14. Which atom is pulled more strongly into a magnetic field?
(a) C
(b) N
(c) O
(d) F
(e) Ne
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Answer: B
Solution:
C has 2 unpaired electrons
N has 3 unpaired electrons
O has 2 unpaired electrons
F has 1 unpaired electrons
Ne has 0 unpaired electrons
Therefore N will more strongly pulled into an in-homogeneous magnetic field.
Problem 15. Complete the following table by identifying the species, indicating whether it is in the ground
state or an excited state and whether it is paramagnetic or diamagnetic. The first line is given as an example.
Atomic
Configuration
Species
State
Paramagnetic
Number
Or Diamagnetic
2

1
1s
Ground
Diamagnetic
H
2
2
3
7
1s 2s 2p
17
1s22s22p63s23p6
11
1s22s22p63p1
22
1s22s22p63s23p64s2
13
1s22s22p6
Answer:
Atomic
Configuration
Species
Number
1
1s2
H
2
2
3
7
1s 2s 2p
N
2
2
6
2
6
17
1s 2s 2p 3s 3p
Cl
11
1s22s22p63p1
Na
2
2
6
2
6
2
22
1s 2s 2p 3s 3p 4s
Ti2+
13
1s22s22p6
Al3+
configuration is correct for the neutral atom but NOT the cation.
Note:
an
state
it would
lose its 4s
first
3d
The given
State
Ground
Ground
Ground
Excited
Excited
Ground
Paramagnetic
Or Diamagnetic
Diamagnetic
Paramagnetic
Diamagnetic
Paramagnetic
Diamagnetic
Diamagnetic
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Ti2+ is in
excited
because
normally
electrons
before its
electrons.
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Problem 16. Give the number of unpaired electrons in the ground states of the following species.
Species
Si
Cr
V3P
Zn2+
Number of unpaired
electrons
Answer:
Species
Number of unpaired
electrons
Si
Cr
V3-
P
Zn2+
2
4
4
3
2
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Periodic Properties Practice Problems
Problem 17. In which atom does the 1s orbital have the highest energy?
(a) He
(b) H
(c) Li
(d) Be
(e) Ne
Answer: B
Solution: H will have the highest 1s orbital energy because it has the least positive charge on the nucleus.
Problem 18. a) Arrange the following in order of increasing radius: Sr2+, Br-, V5+, Ti4+
b) Arrange the following in order of decreasing electron affinity: Cl, Br, I.
Answer:
a) V5+ < Ti4+ < Sr2+ < BrThe electron configurations are...
Sr2+:
[Kr]
[Kr]
5+
V : [Ar]
Ti4+: [Ar]
Br-:
# of protons
38
35
23
22
Sr2+ and Br- has the same number of electrons; however, Zeff is greater for Sr2+ 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-.
V5+ and Ti4+ has the same number of electrons; however, Zeff is greater for V 5+ due to a greater # of protons.
Thus V5+ is smaller than Ti4+.
Sr2+ and Br- are larger than V5+ and Ti4+ because there are more electrons held in a larger subshell.
(smallest) V5+ < Ti4+ < Sr2+ < Br- (largest)
b) Cl > Br > I
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Problem 19.
a) Which element, S or Cl, has the more negative electron affinity? Why?
b) Which element, Li or K, has the larger atomic radius? Why?
c) Which atom, Kr or Xe, has the lower ionization energy? Why?
d) Which atom, B or O, is more electronegative? Why?
Answer:
a) Electron affinity becomes more negative from left to right because Cl has higher Zeff than S.
b) For k, valence electron is in 4s orbital while valence electron in Li is in 2s orbital. 4s orbital is larger than 2s.
1
c) Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron. Ionization energy  2
n
d) Zeff is larger in O than in B.
Problem 20.
a) Circle which of the following has the highest ionization energy: Na Ca
b) Circle which of the following has the largest radius: N
F
P
Mg
c) Place the following in increasing order of electronegativity: O, P, Be, Ca, Cl
Mg
Answers:
a) Mg, Ionization energy increase as you move up a group and it also increases as you move right across a
period.
b) Mg, Radius increase as you move left along a period and as you move down a group.
c)
Ca
Be
P
Cl
O
Electronegativity increases as you up a group and as you move right across a period. F is the most
electronegative element and Cs is the least electronegative element.
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Problem 21. The species Ar,
and have the same electron configuration.
a) Give the electron configuration in the standard spdf notation.
b) In which species is Zeff smallest?
c) Which species has the lowest electron affinity?
d) Which species has the smallest radius?
e) Arrange the species in order of increasing ionization energy?
Ca2+,
15
S2-
Answer:
a) [Ar] or [Ne]3s23p6
b) S2c) S2d) Ca2+
e) S2- < Ar < Ca2+
Solution:
a) [Ar] or [Ne]3s23p6
b) Ar, Ca2+ and S2- all have the same number of electrons; however, Ar has 18 protons, Ca2+ has 20, and S2has 16. Thus S2- has the least Zeff since it has the smallest charge pulling on the electrons.
c) 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).
d) Ca2+ 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.
e) Ionization energy is the energy required to remove an electron. It is most difficult to remove an
electron from Ca2+, as this will involve the removal of a core electron instead of a valence electron, and
Ca2+ has the greatest Zeff. It is easier to remove an electron from S2- versus Ar because S2- has a smaller
Zeff. S2- < Ar < Ca2+
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Problem 22. Which of the following species in their ground states
V3+,
Kr,
O,
Cr,
Al,
Br,
Cs,
He+, C
has a noble gas electron configuration?
has the smallest ionization energy?
has an atomic number Z = 13?
has a half-filled sub-shell with l = 2?
is a hydrogen-like species?
has only one 4s electron?
have two unpaired electrons?
has only two d electrons with n = 3?
is diamagnetic?
has the largest radius?
has only one electron with l = 1?
has the largest number of unpaired electrons?
are transition metal species?
Answer:
has a noble gas electron configuration?
has the smallest ionization energy?
has an atomic number Z = 13?
has a half-filled sub-shell with l = 2?
is a hydrogen-like species?
has only one 4s electron?
have two unpaired electrons?
has only two d electrons with n = 3?
is diamagnetic?
has the largest radius?
has only one electron with l = 1?
has the largest number of unpaired electrons?
are transition metal species?
16
Kr
Cs
Al
Cr
He+
Cr
O, C, V3+
V3+
Kr
Cs
Al
Cr
V3+, Cr
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Sketch the cross-sections of the following orbitals:
i) 3p y
ii) 3d x2  y2
iii) 3s
iv) 3d xy
Answer:
i)
ii)
y
y
17

+
+
x
x

iii)
y
y
iv)
+

x
x

+
Problem 23. For 3d x2  y 2 and 2py orbitals provide the number of nodal planes and the number of additional
degenerate atomic orbitals in the same subshell:
Orbital
Number of nodal planes
Number of additional
degenerate orbitals
Orbital
Number of nodal planes
3d x2  y 2
2
Number of additional
degenerate orbitals
4
2py
1
2
3d x2  y 2
2py
Answer:
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Problem 24. Identify the following atomic orbitals from the cross-sections shown below:
18
Z
Cross – Section
X
Name of Orbital
Value of n
Value of l
Total Number
of Nodal Planes
& Surfaces
Answer:
Cross – Section
Name of Orbital
X
3dz2
3
3dxz
3
2
2 nodal surfaces
2
2 nodal surfaces
Value of n
Value of l
Nodes
2pz orbital
Likely 2 (nodes
will show up if 3
or higher)
1
2 nodal surfaces
(one plane, one
sphere)
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Problem 25. Complete each row of the table by entering the value of l and all possible values for ml for each
orbital appearing in the first column. If one of the orbitals in the first column does not exist, fill each box in
that row with an “X”. To complete the final column, enter the symbol of a neutral atom that in its ground state
configuration contains one and only one half-filled orbital of the type listed in the first column.
Orbital
l
All values of ml
Atom
1s
2p
3d
3f
4p
6s
Answer:
Orbital
1s
2p
3d
3f
4p
6s
l
0
1
2
X
1
0
All values of ml
0
-1, 0, 1
-2, -1, 0, 1, 2
X
-1, 0, 1
0
Atom
H
B or F
Sc
X
Ga or Br
Cs
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